For the following linear programming problem,use the SOB method presented in Sec.6.4 of the textbook to construct its dual problem.Minimize Z = 3 x 1 + 2 x 2 , subject to 2 x 1 + x 2 $\ge$ 10 -3 x 1 + 2 x 2 $\le$ 6 1 x 1 + 1 x 2 $\ge$ 6 and x 1 $\ge$ 0,x 2 $\ge$ 0.

The four steps of the SOB method lead to the following conclusions.Step 1: Since the primal problem is in minimization form,the dual problem will be in maximization form.Step 2: Given the minimization form for the primal problem,its first and third functional constraints in ≥ form are labeled as "sensible," as are its two nonnegativity constraints,but the second functional constraint in ≤ form is labeled as "bizarre." Step 3: Because of this labeling of the functional constraints in the primal problem,the first and third variables in the dual problem will have nonnegativity constraints,y 1 ≥ 0 and y 3 ≥ 0,whereas the second variable will have a nonpositivity constraint, ≤ 0. Step 4: Because of the "sensible" labeling for the nonnegativity constraints in the primal problem,the two functional constraints in the dual problem will have the "sensible" ≤ form.Therefore,after placing the numbers in the primal problem in their usual locations for the dual problem,the dual problem is Maximize 10y 1 + 6 + 6y 3 ,subject to 2y 1 - 3 + y 3 $\le$ 3 y 1 + 2 + y 3 $\le$ 2 and y 1 $\ge$ 0, $\le$ 0,y 3 $\ge$ 0.We can also set the dual variable = - y 2 with y 2 $\ge$ 0.Then we obtain an equivalent dual problem: Maximize 10y 1 - 6y 2 + 6y 3 ,subject to 2y 1 + 3y 2 + y 3 $\le$ 3 y 1 - 2y 2 + y 3 $\le$ 2 and y 1 $\ge$ 0,y 2 $\ge$ 0,y 3 $\ge$ 0

Duality Theory Flashcards

The four steps of the SOB method lead to the following conclusions.Step 1: Since the primal problem is in minimization form,the dual problem will be in maximization form.Step 2: Given the minimization form for the primal problem,its first and third functional constraints in ≥ form are labeled as "sensible," as are its two nonnegativity constraints,but the second functional constraint in ≤ form is labeled as "bizarre." Step 3: Because of this labeling of the functional constraints in the primal problem,the first and third variables in the dual problem will have nonnegativity constraints,y₁ ≥ 0 and y₃ ≥ 0,whereas the second variable will have a nonpositivity constraint, $The four steps of the SOB method lead to the following conclusions.Step 1: Since the primal problem is in minimization form,the dual problem will be in maximization form.Step 2: Given the minimization form for the primal problem,its first and third functional constraints in ≥ form are labeled as sensible, as are its two nonnegativity constraints,but the second functional constraint in ≤ form is labeled as bizarre. Step 3: Because of this labeling of the functional constraints in the primal problem,the first and third variables in the dual problem will have nonnegativity constraints,y1 ≥ 0 and y3 ≥ 0,whereas the second variable will have a nonpositivity constraint, ≤ 0. Step 4: Because of the sensible labeling for the nonnegativity constraints in the primal problem,the two functional constraints in the dual problem will have the sensible ≤ form.Therefore,after placing the numbers in the primal problem in their usual locations for the dual problem,the dual problem is Maximize 10y1 + 6 + 6y3,subject to 2y1 - 3 + y3 \le 3 y1 + 2 + y3 \le 2 and y1 \ge 0, \le 0,y3 \ge 0.We can also set the dual variable = - y2 with y2 \ge 0.Then we obtain an equivalent dual problem: Maximize 10y1 - 6y2 + 6y3,subject to 2y1 + 3y2 + y3 \le 3 y1 - 2y2 + y3 \le 2 and y1 \ge 0,y2 \ge 0,y3 \ge 0$ ≤ 0.
Step 4: Because of the "sensible" labeling for the nonnegativity constraints in the primal problem,the two functional constraints in the dual problem will have the "sensible" ≤ form.Therefore,after placing the numbers in the primal problem in their usual locations for the dual problem,the dual problem is Maximize 10y₁ + 6 $The four steps of the SOB method lead to the following conclusions.Step 1: Since the primal problem is in minimization form,the dual problem will be in maximization form.Step 2: Given the minimization form for the primal problem,its first and third functional constraints in ≥ form are labeled as sensible, as are its two nonnegativity constraints,but the second functional constraint in ≤ form is labeled as bizarre. Step 3: Because of this labeling of the functional constraints in the primal problem,the first and third variables in the dual problem will have nonnegativity constraints,y1 ≥ 0 and y3 ≥ 0,whereas the second variable will have a nonpositivity constraint, ≤ 0. Step 4: Because of the sensible labeling for the nonnegativity constraints in the primal problem,the two functional constraints in the dual problem will have the sensible ≤ form.Therefore,after placing the numbers in the primal problem in their usual locations for the dual problem,the dual problem is Maximize 10y1 + 6 + 6y3,subject to 2y1 - 3 + y3 \le 3 y1 + 2 + y3 \le 2 and y1 \ge 0, \le 0,y3 \ge 0.We can also set the dual variable = - y2 with y2 \ge 0.Then we obtain an equivalent dual problem: Maximize 10y1 - 6y2 + 6y3,subject to 2y1 + 3y2 + y3 \le 3 y1 - 2y2 + y3 \le 2 and y1 \ge 0,y2 \ge 0,y3 \ge 0$ + 6y₃,subject to 2y₁ - 3 $The four steps of the SOB method lead to the following conclusions.Step 1: Since the primal problem is in minimization form,the dual problem will be in maximization form.Step 2: Given the minimization form for the primal problem,its first and third functional constraints in ≥ form are labeled as sensible, as are its two nonnegativity constraints,but the second functional constraint in ≤ form is labeled as bizarre. Step 3: Because of this labeling of the functional constraints in the primal problem,the first and third variables in the dual problem will have nonnegativity constraints,y1 ≥ 0 and y3 ≥ 0,whereas the second variable will have a nonpositivity constraint, ≤ 0. Step 4: Because of the sensible labeling for the nonnegativity constraints in the primal problem,the two functional constraints in the dual problem will have the sensible ≤ form.Therefore,after placing the numbers in the primal problem in their usual locations for the dual problem,the dual problem is Maximize 10y1 + 6 + 6y3,subject to 2y1 - 3 + y3 \le 3 y1 + 2 + y3 \le 2 and y1 \ge 0, \le 0,y3 \ge 0.We can also set the dual variable = - y2 with y2 \ge 0.Then we obtain an equivalent dual problem: Maximize 10y1 - 6y2 + 6y3,subject to 2y1 + 3y2 + y3 \le 3 y1 - 2y2 + y3 \le 2 and y1 \ge 0,y2 \ge 0,y3 \ge 0$ + y₃

\le

3 y₁ + 2 $The four steps of the SOB method lead to the following conclusions.Step 1: Since the primal problem is in minimization form,the dual problem will be in maximization form.Step 2: Given the minimization form for the primal problem,its first and third functional constraints in ≥ form are labeled as sensible, as are its two nonnegativity constraints,but the second functional constraint in ≤ form is labeled as bizarre. Step 3: Because of this labeling of the functional constraints in the primal problem,the first and third variables in the dual problem will have nonnegativity constraints,y1 ≥ 0 and y3 ≥ 0,whereas the second variable will have a nonpositivity constraint, ≤ 0. Step 4: Because of the sensible labeling for the nonnegativity constraints in the primal problem,the two functional constraints in the dual problem will have the sensible ≤ form.Therefore,after placing the numbers in the primal problem in their usual locations for the dual problem,the dual problem is Maximize 10y1 + 6 + 6y3,subject to 2y1 - 3 + y3 \le 3 y1 + 2 + y3 \le 2 and y1 \ge 0, \le 0,y3 \ge 0.We can also set the dual variable = - y2 with y2 \ge 0.Then we obtain an equivalent dual problem: Maximize 10y1 - 6y2 + 6y3,subject to 2y1 + 3y2 + y3 \le 3 y1 - 2y2 + y3 \le 2 and y1 \ge 0,y2 \ge 0,y3 \ge 0$ + y₃

\le

2 and y₁

\ge

0, $The four steps of the SOB method lead to the following conclusions.Step 1: Since the primal problem is in minimization form,the dual problem will be in maximization form.Step 2: Given the minimization form for the primal problem,its first and third functional constraints in ≥ form are labeled as sensible, as are its two nonnegativity constraints,but the second functional constraint in ≤ form is labeled as bizarre. Step 3: Because of this labeling of the functional constraints in the primal problem,the first and third variables in the dual problem will have nonnegativity constraints,y1 ≥ 0 and y3 ≥ 0,whereas the second variable will have a nonpositivity constraint, ≤ 0. Step 4: Because of the sensible labeling for the nonnegativity constraints in the primal problem,the two functional constraints in the dual problem will have the sensible ≤ form.Therefore,after placing the numbers in the primal problem in their usual locations for the dual problem,the dual problem is Maximize 10y1 + 6 + 6y3,subject to 2y1 - 3 + y3 \le 3 y1 + 2 + y3 \le 2 and y1 \ge 0, \le 0,y3 \ge 0.We can also set the dual variable = - y2 with y2 \ge 0.Then we obtain an equivalent dual problem: Maximize 10y1 - 6y2 + 6y3,subject to 2y1 + 3y2 + y3 \le 3 y1 - 2y2 + y3 \le 2 and y1 \ge 0,y2 \ge 0,y3 \ge 0$

\le

0,y₃

\ge

0.We can also set the dual variable $The four steps of the SOB method lead to the following conclusions.Step 1: Since the primal problem is in minimization form,the dual problem will be in maximization form.Step 2: Given the minimization form for the primal problem,its first and third functional constraints in ≥ form are labeled as sensible, as are its two nonnegativity constraints,but the second functional constraint in ≤ form is labeled as bizarre. Step 3: Because of this labeling of the functional constraints in the primal problem,the first and third variables in the dual problem will have nonnegativity constraints,y1 ≥ 0 and y3 ≥ 0,whereas the second variable will have a nonpositivity constraint, ≤ 0. Step 4: Because of the sensible labeling for the nonnegativity constraints in the primal problem,the two functional constraints in the dual problem will have the sensible ≤ form.Therefore,after placing the numbers in the primal problem in their usual locations for the dual problem,the dual problem is Maximize 10y1 + 6 + 6y3,subject to 2y1 - 3 + y3 \le 3 y1 + 2 + y3 \le 2 and y1 \ge 0, \le 0,y3 \ge 0.We can also set the dual variable = - y2 with y2 \ge 0.Then we obtain an equivalent dual problem: Maximize 10y1 - 6y2 + 6y3,subject to 2y1 + 3y2 + y3 \le 3 y1 - 2y2 + y3 \le 2 and y1 \ge 0,y2 \ge 0,y3 \ge 0$ = - y₂ with y₂

\ge

0.Then we obtain an equivalent dual problem: Maximize 10y₁ - 6y₂ + 6y₃,subject to 2y₁ + 3y₂ + y₃

\le

3 y₁ - 2y₂ + y₃

\le

2 and y₁

\ge

0,y₂

\ge

0,y₃

\ge

Deck 6: Duality Theory