The alternative to be selected.

Question

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Answer 1

Question

Chapter 9, Problem 31P

To determine

The alternative to be selected.

Expert Solution & Answer

Answer to Problem 31P

The alternative to be selected is Alternative $B$ .

Explanation of Solution

Given:

MARR is $12%$ .

Useful life is $15$ years.

Calculation:

Write the expression to calculate the present worth of the benefit.

$P=A((1+i)n−1i(1+i)n)$ ....... (I)

Here, the annual benefit is A, the present worth is $P$ , the rate of interest is $i$ and the number of period is $n$ .

Calculate the present worth benefit of the alternative $A$ .

Substitute $$130$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$130((1+0.12)15−10.12(1+0.12)15)=$130(6.81)=$885.3$

Calculate the present worth benefit of the alternative $B$ .

Substitute $$60$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$60((1+0.12)15−10.12(1+0.12)15)=$60(6.81)=$408.6$

Calculate the present worth benefit of the alternative $C$ .

Substitute $$35$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$35((1+0.12)15−10.12(1+0.12)15)=$35(6.81)=$238.35$

Write the expression to calculate the benefit-cost ratio.

$B/C=BC$ ....... (II)

Here, the present worth benefit is $B$ , the present worth cost is $C$ .

For Alternative $A$ .

Substitute $$885.3$ for $B$ , $$800$ for $C$ , Equation (I).

$B/C=$885.3$800=1.11$

For Alternative $B$ .

Substitute $$408.6$ for $B$ , $$300$ for $C$ , Equation (I).

$B/C=$408.6$300=1.362$

For Alternative $C$ .

Substitute $$238.35$ for $B$ , $$150$ for $C$ , Equation (I).

$B/C=$238.35$150=1.589$

Since all the ratios are positive hence we need to make comparison be increment method.

Calculate the $ΔCost$ for alternative $B−C$ .

$ΔCost=(PWC−PWB)$ ...... (III)

Here the incremental cost between the alternative $B$ and $C$ is $ΔCost$ , the present worth of cost for alternative $C$ is $PWC$ and the present worth of cost for alternative $B$ is $PWB$ .

Substitute $$150$ for $PWC$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($300−$150)=$150$

Calculate the $ΔBenefit$ for alternative $B−C$ .

$ΔBenefit=(PC−PB)$ ...... (IV)

Here the incremental benefit between the alternative $B$ and $C$ is $ΔBenefit$ , the present worth of benefit for alternative $C$ is $PC$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$238.35$ for $PC$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$408.6−$238.35=$170.25$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$170.25$ for $ΔBenefit$ and $$150$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$170.25$150=1.135$

Since, $Δ(B/C)$ ratio greater than $1$ therefore alternative $B$ is desirable.

Calculate the $ΔCost$ for alternative $A−B$ .

$ΔCost=(PWA−PWB)$ ...... (VI)

Here the incremental cost between the alternative $B$ and $A$ is $ΔCost$ , the present worth of cost for alternative $B$ is $PWB$ and the present worth of cost for alternative $A$ is $PWA$ .

Substitute $$800$ for $PWA$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($800−$300)=$500$

Calculate the $ΔBenefit$ for alternative $A−B$ .

$ΔBenefit=(PA−PB)$ ...... (VII)

Here the incremental benefit between the alternative $B$ and $A$ is $ΔBenefit$ , the present worth of benefit for alternative $A$ is $PA$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$885.3$ for $PA$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$885.3−$408.6=$476.7$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$476.7$ for $ΔBenefit$ and $$500$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$476.7$500=0.9534$

Since, $Δ(B/C)$ ratio less than $1$ therefore alternative $A$ is not desirable.

Hence, alternative $B$ is desirable.

In both the cases alternative $B$ is desirable.

Hence select alternative $B$ .

Conclusion:

The alternative to be selected is Alternative $B$ .

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Answer 2

Question

Chapter 9, Problem 31P

To determine

The alternative to be selected.

Expert Solution & Answer

Answer to Problem 31P

The alternative to be selected is Alternative $B$ .

Explanation of Solution

Given:

MARR is $12%$ .

Useful life is $15$ years.

Calculation:

Write the expression to calculate the present worth of the benefit.

$P=A((1+i)n−1i(1+i)n)$ ....... (I)

Here, the annual benefit is A, the present worth is $P$ , the rate of interest is $i$ and the number of period is $n$ .

Calculate the present worth benefit of the alternative $A$ .

Substitute $$130$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$130((1+0.12)15−10.12(1+0.12)15)=$130(6.81)=$885.3$

Calculate the present worth benefit of the alternative $B$ .

Substitute $$60$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$60((1+0.12)15−10.12(1+0.12)15)=$60(6.81)=$408.6$

Calculate the present worth benefit of the alternative $C$ .

Substitute $$35$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$35((1+0.12)15−10.12(1+0.12)15)=$35(6.81)=$238.35$

Write the expression to calculate the benefit-cost ratio.

$B/C=BC$ ....... (II)

Here, the present worth benefit is $B$ , the present worth cost is $C$ .

For Alternative $A$ .

Substitute $$885.3$ for $B$ , $$800$ for $C$ , Equation (I).

$B/C=$885.3$800=1.11$

For Alternative $B$ .

Substitute $$408.6$ for $B$ , $$300$ for $C$ , Equation (I).

$B/C=$408.6$300=1.362$

For Alternative $C$ .

Substitute $$238.35$ for $B$ , $$150$ for $C$ , Equation (I).

$B/C=$238.35$150=1.589$

Since all the ratios are positive hence we need to make comparison be increment method.

Calculate the $ΔCost$ for alternative $B−C$ .

$ΔCost=(PWC−PWB)$ ...... (III)

Here the incremental cost between the alternative $B$ and $C$ is $ΔCost$ , the present worth of cost for alternative $C$ is $PWC$ and the present worth of cost for alternative $B$ is $PWB$ .

Substitute $$150$ for $PWC$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($300−$150)=$150$

Calculate the $ΔBenefit$ for alternative $B−C$ .

$ΔBenefit=(PC−PB)$ ...... (IV)

Here the incremental benefit between the alternative $B$ and $C$ is $ΔBenefit$ , the present worth of benefit for alternative $C$ is $PC$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$238.35$ for $PC$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$408.6−$238.35=$170.25$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$170.25$ for $ΔBenefit$ and $$150$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$170.25$150=1.135$

Since, $Δ(B/C)$ ratio greater than $1$ therefore alternative $B$ is desirable.

Calculate the $ΔCost$ for alternative $A−B$ .

$ΔCost=(PWA−PWB)$ ...... (VI)

Here the incremental cost between the alternative $B$ and $A$ is $ΔCost$ , the present worth of cost for alternative $B$ is $PWB$ and the present worth of cost for alternative $A$ is $PWA$ .

Substitute $$800$ for $PWA$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($800−$300)=$500$

Calculate the $ΔBenefit$ for alternative $A−B$ .

$ΔBenefit=(PA−PB)$ ...... (VII)

Here the incremental benefit between the alternative $B$ and $A$ is $ΔBenefit$ , the present worth of benefit for alternative $A$ is $PA$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$885.3$ for $PA$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$885.3−$408.6=$476.7$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$476.7$ for $ΔBenefit$ and $$500$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$476.7$500=0.9534$

Since, $Δ(B/C)$ ratio less than $1$ therefore alternative $A$ is not desirable.

Hence, alternative $B$ is desirable.

In both the cases alternative $B$ is desirable.

Hence select alternative $B$ .

Conclusion:

The alternative to be selected is Alternative $B$ .

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Answer 3

Question

Chapter 9, Problem 31P

To determine

The alternative to be selected.

Expert Solution & Answer

Answer to Problem 31P

The alternative to be selected is Alternative $B$ .

Explanation of Solution

Given:

MARR is $12%$ .

Useful life is $15$ years.

Calculation:

Write the expression to calculate the present worth of the benefit.

$P=A((1+i)n−1i(1+i)n)$ ....... (I)

Here, the annual benefit is A, the present worth is $P$ , the rate of interest is $i$ and the number of period is $n$ .

Calculate the present worth benefit of the alternative $A$ .

Substitute $$130$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$130((1+0.12)15−10.12(1+0.12)15)=$130(6.81)=$885.3$

Calculate the present worth benefit of the alternative $B$ .

Substitute $$60$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$60((1+0.12)15−10.12(1+0.12)15)=$60(6.81)=$408.6$

Calculate the present worth benefit of the alternative $C$ .

Substitute $$35$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$35((1+0.12)15−10.12(1+0.12)15)=$35(6.81)=$238.35$

Write the expression to calculate the benefit-cost ratio.

$B/C=BC$ ....... (II)

Here, the present worth benefit is $B$ , the present worth cost is $C$ .

For Alternative $A$ .

Substitute $$885.3$ for $B$ , $$800$ for $C$ , Equation (I).

$B/C=$885.3$800=1.11$

For Alternative $B$ .

Substitute $$408.6$ for $B$ , $$300$ for $C$ , Equation (I).

$B/C=$408.6$300=1.362$

For Alternative $C$ .

Substitute $$238.35$ for $B$ , $$150$ for $C$ , Equation (I).

$B/C=$238.35$150=1.589$

Since all the ratios are positive hence we need to make comparison be increment method.

Calculate the $ΔCost$ for alternative $B−C$ .

$ΔCost=(PWC−PWB)$ ...... (III)

Here the incremental cost between the alternative $B$ and $C$ is $ΔCost$ , the present worth of cost for alternative $C$ is $PWC$ and the present worth of cost for alternative $B$ is $PWB$ .

Substitute $$150$ for $PWC$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($300−$150)=$150$

Calculate the $ΔBenefit$ for alternative $B−C$ .

$ΔBenefit=(PC−PB)$ ...... (IV)

Here the incremental benefit between the alternative $B$ and $C$ is $ΔBenefit$ , the present worth of benefit for alternative $C$ is $PC$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$238.35$ for $PC$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$408.6−$238.35=$170.25$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$170.25$ for $ΔBenefit$ and $$150$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$170.25$150=1.135$

Since, $Δ(B/C)$ ratio greater than $1$ therefore alternative $B$ is desirable.

Calculate the $ΔCost$ for alternative $A−B$ .

$ΔCost=(PWA−PWB)$ ...... (VI)

Here the incremental cost between the alternative $B$ and $A$ is $ΔCost$ , the present worth of cost for alternative $B$ is $PWB$ and the present worth of cost for alternative $A$ is $PWA$ .

Substitute $$800$ for $PWA$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($800−$300)=$500$

Calculate the $ΔBenefit$ for alternative $A−B$ .

$ΔBenefit=(PA−PB)$ ...... (VII)

Here the incremental benefit between the alternative $B$ and $A$ is $ΔBenefit$ , the present worth of benefit for alternative $A$ is $PA$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$885.3$ for $PA$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$885.3−$408.6=$476.7$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$476.7$ for $ΔBenefit$ and $$500$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$476.7$500=0.9534$

Since, $Δ(B/C)$ ratio less than $1$ therefore alternative $A$ is not desirable.

Hence, alternative $B$ is desirable.

In both the cases alternative $B$ is desirable.

Hence select alternative $B$ .

Conclusion:

The alternative to be selected is Alternative $B$ .

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Answer 4

Question

Chapter 9, Problem 31P

To determine

The alternative to be selected.

Expert Solution & Answer

Answer to Problem 31P

The alternative to be selected is Alternative $B$ .

Explanation of Solution

Given:

MARR is $12%$ .

Useful life is $15$ years.

Calculation:

Write the expression to calculate the present worth of the benefit.

$P=A((1+i)n−1i(1+i)n)$ ....... (I)

Here, the annual benefit is A, the present worth is $P$ , the rate of interest is $i$ and the number of period is $n$ .

Calculate the present worth benefit of the alternative $A$ .

Substitute $$130$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$130((1+0.12)15−10.12(1+0.12)15)=$130(6.81)=$885.3$

Calculate the present worth benefit of the alternative $B$ .

Substitute $$60$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$60((1+0.12)15−10.12(1+0.12)15)=$60(6.81)=$408.6$

Calculate the present worth benefit of the alternative $C$ .

Substitute $$35$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$35((1+0.12)15−10.12(1+0.12)15)=$35(6.81)=$238.35$

Write the expression to calculate the benefit-cost ratio.

$B/C=BC$ ....... (II)

Here, the present worth benefit is $B$ , the present worth cost is $C$ .

For Alternative $A$ .

Substitute $$885.3$ for $B$ , $$800$ for $C$ , Equation (I).

$B/C=$885.3$800=1.11$

For Alternative $B$ .

Substitute $$408.6$ for $B$ , $$300$ for $C$ , Equation (I).

$B/C=$408.6$300=1.362$

For Alternative $C$ .

Substitute $$238.35$ for $B$ , $$150$ for $C$ , Equation (I).

$B/C=$238.35$150=1.589$

Since all the ratios are positive hence we need to make comparison be increment method.

Calculate the $ΔCost$ for alternative $B−C$ .

$ΔCost=(PWC−PWB)$ ...... (III)

Here the incremental cost between the alternative $B$ and $C$ is $ΔCost$ , the present worth of cost for alternative $C$ is $PWC$ and the present worth of cost for alternative $B$ is $PWB$ .

Substitute $$150$ for $PWC$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($300−$150)=$150$

Calculate the $ΔBenefit$ for alternative $B−C$ .

$ΔBenefit=(PC−PB)$ ...... (IV)

Here the incremental benefit between the alternative $B$ and $C$ is $ΔBenefit$ , the present worth of benefit for alternative $C$ is $PC$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$238.35$ for $PC$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$408.6−$238.35=$170.25$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$170.25$ for $ΔBenefit$ and $$150$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$170.25$150=1.135$

Since, $Δ(B/C)$ ratio greater than $1$ therefore alternative $B$ is desirable.

Calculate the $ΔCost$ for alternative $A−B$ .

$ΔCost=(PWA−PWB)$ ...... (VI)

Here the incremental cost between the alternative $B$ and $A$ is $ΔCost$ , the present worth of cost for alternative $B$ is $PWB$ and the present worth of cost for alternative $A$ is $PWA$ .

Substitute $$800$ for $PWA$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($800−$300)=$500$

Calculate the $ΔBenefit$ for alternative $A−B$ .

$ΔBenefit=(PA−PB)$ ...... (VII)

Here the incremental benefit between the alternative $B$ and $A$ is $ΔBenefit$ , the present worth of benefit for alternative $A$ is $PA$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$885.3$ for $PA$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$885.3−$408.6=$476.7$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$476.7$ for $ΔBenefit$ and $$500$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$476.7$500=0.9534$

Since, $Δ(B/C)$ ratio less than $1$ therefore alternative $A$ is not desirable.

Hence, alternative $B$ is desirable.

In both the cases alternative $B$ is desirable.

Hence select alternative $B$ .

Conclusion:

The alternative to be selected is Alternative $B$ .

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Answer 5

Chapter 9, Problem 31P

To determine

The alternative to be selected.

Expert Solution & Answer

Answer to Problem 31P

The alternative to be selected is Alternative $B$ .

Explanation of Solution

Given:

MARR is $12%$ .

Useful life is $15$ years.

Calculation:

Write the expression to calculate the present worth of the benefit.

$P=A((1+i)n−1i(1+i)n)$ ....... (I)

Here, the annual benefit is A, the present worth is $P$ , the rate of interest is $i$ and the number of period is $n$ .

Calculate the present worth benefit of the alternative $A$ .

Substitute $$130$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$130((1+0.12)15−10.12(1+0.12)15)=$130(6.81)=$885.3$

Calculate the present worth benefit of the alternative $B$ .

Substitute $$60$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$60((1+0.12)15−10.12(1+0.12)15)=$60(6.81)=$408.6$

Calculate the present worth benefit of the alternative $C$ .

Substitute $$35$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$35((1+0.12)15−10.12(1+0.12)15)=$35(6.81)=$238.35$

Write the expression to calculate the benefit-cost ratio.

$B/C=BC$ ....... (II)

Here, the present worth benefit is $B$ , the present worth cost is $C$ .

For Alternative $A$ .

Substitute $$885.3$ for $B$ , $$800$ for $C$ , Equation (I).

$B/C=$885.3$800=1.11$

For Alternative $B$ .

Substitute $$408.6$ for $B$ , $$300$ for $C$ , Equation (I).

$B/C=$408.6$300=1.362$

For Alternative $C$ .

Substitute $$238.35$ for $B$ , $$150$ for $C$ , Equation (I).

$B/C=$238.35$150=1.589$

Since all the ratios are positive hence we need to make comparison be increment method.

Calculate the $ΔCost$ for alternative $B−C$ .

$ΔCost=(PWC−PWB)$ ...... (III)

Here the incremental cost between the alternative $B$ and $C$ is $ΔCost$ , the present worth of cost for alternative $C$ is $PWC$ and the present worth of cost for alternative $B$ is $PWB$ .

Substitute $$150$ for $PWC$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($300−$150)=$150$

Calculate the $ΔBenefit$ for alternative $B−C$ .

$ΔBenefit=(PC−PB)$ ...... (IV)

Here the incremental benefit between the alternative $B$ and $C$ is $ΔBenefit$ , the present worth of benefit for alternative $C$ is $PC$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$238.35$ for $PC$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$408.6−$238.35=$170.25$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$170.25$ for $ΔBenefit$ and $$150$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$170.25$150=1.135$

Since, $Δ(B/C)$ ratio greater than $1$ therefore alternative $B$ is desirable.

Calculate the $ΔCost$ for alternative $A−B$ .

$ΔCost=(PWA−PWB)$ ...... (VI)

Here the incremental cost between the alternative $B$ and $A$ is $ΔCost$ , the present worth of cost for alternative $B$ is $PWB$ and the present worth of cost for alternative $A$ is $PWA$ .

Substitute $$800$ for $PWA$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($800−$300)=$500$

Calculate the $ΔBenefit$ for alternative $A−B$ .

$ΔBenefit=(PA−PB)$ ...... (VII)

Here the incremental benefit between the alternative $B$ and $A$ is $ΔBenefit$ , the present worth of benefit for alternative $A$ is $PA$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$885.3$ for $PA$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$885.3−$408.6=$476.7$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$476.7$ for $ΔBenefit$ and $$500$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$476.7$500=0.9534$

Since, $Δ(B/C)$ ratio less than $1$ therefore alternative $A$ is not desirable.

Hence, alternative $B$ is desirable.

In both the cases alternative $B$ is desirable.

Hence select alternative $B$ .

Conclusion:

The alternative to be selected is Alternative $B$ .

Answer 6

Chapter 9, Problem 31P

To determine

The alternative to be selected.

Expert Solution & Answer

Answer to Problem 31P

The alternative to be selected is Alternative $B$ .

Explanation of Solution

Given:

MARR is $12%$ .

Useful life is $15$ years.

Calculation:

Write the expression to calculate the present worth of the benefit.

$P=A((1+i)n−1i(1+i)n)$ ....... (I)

Here, the annual benefit is A, the present worth is $P$ , the rate of interest is $i$ and the number of period is $n$ .

Calculate the present worth benefit of the alternative $A$ .

Substitute $$130$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$130((1+0.12)15−10.12(1+0.12)15)=$130(6.81)=$885.3$

Calculate the present worth benefit of the alternative $B$ .

Substitute $$60$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$60((1+0.12)15−10.12(1+0.12)15)=$60(6.81)=$408.6$

Calculate the present worth benefit of the alternative $C$ .

Substitute $$35$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$35((1+0.12)15−10.12(1+0.12)15)=$35(6.81)=$238.35$

Write the expression to calculate the benefit-cost ratio.

$B/C=BC$ ....... (II)

Here, the present worth benefit is $B$ , the present worth cost is $C$ .

For Alternative $A$ .

Substitute $$885.3$ for $B$ , $$800$ for $C$ , Equation (I).

$B/C=$885.3$800=1.11$

For Alternative $B$ .

Substitute $$408.6$ for $B$ , $$300$ for $C$ , Equation (I).

$B/C=$408.6$300=1.362$

For Alternative $C$ .

Substitute $$238.35$ for $B$ , $$150$ for $C$ , Equation (I).

$B/C=$238.35$150=1.589$

Since all the ratios are positive hence we need to make comparison be increment method.

Calculate the $ΔCost$ for alternative $B−C$ .

$ΔCost=(PWC−PWB)$ ...... (III)

Here the incremental cost between the alternative $B$ and $C$ is $ΔCost$ , the present worth of cost for alternative $C$ is $PWC$ and the present worth of cost for alternative $B$ is $PWB$ .

Substitute $$150$ for $PWC$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($300−$150)=$150$

Calculate the $ΔBenefit$ for alternative $B−C$ .

$ΔBenefit=(PC−PB)$ ...... (IV)

Here the incremental benefit between the alternative $B$ and $C$ is $ΔBenefit$ , the present worth of benefit for alternative $C$ is $PC$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$238.35$ for $PC$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$408.6−$238.35=$170.25$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$170.25$ for $ΔBenefit$ and $$150$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$170.25$150=1.135$

Since, $Δ(B/C)$ ratio greater than $1$ therefore alternative $B$ is desirable.

Calculate the $ΔCost$ for alternative $A−B$ .

$ΔCost=(PWA−PWB)$ ...... (VI)

Here the incremental cost between the alternative $B$ and $A$ is $ΔCost$ , the present worth of cost for alternative $B$ is $PWB$ and the present worth of cost for alternative $A$ is $PWA$ .

Substitute $$800$ for $PWA$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($800−$300)=$500$

Calculate the $ΔBenefit$ for alternative $A−B$ .

$ΔBenefit=(PA−PB)$ ...... (VII)

Here the incremental benefit between the alternative $B$ and $A$ is $ΔBenefit$ , the present worth of benefit for alternative $A$ is $PA$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$885.3$ for $PA$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$885.3−$408.6=$476.7$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$476.7$ for $ΔBenefit$ and $$500$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$476.7$500=0.9534$

Since, $Δ(B/C)$ ratio less than $1$ therefore alternative $A$ is not desirable.

Hence, alternative $B$ is desirable.

In both the cases alternative $B$ is desirable.

Hence select alternative $B$ .

Conclusion:

The alternative to be selected is Alternative $B$ .

Answer 7

Chapter 9, Problem 31P

To determine

The alternative to be selected.

Answer 8

Expert Solution & Answer

Answer to Problem 31P

The alternative to be selected is Alternative $B$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

MARR is $12%$ .

Useful life is $15$ years.

Calculation:

Write the expression to calculate the present worth of the benefit.

$P=A((1+i)n−1i(1+i)n)$ ....... (I)

Here, the annual benefit is A, the present worth is $P$ , the rate of interest is $i$ and the number of period is $n$ .

Calculate the present worth benefit of the alternative $A$ .

Substitute $$130$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$130((1+0.12)15−10.12(1+0.12)15)=$130(6.81)=$885.3$

Calculate the present worth benefit of the alternative $B$ .

Substitute $$60$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$60((1+0.12)15−10.12(1+0.12)15)=$60(6.81)=$408.6$

Calculate the present worth benefit of the alternative $C$ .

Substitute $$35$ for $A$ , $12%$ for $i$ and $15$ for $n$ in Equation (I).

$P=$35((1+0.12)15−10.12(1+0.12)15)=$35(6.81)=$238.35$

Write the expression to calculate the benefit-cost ratio.

$B/C=BC$ ....... (II)

Here, the present worth benefit is $B$ , the present worth cost is $C$ .

For Alternative $A$ .

Substitute $$885.3$ for $B$ , $$800$ for $C$ , Equation (I).

$B/C=$885.3$800=1.11$

For Alternative $B$ .

Substitute $$408.6$ for $B$ , $$300$ for $C$ , Equation (I).

$B/C=$408.6$300=1.362$

For Alternative $C$ .

Substitute $$238.35$ for $B$ , $$150$ for $C$ , Equation (I).

$B/C=$238.35$150=1.589$

Since all the ratios are positive hence we need to make comparison be increment method.

Calculate the $ΔCost$ for alternative $B−C$ .

$ΔCost=(PWC−PWB)$ ...... (III)

Here the incremental cost between the alternative $B$ and $C$ is $ΔCost$ , the present worth of cost for alternative $C$ is $PWC$ and the present worth of cost for alternative $B$ is $PWB$ .

Substitute $$150$ for $PWC$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($300−$150)=$150$

Calculate the $ΔBenefit$ for alternative $B−C$ .

$ΔBenefit=(PC−PB)$ ...... (IV)

Here the incremental benefit between the alternative $B$ and $C$ is $ΔBenefit$ , the present worth of benefit for alternative $C$ is $PC$ and the present worth of benefit for alternative $B$ is $PB$ .

Substitute $$238.35$ for $PC$ and $$408.6$ for $PB$ in Equation (IV).

$ΔBenefit=$408.6−$238.35=$170.25$

Calculate the incremental benefit-cost ratio

$Δ(B/C)=ΔBenefitΔCost$ ...... (V)

Substitute $$170.25$ for $ΔBenefit$ and $$150$ for $ΔCost$ in Equation (V).

$Δ(B/C)=$170.25$150=1.135$

Since, $Δ(B/C)$ ratio greater than $1$ therefore alternative $B$ is desirable.

Calculate the $ΔCost$ for alternative $A−B$ .

$ΔCost=(PWA−PWB)$ ...... (VI)

Here the incremental cost between the alternative $B$ and $A$ is $ΔCost$ , the present worth of cost for alternative $B$ is $PWB$ and the present worth of cost for alternative $A$ is $PWA$ .

Substitute $$800$ for $PWA$ and $$300$ for $PWB$ in Equation (III).

$ΔCost=($800−$300)=$500$

Calculate the $ΔBenefit$ for alternative $A−B$ .

$ΔBenefit=(PA−PB)$ ...... (VII)

Here the incremental benefit between the alternative $B$ and $A$ is $ΔBenefit$ , the present worth of benefit for alternative $A$ is $PA$ and the present worth of benefit for alternative $B$ is $PB$ .