The best design.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 16, Problem 30P

To determine

The best design.

Expert Solution & Answer

Answer to Problem 30P

The best design is alternative $A$ .

Explanation of Solution

Given:

Interest rate is $8%$ .

Project life is $30$ years.

Concept used:

Write the expression to calculate the $B/C$ ratio using annual worth analysis.

$B/C=AWNet BenefitsAWCost=A+F(i)(1+i)n−1Pi(1+i)n(1+i)n−1$ ...... (I)

Here, the net benefit is $A$ , the salvage value is $F$ , the interest rate is $i$ and the number of year is $n$ .

Calculation:

Alternative $A$ .

Calculate the net benefit for alternative $A$ .

$Net Benefit=(Annual Benefit)−(O&M Cost)−(Annual Disbenifit)$ ...... (II)

Substitute $$2200$ for $Annual Benefit$ , $$550$ for $O&M Cost$ and $$350$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2200−$550−$350=$1300$

Calculate the $B/C$ ratio.

Substitute $$1300$ for $A$ , $$1000$ for $F$ , $$9500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1300+($1000(0.08)(1+0.08)30−1)($95000.08(1+0.08)30(1+0.08)30−1)=$1300+$1000×0.0088$9500×0.0888=$1.55$

Thus, the $B/C$ for alternative $A$ is $1.55$ .

Alternative $B$ .

Calculate the net benefit for alternative $B$ .

Substitute $$1500$ for $Annual Benefit$ , $$175$ for $O&M Cost$ and $$150$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1500−$175−$150=$1175$

Calculate the $B/C$ ratio.

Substitute $$1175$ for $A$ , $$6000$ for $F$ , $$12500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1175+($6000(0.08)(1+0.08)30−1)($125000.08(1+0.08)30(1+0.08)30−1)=$1175+$6000×0.0088$12500×0.0888=1.106$

Thus, the $B/C$ for alternative $B$ is $1.55$ .

Alternative $C$ .

Calculate the net benefit for alternative $C$ .

Substitute $$1000$ for $Annual Benefit$ , $$325$ for $O&M Cost$ and $$75$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1000−$325−$75=$600$

Calculate the $B/C$ ratio.

Substitute $$600$ for $A$ , $$3500$ for $F$ , $$14000$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$600+($3500(0.08)(1+0.08)30−1)($140000.08(1+0.08)30(1+0.08)30−1)=$600+$3500×0.0088$14000×0.0888=0.507$

Thus, the $B/C$ for alternative $C$ is $0.507$ .

Alternative $D$ .

Calculate the net benefit for alternative $D$ .

Substitute $$2500$ for $Annual Benefit$ , $$145$ for $O&M Cost$ and $$700$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2500−$145−$700=$1655$

Calculate the $B/C$ ratio.

Substitute $$1655$ for $A$ , $$7500$ for $F$ , $$15750$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1655+($7500(0.08)(1+0.08)30−1)($157500.08(1+0.08)30(1+0.08)30−1)=$1655+$7500×0.0088$15750×0.0888=1.18$

Thus, the $B/C$ for alternative $B$ is $1.18$ .

Conclusion:

The $B/C$ ratio for the alternative $A$ is the highest.

Thus, the best design is alternative $A$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

not use ai please

1. Lisa has $48 per week set aside for coffees (x) and lunches (z). The price of coffee is $4 and lunches are $6. What is Lisa's budget line equation (with z on the left-hand side)? Graph the budget line, and show how it changes when the price of lunches rise to $8 (including intercepts). What is the new budget line equation? 2. Suppose utility for a consumer of movies (x) and golf (z) is U = 20x0.420.5. The consumer has set aside $1000 to consumer movies and golf for a year. a. If the price of movies is $20 and the price of golf is $40, what is the utility-maximizing consumption of movies and golf? b. Show the optimal consumption bundle on a graph, showing a budget line (with intercepts), a tangent indifference curve, and the optimal choice. 3. Sam has set aside $480 for entertainment this month, which is golf (x) and/or bowling (z). A round of golf is $40 and a night of bowling is $30. His utility function is U = 3x + 2z. a. What is his MRS? b. Solve for the optimal choice of golf…

Question Seven There are specific applications of the hidden-action or moral hazard model. Consider employment contracts signed between a firm's owners and a manager who runs the firm on behalf of the owners. The manager is offered an employment contract which they can accept and decide how much effort, e ≥ 0, to exert. Suppose that an increase in effort, e, increases the firm's gross profit, not including payments to the manager, but is personally costly to the manager and the firm's gross profit, Пg, takes the following form: Пg = e +ε, ε~N(0,2). Let s denote the salary, which may depend on effort and/or gross profit, depending on what the owner can observe, offered as part of the contract between the owner and manager. Suppose that the manager is risk averse and has a utility function with respect to salary of the form: Aσ² U(W)=μ- 2 a) Derive the optimal result of the owner's expected net profit where there is full information and state what it implies. b) Suppose now that the…

Answer 2

Question

Chapter 16, Problem 30P

To determine

The best design.

Expert Solution & Answer

Answer to Problem 30P

The best design is alternative $A$ .

Explanation of Solution

Given:

Interest rate is $8%$ .

Project life is $30$ years.

Concept used:

Write the expression to calculate the $B/C$ ratio using annual worth analysis.

$B/C=AWNet BenefitsAWCost=A+F(i)(1+i)n−1Pi(1+i)n(1+i)n−1$ ...... (I)

Here, the net benefit is $A$ , the salvage value is $F$ , the interest rate is $i$ and the number of year is $n$ .

Calculation:

Alternative $A$ .

Calculate the net benefit for alternative $A$ .

$Net Benefit=(Annual Benefit)−(O&M Cost)−(Annual Disbenifit)$ ...... (II)

Substitute $$2200$ for $Annual Benefit$ , $$550$ for $O&M Cost$ and $$350$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2200−$550−$350=$1300$

Calculate the $B/C$ ratio.

Substitute $$1300$ for $A$ , $$1000$ for $F$ , $$9500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1300+($1000(0.08)(1+0.08)30−1)($95000.08(1+0.08)30(1+0.08)30−1)=$1300+$1000×0.0088$9500×0.0888=$1.55$

Thus, the $B/C$ for alternative $A$ is $1.55$ .

Alternative $B$ .

Calculate the net benefit for alternative $B$ .

Substitute $$1500$ for $Annual Benefit$ , $$175$ for $O&M Cost$ and $$150$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1500−$175−$150=$1175$

Calculate the $B/C$ ratio.

Substitute $$1175$ for $A$ , $$6000$ for $F$ , $$12500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1175+($6000(0.08)(1+0.08)30−1)($125000.08(1+0.08)30(1+0.08)30−1)=$1175+$6000×0.0088$12500×0.0888=1.106$

Thus, the $B/C$ for alternative $B$ is $1.55$ .

Alternative $C$ .

Calculate the net benefit for alternative $C$ .

Substitute $$1000$ for $Annual Benefit$ , $$325$ for $O&M Cost$ and $$75$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1000−$325−$75=$600$

Calculate the $B/C$ ratio.

Substitute $$600$ for $A$ , $$3500$ for $F$ , $$14000$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$600+($3500(0.08)(1+0.08)30−1)($140000.08(1+0.08)30(1+0.08)30−1)=$600+$3500×0.0088$14000×0.0888=0.507$

Thus, the $B/C$ for alternative $C$ is $0.507$ .

Alternative $D$ .

Calculate the net benefit for alternative $D$ .

Substitute $$2500$ for $Annual Benefit$ , $$145$ for $O&M Cost$ and $$700$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2500−$145−$700=$1655$

Calculate the $B/C$ ratio.

Substitute $$1655$ for $A$ , $$7500$ for $F$ , $$15750$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1655+($7500(0.08)(1+0.08)30−1)($157500.08(1+0.08)30(1+0.08)30−1)=$1655+$7500×0.0088$15750×0.0888=1.18$

Thus, the $B/C$ for alternative $B$ is $1.18$ .

Conclusion:

The $B/C$ ratio for the alternative $A$ is the highest.

Thus, the best design is alternative $A$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 16, Problem 30P

To determine

The best design.

Expert Solution & Answer

Answer to Problem 30P

The best design is alternative $A$ .

Explanation of Solution

Given:

Interest rate is $8%$ .

Project life is $30$ years.

Concept used:

Write the expression to calculate the $B/C$ ratio using annual worth analysis.

$B/C=AWNet BenefitsAWCost=A+F(i)(1+i)n−1Pi(1+i)n(1+i)n−1$ ...... (I)

Here, the net benefit is $A$ , the salvage value is $F$ , the interest rate is $i$ and the number of year is $n$ .

Calculation:

Alternative $A$ .

Calculate the net benefit for alternative $A$ .

$Net Benefit=(Annual Benefit)−(O&M Cost)−(Annual Disbenifit)$ ...... (II)

Substitute $$2200$ for $Annual Benefit$ , $$550$ for $O&M Cost$ and $$350$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2200−$550−$350=$1300$

Calculate the $B/C$ ratio.

Substitute $$1300$ for $A$ , $$1000$ for $F$ , $$9500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1300+($1000(0.08)(1+0.08)30−1)($95000.08(1+0.08)30(1+0.08)30−1)=$1300+$1000×0.0088$9500×0.0888=$1.55$

Thus, the $B/C$ for alternative $A$ is $1.55$ .

Alternative $B$ .

Calculate the net benefit for alternative $B$ .

Substitute $$1500$ for $Annual Benefit$ , $$175$ for $O&M Cost$ and $$150$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1500−$175−$150=$1175$

Calculate the $B/C$ ratio.

Substitute $$1175$ for $A$ , $$6000$ for $F$ , $$12500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1175+($6000(0.08)(1+0.08)30−1)($125000.08(1+0.08)30(1+0.08)30−1)=$1175+$6000×0.0088$12500×0.0888=1.106$

Thus, the $B/C$ for alternative $B$ is $1.55$ .

Alternative $C$ .

Calculate the net benefit for alternative $C$ .

Substitute $$1000$ for $Annual Benefit$ , $$325$ for $O&M Cost$ and $$75$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1000−$325−$75=$600$

Calculate the $B/C$ ratio.

Substitute $$600$ for $A$ , $$3500$ for $F$ , $$14000$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$600+($3500(0.08)(1+0.08)30−1)($140000.08(1+0.08)30(1+0.08)30−1)=$600+$3500×0.0088$14000×0.0888=0.507$

Thus, the $B/C$ for alternative $C$ is $0.507$ .

Alternative $D$ .

Calculate the net benefit for alternative $D$ .

Substitute $$2500$ for $Annual Benefit$ , $$145$ for $O&M Cost$ and $$700$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2500−$145−$700=$1655$

Calculate the $B/C$ ratio.

Substitute $$1655$ for $A$ , $$7500$ for $F$ , $$15750$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1655+($7500(0.08)(1+0.08)30−1)($157500.08(1+0.08)30(1+0.08)30−1)=$1655+$7500×0.0088$15750×0.0888=1.18$

Thus, the $B/C$ for alternative $B$ is $1.18$ .

Conclusion:

The $B/C$ ratio for the alternative $A$ is the highest.

Thus, the best design is alternative $A$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 16, Problem 30P

To determine

The best design.

Expert Solution & Answer

Answer to Problem 30P

The best design is alternative $A$ .

Explanation of Solution

Given:

Interest rate is $8%$ .

Project life is $30$ years.

Concept used:

Write the expression to calculate the $B/C$ ratio using annual worth analysis.

$B/C=AWNet BenefitsAWCost=A+F(i)(1+i)n−1Pi(1+i)n(1+i)n−1$ ...... (I)

Here, the net benefit is $A$ , the salvage value is $F$ , the interest rate is $i$ and the number of year is $n$ .

Calculation:

Alternative $A$ .

Calculate the net benefit for alternative $A$ .

$Net Benefit=(Annual Benefit)−(O&M Cost)−(Annual Disbenifit)$ ...... (II)

Substitute $$2200$ for $Annual Benefit$ , $$550$ for $O&M Cost$ and $$350$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2200−$550−$350=$1300$

Calculate the $B/C$ ratio.

Substitute $$1300$ for $A$ , $$1000$ for $F$ , $$9500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1300+($1000(0.08)(1+0.08)30−1)($95000.08(1+0.08)30(1+0.08)30−1)=$1300+$1000×0.0088$9500×0.0888=$1.55$

Thus, the $B/C$ for alternative $A$ is $1.55$ .

Alternative $B$ .

Calculate the net benefit for alternative $B$ .

Substitute $$1500$ for $Annual Benefit$ , $$175$ for $O&M Cost$ and $$150$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1500−$175−$150=$1175$

Calculate the $B/C$ ratio.

Substitute $$1175$ for $A$ , $$6000$ for $F$ , $$12500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1175+($6000(0.08)(1+0.08)30−1)($125000.08(1+0.08)30(1+0.08)30−1)=$1175+$6000×0.0088$12500×0.0888=1.106$

Thus, the $B/C$ for alternative $B$ is $1.55$ .

Alternative $C$ .

Calculate the net benefit for alternative $C$ .

Substitute $$1000$ for $Annual Benefit$ , $$325$ for $O&M Cost$ and $$75$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1000−$325−$75=$600$

Calculate the $B/C$ ratio.

Substitute $$600$ for $A$ , $$3500$ for $F$ , $$14000$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$600+($3500(0.08)(1+0.08)30−1)($140000.08(1+0.08)30(1+0.08)30−1)=$600+$3500×0.0088$14000×0.0888=0.507$

Thus, the $B/C$ for alternative $C$ is $0.507$ .

Alternative $D$ .

Calculate the net benefit for alternative $D$ .

Substitute $$2500$ for $Annual Benefit$ , $$145$ for $O&M Cost$ and $$700$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2500−$145−$700=$1655$

Calculate the $B/C$ ratio.

Substitute $$1655$ for $A$ , $$7500$ for $F$ , $$15750$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1655+($7500(0.08)(1+0.08)30−1)($157500.08(1+0.08)30(1+0.08)30−1)=$1655+$7500×0.0088$15750×0.0888=1.18$

Thus, the $B/C$ for alternative $B$ is $1.18$ .

Conclusion:

The $B/C$ ratio for the alternative $A$ is the highest.

Thus, the best design is alternative $A$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 16, Problem 30P

To determine

The best design.

Expert Solution & Answer

Answer to Problem 30P

The best design is alternative $A$ .

Explanation of Solution

Given:

Interest rate is $8%$ .

Project life is $30$ years.

Concept used:

Write the expression to calculate the $B/C$ ratio using annual worth analysis.

$B/C=AWNet BenefitsAWCost=A+F(i)(1+i)n−1Pi(1+i)n(1+i)n−1$ ...... (I)

Here, the net benefit is $A$ , the salvage value is $F$ , the interest rate is $i$ and the number of year is $n$ .

Calculation:

Alternative $A$ .

Calculate the net benefit for alternative $A$ .

$Net Benefit=(Annual Benefit)−(O&M Cost)−(Annual Disbenifit)$ ...... (II)

Substitute $$2200$ for $Annual Benefit$ , $$550$ for $O&M Cost$ and $$350$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2200−$550−$350=$1300$

Calculate the $B/C$ ratio.

Substitute $$1300$ for $A$ , $$1000$ for $F$ , $$9500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1300+($1000(0.08)(1+0.08)30−1)($95000.08(1+0.08)30(1+0.08)30−1)=$1300+$1000×0.0088$9500×0.0888=$1.55$

Thus, the $B/C$ for alternative $A$ is $1.55$ .

Alternative $B$ .

Calculate the net benefit for alternative $B$ .

Substitute $$1500$ for $Annual Benefit$ , $$175$ for $O&M Cost$ and $$150$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1500−$175−$150=$1175$

Calculate the $B/C$ ratio.

Substitute $$1175$ for $A$ , $$6000$ for $F$ , $$12500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1175+($6000(0.08)(1+0.08)30−1)($125000.08(1+0.08)30(1+0.08)30−1)=$1175+$6000×0.0088$12500×0.0888=1.106$

Thus, the $B/C$ for alternative $B$ is $1.55$ .

Alternative $C$ .

Calculate the net benefit for alternative $C$ .

Substitute $$1000$ for $Annual Benefit$ , $$325$ for $O&M Cost$ and $$75$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1000−$325−$75=$600$

Calculate the $B/C$ ratio.

Substitute $$600$ for $A$ , $$3500$ for $F$ , $$14000$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$600+($3500(0.08)(1+0.08)30−1)($140000.08(1+0.08)30(1+0.08)30−1)=$600+$3500×0.0088$14000×0.0888=0.507$

Thus, the $B/C$ for alternative $C$ is $0.507$ .

Alternative $D$ .

Calculate the net benefit for alternative $D$ .

Substitute $$2500$ for $Annual Benefit$ , $$145$ for $O&M Cost$ and $$700$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2500−$145−$700=$1655$

Calculate the $B/C$ ratio.

Substitute $$1655$ for $A$ , $$7500$ for $F$ , $$15750$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1655+($7500(0.08)(1+0.08)30−1)($157500.08(1+0.08)30(1+0.08)30−1)=$1655+$7500×0.0088$15750×0.0888=1.18$

Thus, the $B/C$ for alternative $B$ is $1.18$ .

Conclusion:

The $B/C$ ratio for the alternative $A$ is the highest.

Thus, the best design is alternative $A$ .

Answer 6

Chapter 16, Problem 30P

To determine

The best design.

Expert Solution & Answer

Answer to Problem 30P

The best design is alternative $A$ .

Explanation of Solution

Given:

Interest rate is $8%$ .

Project life is $30$ years.

Concept used:

Write the expression to calculate the $B/C$ ratio using annual worth analysis.

$B/C=AWNet BenefitsAWCost=A+F(i)(1+i)n−1Pi(1+i)n(1+i)n−1$ ...... (I)

Here, the net benefit is $A$ , the salvage value is $F$ , the interest rate is $i$ and the number of year is $n$ .

Calculation:

Alternative $A$ .

Calculate the net benefit for alternative $A$ .

$Net Benefit=(Annual Benefit)−(O&M Cost)−(Annual Disbenifit)$ ...... (II)

Substitute $$2200$ for $Annual Benefit$ , $$550$ for $O&M Cost$ and $$350$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2200−$550−$350=$1300$

Calculate the $B/C$ ratio.

Substitute $$1300$ for $A$ , $$1000$ for $F$ , $$9500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1300+($1000(0.08)(1+0.08)30−1)($95000.08(1+0.08)30(1+0.08)30−1)=$1300+$1000×0.0088$9500×0.0888=$1.55$

Thus, the $B/C$ for alternative $A$ is $1.55$ .

Alternative $B$ .

Calculate the net benefit for alternative $B$ .

Substitute $$1500$ for $Annual Benefit$ , $$175$ for $O&M Cost$ and $$150$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1500−$175−$150=$1175$

Calculate the $B/C$ ratio.

Substitute $$1175$ for $A$ , $$6000$ for $F$ , $$12500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1175+($6000(0.08)(1+0.08)30−1)($125000.08(1+0.08)30(1+0.08)30−1)=$1175+$6000×0.0088$12500×0.0888=1.106$

Thus, the $B/C$ for alternative $B$ is $1.55$ .

Alternative $C$ .

Calculate the net benefit for alternative $C$ .

Substitute $$1000$ for $Annual Benefit$ , $$325$ for $O&M Cost$ and $$75$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1000−$325−$75=$600$

Calculate the $B/C$ ratio.

Substitute $$600$ for $A$ , $$3500$ for $F$ , $$14000$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$600+($3500(0.08)(1+0.08)30−1)($140000.08(1+0.08)30(1+0.08)30−1)=$600+$3500×0.0088$14000×0.0888=0.507$

Thus, the $B/C$ for alternative $C$ is $0.507$ .

Alternative $D$ .

Calculate the net benefit for alternative $D$ .

Substitute $$2500$ for $Annual Benefit$ , $$145$ for $O&M Cost$ and $$700$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2500−$145−$700=$1655$

Calculate the $B/C$ ratio.

Substitute $$1655$ for $A$ , $$7500$ for $F$ , $$15750$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1655+($7500(0.08)(1+0.08)30−1)($157500.08(1+0.08)30(1+0.08)30−1)=$1655+$7500×0.0088$15750×0.0888=1.18$

Thus, the $B/C$ for alternative $B$ is $1.18$ .

Conclusion:

The $B/C$ ratio for the alternative $A$ is the highest.

Thus, the best design is alternative $A$ .

Answer 7

Chapter 16, Problem 30P

To determine

The best design.

Answer 8

Expert Solution & Answer

Answer to Problem 30P

The best design is alternative $A$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

Interest rate is $8%$ .

Project life is $30$ years.

Concept used:

Write the expression to calculate the $B/C$ ratio using annual worth analysis.

$B/C=AWNet BenefitsAWCost=A+F(i)(1+i)n−1Pi(1+i)n(1+i)n−1$ ...... (I)

Here, the net benefit is $A$ , the salvage value is $F$ , the interest rate is $i$ and the number of year is $n$ .

Calculation:

Alternative $A$ .

Calculate the net benefit for alternative $A$ .

$Net Benefit=(Annual Benefit)−(O&M Cost)−(Annual Disbenifit)$ ...... (II)

Substitute $$2200$ for $Annual Benefit$ , $$550$ for $O&M Cost$ and $$350$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2200−$550−$350=$1300$

Calculate the $B/C$ ratio.

Substitute $$1300$ for $A$ , $$1000$ for $F$ , $$9500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1300+($1000(0.08)(1+0.08)30−1)($95000.08(1+0.08)30(1+0.08)30−1)=$1300+$1000×0.0088$9500×0.0888=$1.55$

Thus, the $B/C$ for alternative $A$ is $1.55$ .

Alternative $B$ .

Calculate the net benefit for alternative $B$ .

Substitute $$1500$ for $Annual Benefit$ , $$175$ for $O&M Cost$ and $$150$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1500−$175−$150=$1175$

Calculate the $B/C$ ratio.

Substitute $$1175$ for $A$ , $$6000$ for $F$ , $$12500$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1175+($6000(0.08)(1+0.08)30−1)($125000.08(1+0.08)30(1+0.08)30−1)=$1175+$6000×0.0088$12500×0.0888=1.106$

Thus, the $B/C$ for alternative $B$ is $1.55$ .

Alternative $C$ .

Calculate the net benefit for alternative $C$ .

Substitute $$1000$ for $Annual Benefit$ , $$325$ for $O&M Cost$ and $$75$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$1000−$325−$75=$600$

Calculate the $B/C$ ratio.

Substitute $$600$ for $A$ , $$3500$ for $F$ , $$14000$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$600+($3500(0.08)(1+0.08)30−1)($140000.08(1+0.08)30(1+0.08)30−1)=$600+$3500×0.0088$14000×0.0888=0.507$

Thus, the $B/C$ for alternative $C$ is $0.507$ .

Alternative $D$ .

Calculate the net benefit for alternative $D$ .

Substitute $$2500$ for $Annual Benefit$ , $$145$ for $O&M Cost$ and $$700$ for $Annual Disbenefit$ in Equation (II).

$Net Benefit=$2500−$145−$700=$1655$

Calculate the $B/C$ ratio.

Substitute $$1655$ for $A$ , $$7500$ for $F$ , $$15750$ for P, $0.08$ for $i$ and $30$ for $n$ in Equation (I).

$B/C=$1655+($7500(0.08)(1+0.08)30−1)($157500.08(1+0.08)30(1+0.08)30−1)=$1655+$7500×0.0088$15750×0.0888=1.18$

Thus, the $B/C$ for alternative $B$ is $1.18$ .

Conclusion:

The $B/C$ ratio for the alternative $A$ is the highest.

Thus, the best design is alternative $A$ .

Answer 12

Ch. 16 - Prob. 1QTC Ch. 16 - Prob. 2QTC Ch. 16 - Prob. 3QTC Ch. 16 - Prob. 4QTC Ch. 16 - Prob. 1P Ch. 16 - Prob. 2P Ch. 16 - Prob. 3P Ch. 16 - Prob. 4P Ch. 16 - Prob. 5P Ch. 16 - Prob. 6P

The best design.

Videos

The best design.

Answer to Problem 30P

Explanation of Solution

Want to see more full solutions like this?

Chapter 16 Solutions