Question 1

Given the cash flows in table below. Determine the value of P. i= 6% per year
$$\begin{array} { | l | l | l | l | l | l | l | } 
\hline \text { Year } & 0 & 1 & 2 & 3 & 4 & 5 \
\hline \text { Cash Flow } & - \mathrm { P } & 400 & 800 & 1200 & 1600 & 1800 \
\hline
\end{array}$$

Accepted Answer

A)  $4709.14 
B)  $6125.15 
C)  $5125.63 
D)  $4978.23 
A)  $4709.14 
B)  $6125.15 
C)  $5125.63 
D)  $4978.23 A

Question 2

Determine the better of the two alternatives using the present worth analysis. Use an interest rate of 10%.
 $$\begin{array} { | l | l | l | } 
\hline & \text { Alt. X } & \text { Alt. Y } \
\hline \text { Iritial cost } & \$ 12,500 & \$ 8,900 \
\hline \text { Arrual berefit } & \$ 6,800 & \$ 2,000 \
\hline \text { Salvage value } & \$ 5,000 & \$ 8,900 \
\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } \
\hline \text { MARR } & 10 \% & \
\hline
\end{array}$$

Accepted Answer

The analysis period is 6 years since the LCM of 2 and 3 is 6.
NPW (Alt. X) = 6,800(P/A,10%,6) + 5,000[(P/F, 10%,2) +(P/F,10%,4)+(P/F,10%,6)]
− 12,500[1 + (P/F, 10%, 2) + (P/F, 10%, 4)]
= 6,800(4.355) +5,000[0.8264+0.6830+0.5845] − 12,500[1+0.8264+0.6830] = $8,716
NPW (Alt. Y) = 2,000(P/A, 10%, 6) + 8,900[(P/F, 10%, 3) + (P/F, 10%, 6)]
− 8,900[1+ (P/F, 10%, 3)]
= 2,000(4.355) +8,900 [0.7513 + 0.5645] − 8,900[1 + 0.7513] = $4,834.05
To maximize NPW, choose alt. X.

Question 3

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%
$$\begin{array} { | l | l | l | l | l | l | } 
\hline \text { Year } & 0 & 1 - 10 & 11 - 15 & 16 - 25 & 26 - 30 \
\hline \text { Cash Flow } & - \$ 100 K & 10 K & 20 K & - 5 K & 30 K \
\hline
\end{array}$$

Accepted Answer

A)  $90,030 
B)  $80,914 
C)  $72,916 
D)  $112,200 
A)  $90,030 
B)  $80,914 
C)  $72,916 
D)  $112,200 B

Question 4

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 ________.

Accepted Answer

A)  $18,102.65 
B)  $14,923.86 
C)  $15,355.40 
D)  $16,000 
A)  $18,102.65 
B)  $14,923.86 
C)  $15,355.40 
D)  $16,000

Question 5

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.

Accepted Answer

A)  $396,357 
B)  $416,182 
C)  $411,202 
D)  $399,500 
A)  $396,357 
B)  $416,182 
C)  $411,202 
D)  $399,500

Question 6

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.
 $\begin{array}{|l|l|l|} 
\hline& \text { Machine } X & \text { Machine } Y \
\hline \text { Initial cost } & \$ 120,000 & \$ 96,000 \
\hline \text { Benefits/year } & \begin{array}{l}
\$ 20,000 \text { for the first } 10 \text { years } \
\text { and } \$ 9,000 \text { for the next } 10 \
\text { years }
\end{array} & \begin{array}{l}
\$ 12,000 \text { per year for } 20 \
\text { years. }
\end{array} \
\hline \text { Life } & {20 \text { years }} \
\hline \text { Salvage value } & \$ 40,000 & \$ 20,000 \
\hline \text { MARR } & {8 \%}\\hline
\end{array}$
-The NPW of machine Y is ________________.

Accepted Answer

A)  $42,196 
B)  $29,508 
C)  $26,106 
D)  $32,103 
A)  $42,196 
B)  $29,508 
C)  $26,106 
D)  $32,103

Question 7

Table 5
 $$\begin{array} { | l | l | l | l | } 
\hline & \mathrm { A } & \mathrm { B } & \mathrm { C } \
\hline \text { Initial cost } & \$ 15,000 & \$ 9,000 & \$ 12,000 \
\hline \text { Annual benefit } & \$ 8,000 & \$ 2,000 & \$ 1,800 \
\hline \text { Salvage value } & \$ 5,000 & \$ 9,000 & 0 \
\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } & \text { Infinity } \
\hline \text { MARR } & &&10 \% \
\hline
\end{array}$$
-The NPW of alt. B is __________________.

Accepted Answer

A)  $13,420 
B)  $17,380 
C)  $11,000 
D)  $6,000 
A)  $13,420 
B)  $17,380 
C)  $11,000 
D)  $6,000

Question 8

Table 5
 $$\begin{array} { | l | l | l | l | } 
\hline & \mathrm { A } & \mathrm { B } & \mathrm { C } \
\hline \text { Initial cost } & \$ 15,000 & \$ 9,000 & \$ 12,000 \
\hline \text { Annual benefit } & \$ 8,000 & \$ 2,000 & \$ 1,800 \
\hline \text { Salvage value } & \$ 5,000 & \$ 9,000 & 0 \
\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } & \text { Infinity } \
\hline \text { MARR } & &&10 \% \
\hline
\end{array}$$
-The NPW of alt. C is __________________.

Accepted Answer

A)  $13,420 
B)  $17,380 
C)  $11,000 
D)  $6,000 
A)  $13,420 
B)  $17,380 
C)  $11,000 
D)  $6,000

Question 9

Solve the following problem using the present worth analysis for an interest rate of 8%.
 $$\begin{array} { | l | c | l | l | } 
\hline & \text { Alt. A } & \text { Alt. B } & \text { Alt. C } \
\hline \text { Initial cost } & \$ 1,700 & \$ 2,100 & \$ 3,750 \
\hline \text { Beniefit/year } & 1,000 & 1,000 & 1,000 \
\hline \text { Life in years } & \mathbf { 2 } & 3 & 6 \
\hline
\end{array}$$

Accepted Answer

The answer of Solve the following problem using the present...

Question 10

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.
 $\begin{array}{|l|l|l|} 
\hline& \text { Machine } X & \text { Machine } Y \
\hline \text { Initial cost } & \$ 120,000 & \$ 96,000 \
\hline \text { Benefits/year } & \begin{array}{l}
\$ 20,000 \text { for the first } 10 \text { years } \
\text { and } \$ 9,000 \text { for the next } 10 \
\text { years }
\end{array} & \begin{array}{l}
\$ 12,000 \text { per year for } 20 \
\text { years. }
\end{array} \
\hline \text { Life } & {20 \text { years }} \
\hline \text { Salvage value } & \$ 40,000 & \$ 20,000 \
\hline \text { MARR } & {8 \%}\\hline
\end{array}$
-The NPW of machine X is ________________.

Accepted Answer

A)  $35,158 
B)  $48,192 
C)  $50,752 
D)  $61,239 
A)  $35,158 
B)  $48,192 
C)  $50,752 
D)  $61,239

Question 11

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.
 $\begin{array}{|l|l|l|} 
\hline& \text { Machine } X & \text { Machine } Y \
\hline \text { Initial cost } & \$ 120,000 & \$ 96,000 \
\hline \text { Benefits/year } & \begin{array}{l}
\$ 20,000 \text { for the first } 10 \text { years } \
\text { and } \$ 9,000 \text { for the next } 10 \
\text { years }
\end{array} & \begin{array}{l}
\$ 12,000 \text { per year for } 20 \
\text { years. }
\end{array} \
\hline \text { Life } & {20 \text { years }} \
\hline \text { Salvage value } & \$ 40,000 & \$ 20,000 \
\hline \text { MARR } & {8 \%}\\hline
\end{array}$
-If machine "Y" has no salvage value, what would be the NPW of machine "Y"?

Accepted Answer

A)  $19,891 
B)  $20,626 
C)  $23,478 
D)  $21,816 
A)  $19,891 
B)  $20,626 
C)  $23,478 
D)  $21,816

Question 12

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

Accepted Answer

The answer of How much money would have to be...

Question 13

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.

Accepted Answer

A)  4.58% 
B)  4.76% 
C)  4.39% 
D)  None of these 
A)  4.58% 
B)  4.76% 
C)  4.39% 
D)  None of these

Question 14

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.

Accepted Answer

A) True 
 B)False

Question 15

The NPW of a set of cash flows will decrease as the interest rate is increased.

Accepted Answer

A) True 
 B)False

Question 16

Table 5
 $$\begin{array} { | l | l | l | l | } 
\hline & \mathrm { A } & \mathrm { B } & \mathrm { C } \
\hline \text { Initial cost } & \$ 15,000 & \$ 9,000 & \$ 12,000 \
\hline \text { Annual benefit } & \$ 8,000 & \$ 2,000 & \$ 1,800 \
\hline \text { Salvage value } & \$ 5,000 & \$ 9,000 & 0 \
\hline \text { Life in years } & 2 \text { years } & 3 \text { Years } & \text { Infinity } \
\hline \text { MARR } & &&10 \% \
\hline
\end{array}$$
-The NPW of alt. A is __________________.

Accepted Answer

A)  $13,420 
B)  $17,380 
C)  $11,000 
D)  $6,000 
A)  $13,420 
B)  $17,380 
C)  $11,000 
D)  $6,000

Given the cash flows in table below. Determine the value of P. i= 6% per year Year 0 1 2 3 4 5 Cash Flow - 400 800 1200 1600 1800

Determine the better of the two alternatives using the present worth analysis. Use an interest rate of 10%. Alt. X Alt. Y Iritial cost \ 12,500 \ 8,900 Arrual berefit \ 6,800 \ 2,000 Salvage value \ 5,000 \ 8,900 Life in years 2 years 3 Years MARR 10\%

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 0 1-10 11-15 16-25 26-30 Cash Flow -\ 100K 10K 20K -5K 30K

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 ________.

Table 5 Initial cost \ 15,000 \ 9,000 \ 12,000 Annual benefit \ 8,000 \ 2,000 \ 1,800 Salvage value \ 5,000 \ 9,000 0 Life in years 2 years 3 Years Infinity MARR 10\% -The NPW of alt. B is __________________.

Table 5 Initial cost \ 15,000 \ 9,000 \ 12,000 Annual benefit \ 8,000 \ 2,000 \ 1,800 Salvage value \ 5,000 \ 9,000 0 Life in years 2 years 3 Years Infinity MARR 10\% -The NPW of alt. C is __________________.

Solve the following problem using the present worth analysis for an interest rate of 8%. Alt. A Alt. B Alt. C Initial cost \ 1,700 \ 2,100 \ 3,750 Beniefit/year 1,000 1,000 1,000 Life in years 3 6

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.

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.

The NPW of a set of cash flows will decrease as the interest rate is increased.

Table 5 Initial cost \ 15,000 \ 9,000 \ 12,000 Annual benefit \ 8,000 \ 2,000 \ 1,800 Salvage value \ 5,000 \ 9,000 0 Life in years 2 years 3 Years Infinity MARR 10\% -The NPW of alt. A is __________________.

Making Economic Decisions

Estimating Engineering Costs and Benefits

Interest and Equivalence

Equivalence for Repeated Cash Flows

Annual Cash Flow Analysis

Rate of Return Analysis

Choosing the Best Alternative

Other Analysis Techniques

Uncertainty in Future Events

Depreciation

Income Taxes for Corporations

Replacement Analysis

Inflation and Price Change

Selection of a Minimum Attractive Rate of Return

Economic Analysis in the Public Sector

Accounting and Engineering Economy

Filters

Exam 5: Present Worth Analysis

Given the cash flows in table below. Determine the value of P. i= 6% per year Year 0 1 2 3 4 5 Cash Flow - 400 800 1200 1600 1800

Determine the better of the two alternatives using the present worth analysis. Use an interest rate of 10%. Alt. X Alt. Y Iritial cost \ 12,500 \ 8,900 Arrual berefit \ 6,800 \ 2,000 Salvage value \ 5,000 \ 8,900 Life in years 2 years 3 Years MARR 10\%

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 0 1-10 11-15 16-25 26-30 Cash Flow -\ 100K 10K 20K -5K 30K

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 ________.

Table 5 Initial cost \ 15,000 \ 9,000 \ 12,000 Annual benefit \ 8,000 \ 2,000 \ 1,800 Salvage value \ 5,000 \ 9,000 0 Life in years 2 years 3 Years Infinity MARR 10\% -The NPW of alt. B is __________________.

Table 5 Initial cost \ 15,000 \ 9,000 \ 12,000 Annual benefit \ 8,000 \ 2,000 \ 1,800 Salvage value \ 5,000 \ 9,000 0 Life in years 2 years 3 Years Infinity MARR 10\% -The NPW of alt. C is __________________.

Solve the following problem using the present worth analysis for an interest rate of 8%. Alt. A Alt. B Alt. C Initial cost \ 1,700 \ 2,100 \ 3,750 Beniefit/year 1,000 1,000 1,000 Life in years 3 6

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.

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.

The NPW of a set of cash flows will decrease as the interest rate is increased.

Table 5 Initial cost \ 15,000 \ 9,000 \ 12,000 Annual benefit \ 8,000 \ 2,000 \ 1,800 Salvage value \ 5,000 \ 9,000 0 Life in years 2 years 3 Years Infinity MARR 10\% -The NPW of alt. A is __________________.

Making Economic Decisions

Estimating Engineering Costs and Benefits

Interest and Equivalence

Equivalence for Repeated Cash Flows

Annual Cash Flow Analysis

Rate of Return Analysis

Choosing the Best Alternative

Other Analysis Techniques

Uncertainty in Future Events

Depreciation

Income Taxes for Corporations

Replacement Analysis

Inflation and Price Change

Selection of a Minimum Attractive Rate of Return

Economic Analysis in the Public Sector

Accounting and Engineering Economy

Filters