Deck 11: Dynamic Programming

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سؤال
The management of a company is considering three possible new products for next year's product line.A decision now needs to be made regarding which products to market and at what production levels.Initiating the production of two of these products would require a substantial start-up cost,as shown in the first row of the table below.Once production is under way,the marginal net revenue from each unit produced is shown in the second row.The third row gives the percentage of the available production capacity that would be used for each unit produced. The management of a company is considering three possible new products for next year's product line.A decision now needs to be made regarding which products to market and at what production levels.Initiating the production of two of these products would require a substantial start-up cost,as shown in the first row of the table below.Once production is under way,the marginal net revenue from each unit produced is shown in the second row.The third row gives the percentage of the available production capacity that would be used for each unit produced.   Only 3 units of product 1 could be sold,whereas all units that could be produced of the other two products could be sold.The objective is to determine the number of units of each product to produce in order to maximize the total profit (total net revenue minus start-up costs).(a)Assuming that production quantities must be integers,use dynamic programming to solve this problem.(b)Now consider the case where the divisibility assumption holds so that the variables representing production quantities are treated as continuous variables.Assuming that proportionality holds for both net revenues and capacities used,use dynamic programming to solve this problem.<div style=padding-top: 35px> Only 3 units of product 1 could be sold,whereas all units that could be produced of the other two products could be sold.The objective is to determine the number of units of each product to produce in order to maximize the total profit (total net revenue minus start-up costs).(a)Assuming that production quantities must be integers,use dynamic programming to solve this problem.(b)Now consider the case where the divisibility assumption holds so that the variables representing production quantities are treated as continuous variables.Assuming that proportionality holds for both net revenues and capacities used,use dynamic programming to solve this problem.
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سؤال
A company is planning its advertising strategy for next year for its three major products.Since the three products are quite different,each advertising effort will focus on a single product.In units of millions of dollars,a total of 6 is available for advertising next year,where the advertising expenditure for each product must be an integer greater than or equal to 1.The vice-president for marketing has established the objective: Determine how much to spend on each product in order to maximize total sales.The following table gives the estimated increase in sales (in appropriate units)for the different advertising expenditures. A company is planning its advertising strategy for next year for its three major products.Since the three products are quite different,each advertising effort will focus on a single product.In units of millions of dollars,a total of 6 is available for advertising next year,where the advertising expenditure for each product must be an integer greater than or equal to 1.The vice-president for marketing has established the objective: Determine how much to spend on each product in order to maximize total sales.The following table gives the estimated increase in sales (in appropriate units)for the different advertising expenditures.   Use dynamic programming to solve this problem.<div style=padding-top: 35px> Use dynamic programming to solve this problem.
سؤال
Consider the following integer nonlinear programming problem.Maximize Z = Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ,subject to Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem.<div style=padding-top: 35px> Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ≥ 1, Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ≥ 1, Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ≥ 1,and Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem.<div style=padding-top: 35px> , Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem.<div style=padding-top: 35px> , Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem.<div style=padding-top: 35px> are integers.Use dynamic programming to solve this problem.
سؤال
Consider the following nonlinear programming problem.Maximize Z = Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ,subject to 2 Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> + Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ≤ 13 Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> + Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ≤ 9 and Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ≥ 0, Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ≥ 0.Use dynamic programming to solve this problem.
سؤال
Consider the following nonlinear programming problem.Maximize Z = Consider the following nonlinear programming problem.Maximize Z =   ,subject to   and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ,subject to Consider the following nonlinear programming problem.Maximize Z =   ,subject to   and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> and Consider the following nonlinear programming problem.Maximize Z =   ,subject to   and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ≥ 0, Consider the following nonlinear programming problem.Maximize Z =   ,subject to   and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem.<div style=padding-top: 35px> ≥ 0.Use dynamic programming to solve this problem.
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Deck 11: Dynamic Programming
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The management of a company is considering three possible new products for next year's product line.A decision now needs to be made regarding which products to market and at what production levels.Initiating the production of two of these products would require a substantial start-up cost,as shown in the first row of the table below.Once production is under way,the marginal net revenue from each unit produced is shown in the second row.The third row gives the percentage of the available production capacity that would be used for each unit produced. The management of a company is considering three possible new products for next year's product line.A decision now needs to be made regarding which products to market and at what production levels.Initiating the production of two of these products would require a substantial start-up cost,as shown in the first row of the table below.Once production is under way,the marginal net revenue from each unit produced is shown in the second row.The third row gives the percentage of the available production capacity that would be used for each unit produced.   Only 3 units of product 1 could be sold,whereas all units that could be produced of the other two products could be sold.The objective is to determine the number of units of each product to produce in order to maximize the total profit (total net revenue minus start-up costs).(a)Assuming that production quantities must be integers,use dynamic programming to solve this problem.(b)Now consider the case where the divisibility assumption holds so that the variables representing production quantities are treated as continuous variables.Assuming that proportionality holds for both net revenues and capacities used,use dynamic programming to solve this problem. Only 3 units of product 1 could be sold,whereas all units that could be produced of the other two products could be sold.The objective is to determine the number of units of each product to produce in order to maximize the total profit (total net revenue minus start-up costs).(a)Assuming that production quantities must be integers,use dynamic programming to solve this problem.(b)Now consider the case where the divisibility assumption holds so that the variables representing production quantities are treated as continuous variables.Assuming that proportionality holds for both net revenues and capacities used,use dynamic programming to solve this problem.
Since the decisions to be made are xn = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state sn be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of sn are 0,1,2,3,4,5,where s1 = 5,s2 = 5 - x1,and s3 = s2 - 2x2.Using the cn and rn values given in the above table,the profit from producing xn units of product n is easily calculated as pn(xn)= rnxn - cn,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. , Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. , Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. .Using these expressions for Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3, Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. For n = 2, Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. For n = 1, Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. (b)Now production quantities are treated as continuous variables.Let sn be the percentage of the capacity left for the remaining products.(Note that the units of sn now are different than in part (a),so the possible values of sn now are 0 ≤ sn ≤ 100 rather than 0,1,2,3,4,5.)We now have s1 = 100,s2 = 100 -20x1,and s3 = s2 - 40x2.For product 3: Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. = s3 /20,with Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. = s3 /20.For product 2: Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. ,where Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. .This is linear in x2 except for the Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. term.Therefore,we only need to examine the endpoints of the range for x2. If x2 = 0,we get Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. = s2 /20.If x2 = s2 /40,we get Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. .Equating these two quantities,the breakpoint is s2 =80.So, Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. and Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. .For product 1: Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. ,where Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. .Again,this is linear in x1 except for the Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. term.Therefore,we only need to examine the endpoints of the range for x1. If x1 = 0,we get Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. = 5.5.If x1 = 3,we get Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. = 6 - 3 + Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. (40)= 5.So, Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. = 0 and Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. (100)= 5.5.With Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. = 0,the state at stage 2 becomes s2 = 100,which leads to Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. = 2.5 and s3 = 0,so Since the decisions to be made are x<sub>n</sub> = production level of product n,for n = 1,2,3,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of production capacity still remaining,so this becomes the current state in this formulation.(a)At stage n,let the state s<sub>n</sub> be measured as the number of 20% blocks of production capacity still available for use on the remaining products,so the possible values of s<sub>n</sub> are 0,1,2,3,4,5,where s<sub>1</sub> = 5,s<sub>2</sub> = 5 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Using the c<sub>n</sub> and r<sub>n</sub> values given in the above table,the profit from producing x<sub>n</sub> units of product n is easily calculated as p<sub>n</sub>(x<sub>n</sub>)= r<sub>n</sub>x<sub>n</sub> - c<sub>n</sub>,for n = 1,2,3.Therefore,using the usual dynamic programming notation presented in Sec.11.2 of the textbook,we have   ,   ,   .Using these expressions for   for n = 3,then n = 2,and then n = 1 in turn,the dynamic programming calculations are given below.For n = 3,   For n = 2,   For n = 1,     (b)Now production quantities are treated as continuous variables.Let s<sub>n</sub> be the percentage of the capacity left for the remaining products.(Note that the units of s<sub>n</sub> now are different than in part (a),so the possible values of s<sub>n</sub> now are 0 ≤ s<sub>n</sub> ≤ 100 rather than 0,1,2,3,4,5.)We now have s<sub>1</sub> = 100,s<sub>2</sub> = 100 -20x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 40x<sub>2</sub>.For product 3:   = s<sub>3 </sub>/20,with   = s<sub>3 </sub>/20.For product 2:   ,where   .This is linear in x<sub>2</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>2.</sub> If x<sub>2 </sub>= 0,we get   = s<sub>2</sub> /20.If x<sub>2 </sub>= s<sub>2</sub> /40,we get   .Equating these two quantities,the breakpoint is s<sub>2 </sub>=80.So,   and   .For product 1:   ,where   .Again,this is linear in x<sub>1</sub> except for the   term.Therefore,we only need to examine the endpoints of the range for x<sub>1.</sub> If x<sub>1 </sub>= 0,we get   = 5.5.If x<sub>1 </sub>= 3,we get   = 6 - 3 +   (40)= 5.So,   <sub> </sub>= 0 and   (100)= 5.5.With   = 0,the state at stage 2 becomes s<sub>2</sub> = 100,which leads to   = 2.5 and s<sub>3</sub> = 0,so   = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect. = 0.Therefore,the optimal solution is to produce 2.5 units of product 2,but none of products 1 and 3,with a profit of 5.5.This is better than with the integer restriction in part (a),as we would expect.
2
A company is planning its advertising strategy for next year for its three major products.Since the three products are quite different,each advertising effort will focus on a single product.In units of millions of dollars,a total of 6 is available for advertising next year,where the advertising expenditure for each product must be an integer greater than or equal to 1.The vice-president for marketing has established the objective: Determine how much to spend on each product in order to maximize total sales.The following table gives the estimated increase in sales (in appropriate units)for the different advertising expenditures. A company is planning its advertising strategy for next year for its three major products.Since the three products are quite different,each advertising effort will focus on a single product.In units of millions of dollars,a total of 6 is available for advertising next year,where the advertising expenditure for each product must be an integer greater than or equal to 1.The vice-president for marketing has established the objective: Determine how much to spend on each product in order to maximize total sales.The following table gives the estimated increase in sales (in appropriate units)for the different advertising expenditures.   Use dynamic programming to solve this problem. Use dynamic programming to solve this problem.
Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let xn be the advertising dollars (in millions)spent on product n.Let sn be the amount of advertising budget remaining.Let Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let x<sub>n</sub> be the advertising dollars (in millions)spent on product n.Let s<sub>n</sub> be the amount of advertising budget remaining.Let   be the increase in sales of product n when x<sub>n</sub> million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is   .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3,   For n = 2,   For n = 1,   Hence,since s<sub>2</sub> = 6 -   and s<sub>3</sub> = s<sub>2</sub> -   ,there are two optimal plans as given in the table below.  be the increase in sales of product n when xn million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let x<sub>n</sub> be the advertising dollars (in millions)spent on product n.Let s<sub>n</sub> be the amount of advertising budget remaining.Let   be the increase in sales of product n when x<sub>n</sub> million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is   .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3,   For n = 2,   For n = 1,   Hence,since s<sub>2</sub> = 6 -   and s<sub>3</sub> = s<sub>2</sub> -   ,there are two optimal plans as given in the table below.  .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3, Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let x<sub>n</sub> be the advertising dollars (in millions)spent on product n.Let s<sub>n</sub> be the amount of advertising budget remaining.Let   be the increase in sales of product n when x<sub>n</sub> million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is   .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3,   For n = 2,   For n = 1,   Hence,since s<sub>2</sub> = 6 -   and s<sub>3</sub> = s<sub>2</sub> -   ,there are two optimal plans as given in the table below.  For n = 2, Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let x<sub>n</sub> be the advertising dollars (in millions)spent on product n.Let s<sub>n</sub> be the amount of advertising budget remaining.Let   be the increase in sales of product n when x<sub>n</sub> million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is   .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3,   For n = 2,   For n = 1,   Hence,since s<sub>2</sub> = 6 -   and s<sub>3</sub> = s<sub>2</sub> -   ,there are two optimal plans as given in the table below.  For n = 1, Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let x<sub>n</sub> be the advertising dollars (in millions)spent on product n.Let s<sub>n</sub> be the amount of advertising budget remaining.Let   be the increase in sales of product n when x<sub>n</sub> million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is   .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3,   For n = 2,   For n = 1,   Hence,since s<sub>2</sub> = 6 -   and s<sub>3</sub> = s<sub>2</sub> -   ,there are two optimal plans as given in the table below.  Hence,since s2 = 6 - Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let x<sub>n</sub> be the advertising dollars (in millions)spent on product n.Let s<sub>n</sub> be the amount of advertising budget remaining.Let   be the increase in sales of product n when x<sub>n</sub> million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is   .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3,   For n = 2,   For n = 1,   Hence,since s<sub>2</sub> = 6 -   and s<sub>3</sub> = s<sub>2</sub> -   ,there are two optimal plans as given in the table below.  and s3 = s2 - Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let x<sub>n</sub> be the advertising dollars (in millions)spent on product n.Let s<sub>n</sub> be the amount of advertising budget remaining.Let   be the increase in sales of product n when x<sub>n</sub> million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is   .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3,   For n = 2,   For n = 1,   Hence,since s<sub>2</sub> = 6 -   and s<sub>3</sub> = s<sub>2</sub> -   ,there are two optimal plans as given in the table below.  ,there are two optimal plans as given in the table below. Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let x<sub>n</sub> be the advertising dollars (in millions)spent on product n.Let s<sub>n</sub> be the amount of advertising budget remaining.Let   be the increase in sales of product n when x<sub>n</sub> million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is   .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3,   For n = 2,   For n = 1,   Hence,since s<sub>2</sub> = 6 -   and s<sub>3</sub> = s<sub>2</sub> -   ,there are two optimal plans as given in the table below.
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Consider the following integer nonlinear programming problem.Maximize Z = Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem. ,subject to Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem. Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem. ≥ 1, Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem. ≥ 1, Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem. ≥ 1,and Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem. , Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem. , Consider the following integer nonlinear programming problem.Maximize Z =   ,subject to     ≥ 1,   ≥ 1,   ≥ 1,and   ,   ,   are integers.Use dynamic programming to solve this problem. are integers.Use dynamic programming to solve this problem.
The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state sn be the remaining slack in the constraint The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. ,so s1 = 10,s2 = 10 - x1,and s3 = s2 - 2x2.Let The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. .Then The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. ,and The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. for n = 1,2.For n = 3, The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. For n = 2, The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. For n = 1, The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. Since The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. = 2 leads to s2 = 10 - The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. = 8 and The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. = 1,which then leads to s3 = s2 - 2 The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. = 6 and The three stages involve making the decisions on the values of the three variables in order.At stage n,let the state s<sub>n</sub> be the remaining slack in the constraint   ,so s<sub>1</sub> = 10,s<sub>2</sub> = 10 - x<sub>1</sub>,and s<sub>3</sub> = s<sub>2</sub> - 2x<sub>2</sub>.Let   .Then   ,and   for n = 1,2.For n = 3,   For n = 2,   For n = 1,   Since   = 2 leads to s<sub>2</sub> = 10 -   = 8 and   = 1,which then leads to s<sub>3</sub> = s<sub>2</sub> - 2   = 6 and   = 2,it follows that (x<sub>1</sub>,x<sub>2</sub>,x<sub>3</sub>)= (2,1,2)is optimal with Z = 16. = 2,it follows that (x1,x2,x3)= (2,1,2)is optimal with Z = 16.
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Consider the following nonlinear programming problem.Maximize Z = Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. ,subject to 2 Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. + Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. ≤ 13 Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. + Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. ≤ 9 and Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. ≥ 0, Consider the following nonlinear programming problem.Maximize Z =   ,subject to 2   +   ≤ 13   +   ≤ 9 and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. ≥ 0.Use dynamic programming to solve this problem.
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Consider the following nonlinear programming problem.Maximize Z = Consider the following nonlinear programming problem.Maximize Z =   ,subject to   and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. ,subject to Consider the following nonlinear programming problem.Maximize Z =   ,subject to   and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. and Consider the following nonlinear programming problem.Maximize Z =   ,subject to   and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. ≥ 0, Consider the following nonlinear programming problem.Maximize Z =   ,subject to   and   ≥ 0,   ≥ 0.Use dynamic programming to solve this problem. ≥ 0.Use dynamic programming to solve this problem.
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