Compare the projects based on the present worth.

Question

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Answer 1

Question

Chapter 5, Problem 19P

(a)

To determine

Compare the projects based on the present worth.

(a)

Expert Solution

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	1	2
First cost (C)	400,000	600,000
Maintenance cost (MO) per year	140,000	100,000
Salvage value (SV)	10% to C	10% to C
Time period (n)	3	6

MARR (i) is 15%.

The time period for project 1 should equate with project time period two. Thus, all the cash flows are repeated for other three years. The time period (n1) is six years and time period 2 (n2) is three years.

Present worth of the project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n1−1i(1+i)n1)+−C+(C×0.1)(1+i)n2+(C×0.1)(1+i)n1=(−400,000−140,000((1+0.15)6−10.15(1+0.15)6)+−400,000+(400,000×0.1)(1+0.15)3+(400,000×0.1)(1+0.15)6)=(−400,000−140,000(2.313061−10.15(2.313061))+−400,000+(400,000×0.1)1.520875+(400,000×0.1)2.313061)=(−400,000−140,000(1.3130610.346959)+−360,0001.520875+40,0002.313061)=(−400,000−140,000(3.784485)−236,705.84+17,293.1)=(−400,000−529,827.9−236,705.84+17,293.1)=−1,149,240.64

The present worth of project 1 is -$1,149,240.64.

The present worth of the project 2 (PW2) can be calculated as follows:

PW2=C+MO((1+i)n−1i(1+i)n)+C×0.1(1+i)n=−600,000−100,000((1+0.15)6−10.15(1+0.15)6)+600,000×0.1(1+0.15)6=−600,000−100,000(2.313061−10.15(2.313061))+60,0002.313061=−600,000−100,000(1.3130610.346959)+25,939.65=−600,000−100,000(3.784485)+25,939.65=−600,000−378,448.5+13,660.27=−952,508.85

The present worth of project 2 is -$952,508.85. Since the present worth of the project 2 is greater than project 1, select project 2.

(b)

To determine

Compare the projects based on the present worth.

(b)

Expert Solution

Explanation of Solution

Present worth of project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n−1i(1+i)n)+(C×0.1)(1+i)n=(−400,000−140,000((1+0.15)3−10.15(1+0.15)3)+(400,000×0.1)(1+0.15)3)=(−400,000−140,000(1.520875−10.15(1.520875))+(400,000×0.1)1.520875)=(−400,000−140,000(0.52087510.2281313)+40,0001.520875)=(−400,000−140,000(2.283225)+26,300.65)=(−400,000−319,651.5+26,300.65)=−693,350.85

The present worth of project 1 with the wrong method is -$693,350.85. Since the present worth of project 1 is greater than project 2, select project 1. This decision is economically incorrect due to the wrong life time for project 1. When comparing the project, the life time of both projects should be equal.

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Answer 2

Question

Chapter 5, Problem 19P

(a)

To determine

Compare the projects based on the present worth.

(a)

Expert Solution

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	1	2
First cost (C)	400,000	600,000
Maintenance cost (MO) per year	140,000	100,000
Salvage value (SV)	10% to C	10% to C
Time period (n)	3	6

MARR (i) is 15%.

The time period for project 1 should equate with project time period two. Thus, all the cash flows are repeated for other three years. The time period (n1) is six years and time period 2 (n2) is three years.

Present worth of the project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n1−1i(1+i)n1)+−C+(C×0.1)(1+i)n2+(C×0.1)(1+i)n1=(−400,000−140,000((1+0.15)6−10.15(1+0.15)6)+−400,000+(400,000×0.1)(1+0.15)3+(400,000×0.1)(1+0.15)6)=(−400,000−140,000(2.313061−10.15(2.313061))+−400,000+(400,000×0.1)1.520875+(400,000×0.1)2.313061)=(−400,000−140,000(1.3130610.346959)+−360,0001.520875+40,0002.313061)=(−400,000−140,000(3.784485)−236,705.84+17,293.1)=(−400,000−529,827.9−236,705.84+17,293.1)=−1,149,240.64

The present worth of project 1 is -$1,149,240.64.

The present worth of the project 2 (PW2) can be calculated as follows:

PW2=C+MO((1+i)n−1i(1+i)n)+C×0.1(1+i)n=−600,000−100,000((1+0.15)6−10.15(1+0.15)6)+600,000×0.1(1+0.15)6=−600,000−100,000(2.313061−10.15(2.313061))+60,0002.313061=−600,000−100,000(1.3130610.346959)+25,939.65=−600,000−100,000(3.784485)+25,939.65=−600,000−378,448.5+13,660.27=−952,508.85

The present worth of project 2 is -$952,508.85. Since the present worth of the project 2 is greater than project 1, select project 2.

(b)

To determine

Compare the projects based on the present worth.

(b)

Expert Solution

Explanation of Solution

Present worth of project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n−1i(1+i)n)+(C×0.1)(1+i)n=(−400,000−140,000((1+0.15)3−10.15(1+0.15)3)+(400,000×0.1)(1+0.15)3)=(−400,000−140,000(1.520875−10.15(1.520875))+(400,000×0.1)1.520875)=(−400,000−140,000(0.52087510.2281313)+40,0001.520875)=(−400,000−140,000(2.283225)+26,300.65)=(−400,000−319,651.5+26,300.65)=−693,350.85

The present worth of project 1 with the wrong method is -$693,350.85. Since the present worth of project 1 is greater than project 2, select project 1. This decision is economically incorrect due to the wrong life time for project 1. When comparing the project, the life time of both projects should be equal.

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Answer 3

Question

Chapter 5, Problem 19P

(a)

To determine

Compare the projects based on the present worth.

(a)

Expert Solution

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	1	2
First cost (C)	400,000	600,000
Maintenance cost (MO) per year	140,000	100,000
Salvage value (SV)	10% to C	10% to C
Time period (n)	3	6

MARR (i) is 15%.

The time period for project 1 should equate with project time period two. Thus, all the cash flows are repeated for other three years. The time period (n1) is six years and time period 2 (n2) is three years.

Present worth of the project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n1−1i(1+i)n1)+−C+(C×0.1)(1+i)n2+(C×0.1)(1+i)n1=(−400,000−140,000((1+0.15)6−10.15(1+0.15)6)+−400,000+(400,000×0.1)(1+0.15)3+(400,000×0.1)(1+0.15)6)=(−400,000−140,000(2.313061−10.15(2.313061))+−400,000+(400,000×0.1)1.520875+(400,000×0.1)2.313061)=(−400,000−140,000(1.3130610.346959)+−360,0001.520875+40,0002.313061)=(−400,000−140,000(3.784485)−236,705.84+17,293.1)=(−400,000−529,827.9−236,705.84+17,293.1)=−1,149,240.64

The present worth of project 1 is -$1,149,240.64.

The present worth of the project 2 (PW2) can be calculated as follows:

PW2=C+MO((1+i)n−1i(1+i)n)+C×0.1(1+i)n=−600,000−100,000((1+0.15)6−10.15(1+0.15)6)+600,000×0.1(1+0.15)6=−600,000−100,000(2.313061−10.15(2.313061))+60,0002.313061=−600,000−100,000(1.3130610.346959)+25,939.65=−600,000−100,000(3.784485)+25,939.65=−600,000−378,448.5+13,660.27=−952,508.85

The present worth of project 2 is -$952,508.85. Since the present worth of the project 2 is greater than project 1, select project 2.

(b)

To determine

Compare the projects based on the present worth.

(b)

Expert Solution

Explanation of Solution

Present worth of project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n−1i(1+i)n)+(C×0.1)(1+i)n=(−400,000−140,000((1+0.15)3−10.15(1+0.15)3)+(400,000×0.1)(1+0.15)3)=(−400,000−140,000(1.520875−10.15(1.520875))+(400,000×0.1)1.520875)=(−400,000−140,000(0.52087510.2281313)+40,0001.520875)=(−400,000−140,000(2.283225)+26,300.65)=(−400,000−319,651.5+26,300.65)=−693,350.85

The present worth of project 1 with the wrong method is -$693,350.85. Since the present worth of project 1 is greater than project 2, select project 1. This decision is economically incorrect due to the wrong life time for project 1. When comparing the project, the life time of both projects should be equal.

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Answer 4

Question

Chapter 5, Problem 19P

(a)

To determine

Compare the projects based on the present worth.

(a)

Expert Solution

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	1	2
First cost (C)	400,000	600,000
Maintenance cost (MO) per year	140,000	100,000
Salvage value (SV)	10% to C	10% to C
Time period (n)	3	6

MARR (i) is 15%.

The time period for project 1 should equate with project time period two. Thus, all the cash flows are repeated for other three years. The time period (n1) is six years and time period 2 (n2) is three years.

Present worth of the project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n1−1i(1+i)n1)+−C+(C×0.1)(1+i)n2+(C×0.1)(1+i)n1=(−400,000−140,000((1+0.15)6−10.15(1+0.15)6)+−400,000+(400,000×0.1)(1+0.15)3+(400,000×0.1)(1+0.15)6)=(−400,000−140,000(2.313061−10.15(2.313061))+−400,000+(400,000×0.1)1.520875+(400,000×0.1)2.313061)=(−400,000−140,000(1.3130610.346959)+−360,0001.520875+40,0002.313061)=(−400,000−140,000(3.784485)−236,705.84+17,293.1)=(−400,000−529,827.9−236,705.84+17,293.1)=−1,149,240.64

The present worth of project 1 is -$1,149,240.64.

The present worth of the project 2 (PW2) can be calculated as follows:

PW2=C+MO((1+i)n−1i(1+i)n)+C×0.1(1+i)n=−600,000−100,000((1+0.15)6−10.15(1+0.15)6)+600,000×0.1(1+0.15)6=−600,000−100,000(2.313061−10.15(2.313061))+60,0002.313061=−600,000−100,000(1.3130610.346959)+25,939.65=−600,000−100,000(3.784485)+25,939.65=−600,000−378,448.5+13,660.27=−952,508.85

The present worth of project 2 is -$952,508.85. Since the present worth of the project 2 is greater than project 1, select project 2.

(b)

To determine

Compare the projects based on the present worth.

(b)

Expert Solution

Explanation of Solution

Present worth of project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n−1i(1+i)n)+(C×0.1)(1+i)n=(−400,000−140,000((1+0.15)3−10.15(1+0.15)3)+(400,000×0.1)(1+0.15)3)=(−400,000−140,000(1.520875−10.15(1.520875))+(400,000×0.1)1.520875)=(−400,000−140,000(0.52087510.2281313)+40,0001.520875)=(−400,000−140,000(2.283225)+26,300.65)=(−400,000−319,651.5+26,300.65)=−693,350.85

The present worth of project 1 with the wrong method is -$693,350.85. Since the present worth of project 1 is greater than project 2, select project 1. This decision is economically incorrect due to the wrong life time for project 1. When comparing the project, the life time of both projects should be equal.

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Answer 5

Chapter 5, Problem 19P

(a)

To determine

Compare the projects based on the present worth.

(a)

Expert Solution

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	1	2
First cost (C)	400,000	600,000
Maintenance cost (MO) per year	140,000	100,000
Salvage value (SV)	10% to C	10% to C
Time period (n)	3	6

MARR (i) is 15%.

The time period for project 1 should equate with project time period two. Thus, all the cash flows are repeated for other three years. The time period (n1) is six years and time period 2 (n2) is three years.

Present worth of the project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n1−1i(1+i)n1)+−C+(C×0.1)(1+i)n2+(C×0.1)(1+i)n1=(−400,000−140,000((1+0.15)6−10.15(1+0.15)6)+−400,000+(400,000×0.1)(1+0.15)3+(400,000×0.1)(1+0.15)6)=(−400,000−140,000(2.313061−10.15(2.313061))+−400,000+(400,000×0.1)1.520875+(400,000×0.1)2.313061)=(−400,000−140,000(1.3130610.346959)+−360,0001.520875+40,0002.313061)=(−400,000−140,000(3.784485)−236,705.84+17,293.1)=(−400,000−529,827.9−236,705.84+17,293.1)=−1,149,240.64

The present worth of project 1 is -$1,149,240.64.

The present worth of the project 2 (PW2) can be calculated as follows:

PW2=C+MO((1+i)n−1i(1+i)n)+C×0.1(1+i)n=−600,000−100,000((1+0.15)6−10.15(1+0.15)6)+600,000×0.1(1+0.15)6=−600,000−100,000(2.313061−10.15(2.313061))+60,0002.313061=−600,000−100,000(1.3130610.346959)+25,939.65=−600,000−100,000(3.784485)+25,939.65=−600,000−378,448.5+13,660.27=−952,508.85

The present worth of project 2 is -$952,508.85. Since the present worth of the project 2 is greater than project 1, select project 2.

(b)

To determine

Compare the projects based on the present worth.

(b)

Expert Solution

Explanation of Solution

Present worth of project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n−1i(1+i)n)+(C×0.1)(1+i)n=(−400,000−140,000((1+0.15)3−10.15(1+0.15)3)+(400,000×0.1)(1+0.15)3)=(−400,000−140,000(1.520875−10.15(1.520875))+(400,000×0.1)1.520875)=(−400,000−140,000(0.52087510.2281313)+40,0001.520875)=(−400,000−140,000(2.283225)+26,300.65)=(−400,000−319,651.5+26,300.65)=−693,350.85

The present worth of project 1 with the wrong method is -$693,350.85. Since the present worth of project 1 is greater than project 2, select project 1. This decision is economically incorrect due to the wrong life time for project 1. When comparing the project, the life time of both projects should be equal.

Answer 6

Chapter 5, Problem 19P

(a)

To determine

Compare the projects based on the present worth.

(a)

Expert Solution

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	1	2
First cost (C)	400,000	600,000
Maintenance cost (MO) per year	140,000	100,000
Salvage value (SV)	10% to C	10% to C
Time period (n)	3	6

MARR (i) is 15%.

The time period for project 1 should equate with project time period two. Thus, all the cash flows are repeated for other three years. The time period (n1) is six years and time period 2 (n2) is three years.

Present worth of the project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n1−1i(1+i)n1)+−C+(C×0.1)(1+i)n2+(C×0.1)(1+i)n1=(−400,000−140,000((1+0.15)6−10.15(1+0.15)6)+−400,000+(400,000×0.1)(1+0.15)3+(400,000×0.1)(1+0.15)6)=(−400,000−140,000(2.313061−10.15(2.313061))+−400,000+(400,000×0.1)1.520875+(400,000×0.1)2.313061)=(−400,000−140,000(1.3130610.346959)+−360,0001.520875+40,0002.313061)=(−400,000−140,000(3.784485)−236,705.84+17,293.1)=(−400,000−529,827.9−236,705.84+17,293.1)=−1,149,240.64

The present worth of project 1 is -$1,149,240.64.

The present worth of the project 2 (PW2) can be calculated as follows:

PW2=C+MO((1+i)n−1i(1+i)n)+C×0.1(1+i)n=−600,000−100,000((1+0.15)6−10.15(1+0.15)6)+600,000×0.1(1+0.15)6=−600,000−100,000(2.313061−10.15(2.313061))+60,0002.313061=−600,000−100,000(1.3130610.346959)+25,939.65=−600,000−100,000(3.784485)+25,939.65=−600,000−378,448.5+13,660.27=−952,508.85

The present worth of project 2 is -$952,508.85. Since the present worth of the project 2 is greater than project 1, select project 2.

Answer 7

Chapter 5, Problem 19P

(a)

To determine

Compare the projects based on the present worth.

Answer 8

(a)

Expert Solution

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	1	2
First cost (C)	400,000	600,000
Maintenance cost (MO) per year	140,000	100,000
Salvage value (SV)	10% to C	10% to C
Time period (n)	3	6

MARR (i) is 15%.

The time period for project 1 should equate with project time period two. Thus, all the cash flows are repeated for other three years. The time period (n1) is six years and time period 2 (n2) is three years.

Present worth of the project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n1−1i(1+i)n1)+−C+(C×0.1)(1+i)n2+(C×0.1)(1+i)n1=(−400,000−140,000((1+0.15)6−10.15(1+0.15)6)+−400,000+(400,000×0.1)(1+0.15)3+(400,000×0.1)(1+0.15)6)=(−400,000−140,000(2.313061−10.15(2.313061))+−400,000+(400,000×0.1)1.520875+(400,000×0.1)2.313061)=(−400,000−140,000(1.3130610.346959)+−360,0001.520875+40,0002.313061)=(−400,000−140,000(3.784485)−236,705.84+17,293.1)=(−400,000−529,827.9−236,705.84+17,293.1)=−1,149,240.64

The present worth of project 1 is -$1,149,240.64.

The present worth of the project 2 (PW2) can be calculated as follows:

PW2=C+MO((1+i)n−1i(1+i)n)+C×0.1(1+i)n=−600,000−100,000((1+0.15)6−10.15(1+0.15)6)+600,000×0.1(1+0.15)6=−600,000−100,000(2.313061−10.15(2.313061))+60,0002.313061=−600,000−100,000(1.3130610.346959)+25,939.65=−600,000−100,000(3.784485)+25,939.65=−600,000−378,448.5+13,660.27=−952,508.85

The present worth of project 2 is -$952,508.85. Since the present worth of the project 2 is greater than project 1, select project 2.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

Compare the projects based on the present worth.

(b)

Expert Solution

Explanation of Solution

Present worth of project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n−1i(1+i)n)+(C×0.1)(1+i)n=(−400,000−140,000((1+0.15)3−10.15(1+0.15)3)+(400,000×0.1)(1+0.15)3)=(−400,000−140,000(1.520875−10.15(1.520875))+(400,000×0.1)1.520875)=(−400,000−140,000(0.52087510.2281313)+40,0001.520875)=(−400,000−140,000(2.283225)+26,300.65)=(−400,000−319,651.5+26,300.65)=−693,350.85

The present worth of project 1 with the wrong method is -$693,350.85. Since the present worth of project 1 is greater than project 2, select project 1. This decision is economically incorrect due to the wrong life time for project 1. When comparing the project, the life time of both projects should be equal.

Answer 13

(b)

To determine

Compare the projects based on the present worth.

Answer 14

(b)

Expert Solution

Explanation of Solution

Present worth of project 1 (PW1) can be calculated as follows:

PW1=−C−MO((1+i)n−1i(1+i)n)+(C×0.1)(1+i)n=(−400,000−140,000((1+0.15)3−10.15(1+0.15)3)+(400,000×0.1)(1+0.15)3)=(−400,000−140,000(1.520875−10.15(1.520875))+(400,000×0.1)1.520875)=(−400,000−140,000(0.52087510.2281313)+40,0001.520875)=(−400,000−140,000(2.283225)+26,300.65)=(−400,000−319,651.5+26,300.65)=−693,350.85

The present worth of project 1 with the wrong method is -$693,350.85. Since the present worth of project 1 is greater than project 2, select project 1. This decision is economically incorrect due to the wrong life time for project 1. When comparing the project, the life time of both projects should be equal.