Compare the projects based on the present worth.

Question

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Answer 1

Question

Chapter 5, Problem 23P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	C	D
First cost (C)	-40,000	-32,000
Operating cost (MO) per year	-7,000	-3,000
Increasing operating cost (IO) per year	-1,000	-
Salvage value (SV)	9,000	500
Time period (n)	10	5

MARR (i) is 10%.

The present worth of project C (PWC) can be calculated as follows:

PW=C+(MO((1+i)n−1i(1+i)n)+IO1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−40,000−(7,000((1+0.1)10−10.1(1+0.1)10)+(−1,000)10.1((1+0.1)10−10.1(1+0.1)10−10(1+0.1)10))+9,000(1+0.1)10=−40,000−(7,000(2.593742−10.1(2.593742))+(1,000)10(2.593742−10.1(2.593742)−102.593742))+9,0002.593742=−40,000−(7,000(1.5937420.259374)+10,000(1.5937420.259374−3.855434))+3,469.8902=−36,530.1098−(7,000(6.144571)+10,000(6.144571−3.855434))=−36,530.1098−(43,011.997+10,000(2.289137))=−36,530.1098−(43,011.997+22,891.37)=−102,433.48

The present worth of project C is -$102,433.48.

The time period for project C should equate with project time period B. Thus, all the cash flows are repeated for other five years. The time period (n1) is 10 years and time period 2 (n2) is five years.

Present worth of project D (PWD) can be calculated as follows:

PWD=C+MO((1+i)n1−1i(1+i)n1)+C+SV(1+i)n2+SV(1+i)n1=−32,000−3,000((1+0.1)10−10.1(1+0.1)10)+−32,000+500(1+0.1)5+500(1+0.1)10=−32,000−3,000(2.593742−10.1(2.593742))−31,5001.61051+5002.593742=−32,000−3,000(1.5937420.259374)−19,559.2017+192.7717=−51,366.43−3,000(6.144571)=−51,366.43−18,433.71=−69,800.14

Present worth of project D is -69,800.14. Since the present worth of project D is greater than project C, select project D.

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Answer 2

Question

Chapter 5, Problem 23P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	C	D
First cost (C)	-40,000	-32,000
Operating cost (MO) per year	-7,000	-3,000
Increasing operating cost (IO) per year	-1,000	-
Salvage value (SV)	9,000	500
Time period (n)	10	5

MARR (i) is 10%.

The present worth of project C (PWC) can be calculated as follows:

PW=C+(MO((1+i)n−1i(1+i)n)+IO1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−40,000−(7,000((1+0.1)10−10.1(1+0.1)10)+(−1,000)10.1((1+0.1)10−10.1(1+0.1)10−10(1+0.1)10))+9,000(1+0.1)10=−40,000−(7,000(2.593742−10.1(2.593742))+(1,000)10(2.593742−10.1(2.593742)−102.593742))+9,0002.593742=−40,000−(7,000(1.5937420.259374)+10,000(1.5937420.259374−3.855434))+3,469.8902=−36,530.1098−(7,000(6.144571)+10,000(6.144571−3.855434))=−36,530.1098−(43,011.997+10,000(2.289137))=−36,530.1098−(43,011.997+22,891.37)=−102,433.48

The present worth of project C is -$102,433.48.

The time period for project C should equate with project time period B. Thus, all the cash flows are repeated for other five years. The time period (n1) is 10 years and time period 2 (n2) is five years.

Present worth of project D (PWD) can be calculated as follows:

PWD=C+MO((1+i)n1−1i(1+i)n1)+C+SV(1+i)n2+SV(1+i)n1=−32,000−3,000((1+0.1)10−10.1(1+0.1)10)+−32,000+500(1+0.1)5+500(1+0.1)10=−32,000−3,000(2.593742−10.1(2.593742))−31,5001.61051+5002.593742=−32,000−3,000(1.5937420.259374)−19,559.2017+192.7717=−51,366.43−3,000(6.144571)=−51,366.43−18,433.71=−69,800.14

Present worth of project D is -69,800.14. Since the present worth of project D is greater than project C, select project D.

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Answer 3

Question

Chapter 5, Problem 23P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	C	D
First cost (C)	-40,000	-32,000
Operating cost (MO) per year	-7,000	-3,000
Increasing operating cost (IO) per year	-1,000	-
Salvage value (SV)	9,000	500
Time period (n)	10	5

MARR (i) is 10%.

The present worth of project C (PWC) can be calculated as follows:

PW=C+(MO((1+i)n−1i(1+i)n)+IO1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−40,000−(7,000((1+0.1)10−10.1(1+0.1)10)+(−1,000)10.1((1+0.1)10−10.1(1+0.1)10−10(1+0.1)10))+9,000(1+0.1)10=−40,000−(7,000(2.593742−10.1(2.593742))+(1,000)10(2.593742−10.1(2.593742)−102.593742))+9,0002.593742=−40,000−(7,000(1.5937420.259374)+10,000(1.5937420.259374−3.855434))+3,469.8902=−36,530.1098−(7,000(6.144571)+10,000(6.144571−3.855434))=−36,530.1098−(43,011.997+10,000(2.289137))=−36,530.1098−(43,011.997+22,891.37)=−102,433.48

The present worth of project C is -$102,433.48.

The time period for project C should equate with project time period B. Thus, all the cash flows are repeated for other five years. The time period (n1) is 10 years and time period 2 (n2) is five years.

Present worth of project D (PWD) can be calculated as follows:

PWD=C+MO((1+i)n1−1i(1+i)n1)+C+SV(1+i)n2+SV(1+i)n1=−32,000−3,000((1+0.1)10−10.1(1+0.1)10)+−32,000+500(1+0.1)5+500(1+0.1)10=−32,000−3,000(2.593742−10.1(2.593742))−31,5001.61051+5002.593742=−32,000−3,000(1.5937420.259374)−19,559.2017+192.7717=−51,366.43−3,000(6.144571)=−51,366.43−18,433.71=−69,800.14

Present worth of project D is -69,800.14. Since the present worth of project D is greater than project C, select project D.

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Answer 4

Question

Chapter 5, Problem 23P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	C	D
First cost (C)	-40,000	-32,000
Operating cost (MO) per year	-7,000	-3,000
Increasing operating cost (IO) per year	-1,000	-
Salvage value (SV)	9,000	500
Time period (n)	10	5

MARR (i) is 10%.

The present worth of project C (PWC) can be calculated as follows:

PW=C+(MO((1+i)n−1i(1+i)n)+IO1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−40,000−(7,000((1+0.1)10−10.1(1+0.1)10)+(−1,000)10.1((1+0.1)10−10.1(1+0.1)10−10(1+0.1)10))+9,000(1+0.1)10=−40,000−(7,000(2.593742−10.1(2.593742))+(1,000)10(2.593742−10.1(2.593742)−102.593742))+9,0002.593742=−40,000−(7,000(1.5937420.259374)+10,000(1.5937420.259374−3.855434))+3,469.8902=−36,530.1098−(7,000(6.144571)+10,000(6.144571−3.855434))=−36,530.1098−(43,011.997+10,000(2.289137))=−36,530.1098−(43,011.997+22,891.37)=−102,433.48

The present worth of project C is -$102,433.48.

The time period for project C should equate with project time period B. Thus, all the cash flows are repeated for other five years. The time period (n1) is 10 years and time period 2 (n2) is five years.

Present worth of project D (PWD) can be calculated as follows:

PWD=C+MO((1+i)n1−1i(1+i)n1)+C+SV(1+i)n2+SV(1+i)n1=−32,000−3,000((1+0.1)10−10.1(1+0.1)10)+−32,000+500(1+0.1)5+500(1+0.1)10=−32,000−3,000(2.593742−10.1(2.593742))−31,5001.61051+5002.593742=−32,000−3,000(1.5937420.259374)−19,559.2017+192.7717=−51,366.43−3,000(6.144571)=−51,366.43−18,433.71=−69,800.14

Present worth of project D is -69,800.14. Since the present worth of project D is greater than project C, select project D.

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Answer 5

Chapter 5, Problem 23P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	C	D
First cost (C)	-40,000	-32,000
Operating cost (MO) per year	-7,000	-3,000
Increasing operating cost (IO) per year	-1,000	-
Salvage value (SV)	9,000	500
Time period (n)	10	5

MARR (i) is 10%.

The present worth of project C (PWC) can be calculated as follows:

PW=C+(MO((1+i)n−1i(1+i)n)+IO1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−40,000−(7,000((1+0.1)10−10.1(1+0.1)10)+(−1,000)10.1((1+0.1)10−10.1(1+0.1)10−10(1+0.1)10))+9,000(1+0.1)10=−40,000−(7,000(2.593742−10.1(2.593742))+(1,000)10(2.593742−10.1(2.593742)−102.593742))+9,0002.593742=−40,000−(7,000(1.5937420.259374)+10,000(1.5937420.259374−3.855434))+3,469.8902=−36,530.1098−(7,000(6.144571)+10,000(6.144571−3.855434))=−36,530.1098−(43,011.997+10,000(2.289137))=−36,530.1098−(43,011.997+22,891.37)=−102,433.48

The present worth of project C is -$102,433.48.

The time period for project C should equate with project time period B. Thus, all the cash flows are repeated for other five years. The time period (n1) is 10 years and time period 2 (n2) is five years.

Present worth of project D (PWD) can be calculated as follows:

PWD=C+MO((1+i)n1−1i(1+i)n1)+C+SV(1+i)n2+SV(1+i)n1=−32,000−3,000((1+0.1)10−10.1(1+0.1)10)+−32,000+500(1+0.1)5+500(1+0.1)10=−32,000−3,000(2.593742−10.1(2.593742))−31,5001.61051+5002.593742=−32,000−3,000(1.5937420.259374)−19,559.2017+192.7717=−51,366.43−3,000(6.144571)=−51,366.43−18,433.71=−69,800.14

Present worth of project D is -69,800.14. Since the present worth of project D is greater than project C, select project D.

Answer 6

Chapter 5, Problem 23P

To determine

Compare the projects based on the present worth.

Expert Solution & Answer

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	C	D
First cost (C)	-40,000	-32,000
Operating cost (MO) per year	-7,000	-3,000
Increasing operating cost (IO) per year	-1,000	-
Salvage value (SV)	9,000	500
Time period (n)	10	5

MARR (i) is 10%.

The present worth of project C (PWC) can be calculated as follows:

PW=C+(MO((1+i)n−1i(1+i)n)+IO1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−40,000−(7,000((1+0.1)10−10.1(1+0.1)10)+(−1,000)10.1((1+0.1)10−10.1(1+0.1)10−10(1+0.1)10))+9,000(1+0.1)10=−40,000−(7,000(2.593742−10.1(2.593742))+(1,000)10(2.593742−10.1(2.593742)−102.593742))+9,0002.593742=−40,000−(7,000(1.5937420.259374)+10,000(1.5937420.259374−3.855434))+3,469.8902=−36,530.1098−(7,000(6.144571)+10,000(6.144571−3.855434))=−36,530.1098−(43,011.997+10,000(2.289137))=−36,530.1098−(43,011.997+22,891.37)=−102,433.48

The present worth of project C is -$102,433.48.

The time period for project C should equate with project time period B. Thus, all the cash flows are repeated for other five years. The time period (n1) is 10 years and time period 2 (n2) is five years.

Present worth of project D (PWD) can be calculated as follows:

PWD=C+MO((1+i)n1−1i(1+i)n1)+C+SV(1+i)n2+SV(1+i)n1=−32,000−3,000((1+0.1)10−10.1(1+0.1)10)+−32,000+500(1+0.1)5+500(1+0.1)10=−32,000−3,000(2.593742−10.1(2.593742))−31,5001.61051+5002.593742=−32,000−3,000(1.5937420.259374)−19,559.2017+192.7717=−51,366.43−3,000(6.144571)=−51,366.43−18,433.71=−69,800.14

Present worth of project D is -69,800.14. Since the present worth of project D is greater than project C, select project D.

Answer 7

Chapter 5, Problem 23P

To determine

Compare the projects based on the present worth.

Answer 8

Expert Solution & Answer

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	C	D
First cost (C)	-40,000	-32,000
Operating cost (MO) per year	-7,000	-3,000
Increasing operating cost (IO) per year	-1,000	-
Salvage value (SV)	9,000	500
Time period (n)	10	5

MARR (i) is 10%.

The present worth of project C (PWC) can be calculated as follows:

PW=C+(MO((1+i)n−1i(1+i)n)+IO1i((1+i)n−1i(1+i)n−n(1+i)n))+SV(1+i)n=−40,000−(7,000((1+0.1)10−10.1(1+0.1)10)+(−1,000)10.1((1+0.1)10−10.1(1+0.1)10−10(1+0.1)10))+9,000(1+0.1)10=−40,000−(7,000(2.593742−10.1(2.593742))+(1,000)10(2.593742−10.1(2.593742)−102.593742))+9,0002.593742=−40,000−(7,000(1.5937420.259374)+10,000(1.5937420.259374−3.855434))+3,469.8902=−36,530.1098−(7,000(6.144571)+10,000(6.144571−3.855434))=−36,530.1098−(43,011.997+10,000(2.289137))=−36,530.1098−(43,011.997+22,891.37)=−102,433.48

The present worth of project C is -$102,433.48.

The time period for project C should equate with project time period B. Thus, all the cash flows are repeated for other five years. The time period (n1) is 10 years and time period 2 (n2) is five years.

Present worth of project D (PWD) can be calculated as follows:

PWD=C+MO((1+i)n1−1i(1+i)n1)+C+SV(1+i)n2+SV(1+i)n1=−32,000−3,000((1+0.1)10−10.1(1+0.1)10)+−32,000+500(1+0.1)5+500(1+0.1)10=−32,000−3,000(2.593742−10.1(2.593742))−31,5001.61051+5002.593742=−32,000−3,000(1.5937420.259374)−19,559.2017+192.7717=−51,366.43−3,000(6.144571)=−51,366.43−18,433.71=−69,800.14