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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Students have asked these similar questions

Please answer questions D-H, I have already answered A , B,C but it may help you to still solve them yourself. Thank you!

2. A firm’s production function is given by:Q = 10KLThe unit capital and labour costs are 2 and 1 pounds respectively. The firm is contracted to produce2000 units.(a) Write out the optimisation problem of the firm. (b) Express this problem using a Lagrangian function. (c) Find values of K and L which fulfil the contract with minimal cost to the firm. (d) Calculate the total cost to the firm.

3. Consider the following estimated regression equation, estimated using a sample of firms, where RDis total firm spending on research and development in USD ($), Revenue is total firm revenuein USD ($), and W ages is the firms’ total spending on wages in USD ($) (standard errors inparentheses):RDd = 1000(600)+ 0.5(0.1)Revenue + 1.5(0.5)W ages,(a) Interpret the coefficients on each of the explanatory variables. (b) Which of the three coefficients are statistically significant at the 5% level of significance? Howdo you know? A researcher runs a two-sided statistical test of the null hypothesis that both the coefficients onthe explanatory variables above are jointly equal to 0.25 (mathematically, that β1 = β2 = 0.25),and reports a p-value of 0.045.(c) What does this p-value mean for the outcome of the test? (d) What would an appropriate two-sided alternative hypothesis look like?

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.