Deck 5: Present Worth Analysis

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Question
The annual maintenance cost of a monument in the state capital is estimated to be $4850. A perpetual i fund of $100,000 is set up to pay for this maintenance expenditure. Determine the interest rate this fund earns if the interest is compounded quarterly.

A) 4.58%
B) 4.76%
C) 4.39%
D) None of these
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Case Study 5
Dunn Manufacturing is considering the following two alternatives. The cost information for the two proposals for replacing an equipment are provided are in table below.
 Machine X Machine Y Initial cost $120,000$96,000 Benefits/year $20,000 for the first 10 years  and $9,000 for the next 10 years $12,000 per year for 20 years.  Life 20 years  Salvage value $40,000$20,000 MARR 8%\begin{array}{|l|l|l|} \hline& \text { Machine } X & \text { Machine } Y \\\hline \text { Initial cost } & \$ 120,000 & \$ 96,000 \\\hline \text { Benefits/year } & \begin{array}{l}\$ 20,000 \text { for the first } 10 \text { years } \\\text { and } \$ 9,000 \text { for the next } 10 \\\text { years }\end{array} & \begin{array}{l}\$ 12,000 \text { per year for } 20 \\\text { years. }\end{array} \\\hline \text { Life } & {20 \text { years }} \\\hline \text { Salvage value } & \$ 40,000 & \$ 20,000 \\\hline \text { MARR } & {8 \%}\\\hline\end{array}

-The NPW of machine X is ________________.

A) $35,158
B) $48,192
C) $50,752
D) $61,239
Question
Case Study 5
Dunn Manufacturing is considering the following two alternatives. The cost information for the two proposals for replacing an equipment are provided are in table below.
 Machine X Machine Y Initial cost $120,000$96,000 Benefits/year $20,000 for the first 10 years  and $9,000 for the next 10 years $12,000 per year for 20 years.  Life 20 years  Salvage value $40,000$20,000 MARR 8%\begin{array}{|l|l|l|} \hline& \text { Machine } X & \text { Machine } Y \\\hline \text { Initial cost } & \$ 120,000 & \$ 96,000 \\\hline \text { Benefits/year } & \begin{array}{l}\$ 20,000 \text { for the first } 10 \text { years } \\\text { and } \$ 9,000 \text { for the next } 10 \\\text { years }\end{array} & \begin{array}{l}\$ 12,000 \text { per year for } 20 \\\text { years. }\end{array} \\\hline \text { Life } & {20 \text { years }} \\\hline \text { Salvage value } & \$ 40,000 & \$ 20,000 \\\hline \text { MARR } & {8 \%}\\\hline\end{array}

-The NPW of machine Y is ________________.

A) $42,196
B) $29,508
C) $26,106
D) $32,103
Question
Case Study 5
Dunn Manufacturing is considering the following two alternatives. The cost information for the two proposals for replacing an equipment are provided are in table below.
 Machine X Machine Y Initial cost $120,000$96,000 Benefits/year $20,000 for the first 10 years  and $9,000 for the next 10 years $12,000 per year for 20 years.  Life 20 years  Salvage value $40,000$20,000 MARR 8%\begin{array}{|l|l|l|} \hline& \text { Machine } X & \text { Machine } Y \\\hline \text { Initial cost } & \$ 120,000 & \$ 96,000 \\\hline \text { Benefits/year } & \begin{array}{l}\$ 20,000 \text { for the first } 10 \text { years } \\\text { and } \$ 9,000 \text { for the next } 10 \\\text { years }\end{array} & \begin{array}{l}\$ 12,000 \text { per year for } 20 \\\text { years. }\end{array} \\\hline \text { Life } & {20 \text { years }} \\\hline \text { Salvage value } & \$ 40,000 & \$ 20,000 \\\hline \text { MARR } & {8 \%}\\\hline\end{array}

-If machine "Y" has no salvage value, what would be the NPW of machine "Y"?

A) $19,891
B) $20,626
C) $23,478
D) $21,816
Question
How much money would have to be placed in a sinking fund each year to replace a machine costing $20,000 today at the end of 10 years if the fund yields 9% annual interest rate compounded yearly and if the cost of the machine is assumed to increase at 5% annually for the next 10 years? Assume the salvage value is zero.

A) $2,143.63
B) $2150
C) $2,340
D) $1923
Question
Given the cash flows in table below. Determine the value of P. i= 6% per year
 Year 012345 Cash Flow P400800120016001800\begin{array} { | l | l | l | l | l | l | l | } \hline \text { Year } & 0 & 1 & 2 & 3 & 4 & 5 \\\hline \text { Cash Flow } & - \mathrm { P } & 400 & 800 & 1200 & 1600 & 1800 \\\hline\end{array}

A) $4709.14
B) $6125.15
C) $5125.63
D) $4978.23
Question
Determine the net present worth (NPW) of the cash flows given in table below for an investment opportunity being presented to a company. MARR =12%
 Year 0110111516252630 Cash Flow $100K10K20K5K30K\begin{array} { | l | l | l | l | l | l | } \hline \text { Year } & 0 & 1 - 10 & 11 - 15 & 16 - 25 & 26 - 30 \\\hline \text { Cash Flow } & - \$ 100 K & 10 K & 20 K & - 5 K & 30 K \\\hline\end{array}

A) $90,030
B) $80,914
C) $72,916
D) $112,200
Question
Kal Tech Engineering Systems is considering buying a CNC machining center for its operation in Tennessee. The net benefits in the first year is estimated to be $40,000 and increasing at the rate $5,000 for the next four years and stays at the same level as that of year 5 for the next 5 years. If MARR is 8%, determine the amount of money that the company can invest justifying on this machining center. A salvage value of 20% of the initial cost is reasonable to assume at the end of year 10.

A) $396,357
B) $416,182
C) $411,202
D) $399,500
Question
Sam is considering investing in a bond with a face value of $20,000. The bond pays an interest of 4% payable quarterly. If he expects to make a 1 ½ % return per quarter on this investment with a maturity of 20 years, determine the most he can pay for the bond ________.

A) $18,102.65
B) $14,923.86
C) $15,355.40
D) $16,000
Question
Table 5
ABC Initial cost $15,000$9,000$12,000 Annual benefit $8,000$2,000$1,800 Salvage value $5,000$9,0000 Life in years 2 years 3 Years  Infinity  MARR 10%\begin{array} { | l | l | l | l | } \hline & \mathrm { A } & \mathrm { B } & \mathrm { C } \\\hline \text { Initial cost } & \$ 15,000 & \$ 9,000 & \$ 12,000 \\\hline \text { Annual benefit } & \$ 8,000 & \$ 2,000 & \$ 1,800 \\\hline \text { Salvage value } & \$ 5,000 & \$ 9,000 & 0 \\\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } & \text { Infinity } \\\hline \text { MARR } & &&10 \% \\\hline\end{array}

-The NPW of alt. A is __________________.

A) $13,420
B) $17,380
C) $11,000
D) $6,000
Question
Table 5
ABC Initial cost $15,000$9,000$12,000 Annual benefit $8,000$2,000$1,800 Salvage value $5,000$9,0000 Life in years 2 years 3 Years  Infinity  MARR 10%\begin{array} { | l | l | l | l | } \hline & \mathrm { A } & \mathrm { B } & \mathrm { C } \\\hline \text { Initial cost } & \$ 15,000 & \$ 9,000 & \$ 12,000 \\\hline \text { Annual benefit } & \$ 8,000 & \$ 2,000 & \$ 1,800 \\\hline \text { Salvage value } & \$ 5,000 & \$ 9,000 & 0 \\\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } & \text { Infinity } \\\hline \text { MARR } & &&10 \% \\\hline\end{array}

-The NPW of alt. B is __________________.

A) $13,420
B) $17,380
C) $11,000
D) $6,000
Question
Table 5
ABC Initial cost $15,000$9,000$12,000 Annual benefit $8,000$2,000$1,800 Salvage value $5,000$9,0000 Life in years 2 years 3 Years  Infinity  MARR 10%\begin{array} { | l | l | l | l | } \hline & \mathrm { A } & \mathrm { B } & \mathrm { C } \\\hline \text { Initial cost } & \$ 15,000 & \$ 9,000 & \$ 12,000 \\\hline \text { Annual benefit } & \$ 8,000 & \$ 2,000 & \$ 1,800 \\\hline \text { Salvage value } & \$ 5,000 & \$ 9,000 & 0 \\\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } & \text { Infinity } \\\hline \text { MARR } & &&10 \% \\\hline\end{array}

-The NPW of alt. C is __________________.

A) $13,420
B) $17,380
C) $11,000
D) $6,000
Question
Determine the better of the two alternatives using the present worth analysis. Use an interest rate of 10%.
Determine the better of the two alternatives using the present worth analysis. Use an interest rate of 10%.  <div style=padding-top: 35px>
Question
Solve the following problem using the present worth analysis for an interest rate of 8%.
Solve the following problem using the present worth analysis for an interest rate of 8%.  <div style=padding-top: 35px>
Question
The capitalized cost of any investment may be determined using the equation P = A/i where P is the capitalized cost, A is the annual amount and i is the interest rate.
Question
The NPW of a set of cash flows will decrease as the interest rate is increased.
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Deck 5: Present Worth Analysis
1
The annual maintenance cost of a monument in the state capital is estimated to be $4850. A perpetual i fund of $100,000 is set up to pay for this maintenance expenditure. Determine the interest rate this fund earns if the interest is compounded quarterly.

A) 4.58%
B) 4.76%
C) 4.39%
D) None of these
4.76%
2
Case Study 5
Dunn Manufacturing is considering the following two alternatives. The cost information for the two proposals for replacing an equipment are provided are in table below.
 Machine X Machine Y Initial cost $120,000$96,000 Benefits/year $20,000 for the first 10 years  and $9,000 for the next 10 years $12,000 per year for 20 years.  Life 20 years  Salvage value $40,000$20,000 MARR 8%\begin{array}{|l|l|l|} \hline& \text { Machine } X & \text { Machine } Y \\\hline \text { Initial cost } & \$ 120,000 & \$ 96,000 \\\hline \text { Benefits/year } & \begin{array}{l}\$ 20,000 \text { for the first } 10 \text { years } \\\text { and } \$ 9,000 \text { for the next } 10 \\\text { years }\end{array} & \begin{array}{l}\$ 12,000 \text { per year for } 20 \\\text { years. }\end{array} \\\hline \text { Life } & {20 \text { years }} \\\hline \text { Salvage value } & \$ 40,000 & \$ 20,000 \\\hline \text { MARR } & {8 \%}\\\hline\end{array}

-The NPW of machine X is ________________.

A) $35,158
B) $48,192
C) $50,752
D) $61,239
$50,752
3
Case Study 5
Dunn Manufacturing is considering the following two alternatives. The cost information for the two proposals for replacing an equipment are provided are in table below.
 Machine X Machine Y Initial cost $120,000$96,000 Benefits/year $20,000 for the first 10 years  and $9,000 for the next 10 years $12,000 per year for 20 years.  Life 20 years  Salvage value $40,000$20,000 MARR 8%\begin{array}{|l|l|l|} \hline& \text { Machine } X & \text { Machine } Y \\\hline \text { Initial cost } & \$ 120,000 & \$ 96,000 \\\hline \text { Benefits/year } & \begin{array}{l}\$ 20,000 \text { for the first } 10 \text { years } \\\text { and } \$ 9,000 \text { for the next } 10 \\\text { years }\end{array} & \begin{array}{l}\$ 12,000 \text { per year for } 20 \\\text { years. }\end{array} \\\hline \text { Life } & {20 \text { years }} \\\hline \text { Salvage value } & \$ 40,000 & \$ 20,000 \\\hline \text { MARR } & {8 \%}\\\hline\end{array}

-The NPW of machine Y is ________________.

A) $42,196
B) $29,508
C) $26,106
D) $32,103
$26,106
4
Case Study 5
Dunn Manufacturing is considering the following two alternatives. The cost information for the two proposals for replacing an equipment are provided are in table below.
 Machine X Machine Y Initial cost $120,000$96,000 Benefits/year $20,000 for the first 10 years  and $9,000 for the next 10 years $12,000 per year for 20 years.  Life 20 years  Salvage value $40,000$20,000 MARR 8%\begin{array}{|l|l|l|} \hline& \text { Machine } X & \text { Machine } Y \\\hline \text { Initial cost } & \$ 120,000 & \$ 96,000 \\\hline \text { Benefits/year } & \begin{array}{l}\$ 20,000 \text { for the first } 10 \text { years } \\\text { and } \$ 9,000 \text { for the next } 10 \\\text { years }\end{array} & \begin{array}{l}\$ 12,000 \text { per year for } 20 \\\text { years. }\end{array} \\\hline \text { Life } & {20 \text { years }} \\\hline \text { Salvage value } & \$ 40,000 & \$ 20,000 \\\hline \text { MARR } & {8 \%}\\\hline\end{array}

-If machine "Y" has no salvage value, what would be the NPW of machine "Y"?

A) $19,891
B) $20,626
C) $23,478
D) $21,816
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5
How much money would have to be placed in a sinking fund each year to replace a machine costing $20,000 today at the end of 10 years if the fund yields 9% annual interest rate compounded yearly and if the cost of the machine is assumed to increase at 5% annually for the next 10 years? Assume the salvage value is zero.

A) $2,143.63
B) $2150
C) $2,340
D) $1923
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6
Given the cash flows in table below. Determine the value of P. i= 6% per year
 Year 012345 Cash Flow P400800120016001800\begin{array} { | l | l | l | l | l | l | l | } \hline \text { Year } & 0 & 1 & 2 & 3 & 4 & 5 \\\hline \text { Cash Flow } & - \mathrm { P } & 400 & 800 & 1200 & 1600 & 1800 \\\hline\end{array}

A) $4709.14
B) $6125.15
C) $5125.63
D) $4978.23
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7
Determine the net present worth (NPW) of the cash flows given in table below for an investment opportunity being presented to a company. MARR =12%
 Year 0110111516252630 Cash Flow $100K10K20K5K30K\begin{array} { | l | l | l | l | l | l | } \hline \text { Year } & 0 & 1 - 10 & 11 - 15 & 16 - 25 & 26 - 30 \\\hline \text { Cash Flow } & - \$ 100 K & 10 K & 20 K & - 5 K & 30 K \\\hline\end{array}

A) $90,030
B) $80,914
C) $72,916
D) $112,200
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8
Kal Tech Engineering Systems is considering buying a CNC machining center for its operation in Tennessee. The net benefits in the first year is estimated to be $40,000 and increasing at the rate $5,000 for the next four years and stays at the same level as that of year 5 for the next 5 years. If MARR is 8%, determine the amount of money that the company can invest justifying on this machining center. A salvage value of 20% of the initial cost is reasonable to assume at the end of year 10.

A) $396,357
B) $416,182
C) $411,202
D) $399,500
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9
Sam is considering investing in a bond with a face value of $20,000. The bond pays an interest of 4% payable quarterly. If he expects to make a 1 ½ % return per quarter on this investment with a maturity of 20 years, determine the most he can pay for the bond ________.

A) $18,102.65
B) $14,923.86
C) $15,355.40
D) $16,000
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10
Table 5
ABC Initial cost $15,000$9,000$12,000 Annual benefit $8,000$2,000$1,800 Salvage value $5,000$9,0000 Life in years 2 years 3 Years  Infinity  MARR 10%\begin{array} { | l | l | l | l | } \hline & \mathrm { A } & \mathrm { B } & \mathrm { C } \\\hline \text { Initial cost } & \$ 15,000 & \$ 9,000 & \$ 12,000 \\\hline \text { Annual benefit } & \$ 8,000 & \$ 2,000 & \$ 1,800 \\\hline \text { Salvage value } & \$ 5,000 & \$ 9,000 & 0 \\\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } & \text { Infinity } \\\hline \text { MARR } & &&10 \% \\\hline\end{array}

-The NPW of alt. A is __________________.

A) $13,420
B) $17,380
C) $11,000
D) $6,000
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11
Table 5
ABC Initial cost $15,000$9,000$12,000 Annual benefit $8,000$2,000$1,800 Salvage value $5,000$9,0000 Life in years 2 years 3 Years  Infinity  MARR 10%\begin{array} { | l | l | l | l | } \hline & \mathrm { A } & \mathrm { B } & \mathrm { C } \\\hline \text { Initial cost } & \$ 15,000 & \$ 9,000 & \$ 12,000 \\\hline \text { Annual benefit } & \$ 8,000 & \$ 2,000 & \$ 1,800 \\\hline \text { Salvage value } & \$ 5,000 & \$ 9,000 & 0 \\\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } & \text { Infinity } \\\hline \text { MARR } & &&10 \% \\\hline\end{array}

-The NPW of alt. B is __________________.

A) $13,420
B) $17,380
C) $11,000
D) $6,000
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12
Table 5
ABC Initial cost $15,000$9,000$12,000 Annual benefit $8,000$2,000$1,800 Salvage value $5,000$9,0000 Life in years 2 years 3 Years  Infinity  MARR 10%\begin{array} { | l | l | l | l | } \hline & \mathrm { A } & \mathrm { B } & \mathrm { C } \\\hline \text { Initial cost } & \$ 15,000 & \$ 9,000 & \$ 12,000 \\\hline \text { Annual benefit } & \$ 8,000 & \$ 2,000 & \$ 1,800 \\\hline \text { Salvage value } & \$ 5,000 & \$ 9,000 & 0 \\\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } & \text { Infinity } \\\hline \text { MARR } & &&10 \% \\\hline\end{array}

-The NPW of alt. C is __________________.

A) $13,420
B) $17,380
C) $11,000
D) $6,000
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13
Determine the better of the two alternatives using the present worth analysis. Use an interest rate of 10%.
Determine the better of the two alternatives using the present worth analysis. Use an interest rate of 10%.
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14
Solve the following problem using the present worth analysis for an interest rate of 8%.
Solve the following problem using the present worth analysis for an interest rate of 8%.
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15
The capitalized cost of any investment may be determined using the equation P = A/i where P is the capitalized cost, A is the annual amount and i is the interest rate.
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16
The NPW of a set of cash flows will decrease as the interest rate is increased.
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