Engineering Economic Analysis
Engineering Economic Analysis
13th Edition
ISBN: 9780190296902
Author: Donald G. Newnan, Ted G. Eschenbach, Jerome P. Lavelle
Publisher: Oxford University Press
Question
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Chapter 9, Problem 55P
To determine

(a)

The alternative that should be selected.

Expert Solution
Check Mark

Answer to Problem 55P

The alternative F should be selected.

Explanation of Solution

Calculations:

The table for cash flow and cumulative cash flow for project E is shown below.

Year Cash flow Cumulative cash flow
0 $90 $90
1 $20 $70
2 $20 $50
3 $20 $30
4 $20 $10
5 $20 $10
6 $20 $30

Table (1)

Here, the cumulative cash flow becomes positive in 5th year. At the end of 4th year the cash inflow is only $10 less than the cash outflow. Thus $10 is required to equal the benefits and costs. In the 5th year total cash inflow is $20.

Calculate the payback period for project E.

Payback period=(4+ $10 $20)=(4+0.5)=4.5 years

The table for cash flow and cumulative cash flow for project F is shown below.

Year Cash flow Cumulative cash flow
0 $110 $110
1 $35 $75
2 $35 $40
3 $35 $5
4 $35 $30
5 $0 $30
6 $0 $30

Table (2)

Here, the cumulative cash flow becomes positive in 4th year. At the end of 3rd year the cash inflow is only $5 less than the cash outflow. Thus $5 is required to equal the benefits and costs. In 4th year total cash inflow is $35.

Calculate the payback period for project F.

Payback period=(3+ $5 $35)=(3+0.14)=3.14 years

The table for cash flow and cumulative cash flow for project G is shown below.

Year Cash flow Cumulative cash flow
0 $100 $100
1 $0 $100
2 $10 $90
3 $20 $70
4 $30 $40
5 $40 $0
6 $50 $50

Table (3)

Here, the cumulative cash flow becomes NIL in 5th year. The profit equals cost in 5th year. The payback period will be 5th year.

The table for cash flow and cumulative cash flow for project H is shown below.

Year Cash flow Cumulative cash flow
0 $120 $120
1 $0 $120
2 $0 $120
3 $0 $120
4 $0 $120
5 $0 $120
6 $180 $60

Table (4)

Here, the cumulative cash flow becomes positive in 6th year. At the end of 5th year the cash inflow is only $120 less than the cash outflow. Thus $120 is required to equal the benefits and costs. In 6th year total cash inflow is $180.

Calculate the payback period for project H.

Payback period=(5+ $120 $180)=(5+0.6)=5.6 years

The payback period for project F is minimum which is equal to 3.1 years.

Thus, the alternative F should be selected.

Conclusion:

The alternative F should be selected.

To determine

(b)

The alternative that should be selected.

Expert Solution
Check Mark

Answer to Problem 55P

The alternative G should be selected.

Explanation of Solution

Given:

The interest rate is 5%.

Concept used:

Write the expression for future worth factor.

factor=(1+i)n ...... (I)

Here, the time period is n and the interest rate is i.

Calculations:

The table for future worth analysis of alternatives is shown below.

Year Future worth factor (1.05)n 1 Project (E) Project (F) Project (G) Project (H)
Cash flow2 Future worth3 (1×2) Cash flow4 Future worth5 (1×4) Cash flow6 Future worth7 (1×6) Cash flow8 Future worth9 (1×8)
0 1.3401 $90 $120.61 $110 $147.411 $100 $134 $120 $160.8
1 1.2762 $20 $25.52 $35 $46.67 $0 $0 $0 $0
2 1.2155 $20 $24.31 $35 $42.5425 $10 $12.155 $0 $0
3 1.57625 $20 $31.525 $35 $55.17 $20 $31.52 $0 $0
4 1.1025 $20 $22.05 $35 $38.587 $30 $30.75 $0 $0
5 1.05 $20 $21 $0 $0 $40 $42 $0 $0
6 1.00 $20 $20 $0 $0 $50 $50 $180 $180
Total $23.8 $6.45 $32.425 $19.2

Table (5)

The alternative G has the maximum future worth.

Thus, thee alternative G should be selected.

Conclusion:

The alternative G should be selected.

To determine

(c)

The alternative that should be selected.

Expert Solution
Check Mark

Answer to Problem 55P

The alternative G should be selected.

Explanation of Solution

Given:

The interest rate is 20%.

Calculations:

The table for future worth analysis of alternatives is shown below.

Year Future worth factor (1.05)n 1 Project (E) Project (F) Project (G) Project (H)
Cash flow 2 Future worth 3 (1×2) Cash flow 4 Future worth 5 (1×4) Cash flow 6 Future worth 7 (1×6) Cash flow 8 Future worth 9 (1×8)
0 2.986 $90 $268.74 $110 $328.46 $100 $298.6 $120 $358.32
1 2.488 $20 $49.76 $35 $87.08 $0 $0 $0 $0
2 2.073 $20 $41.47 $35 $72.57 $10 $20.73 $0 $0
3 1.728 $20 $34.56 $35 $60.48 $20 $34.56 $0 $0
4 1.44 $20 $28.8 $35 $50.04 $30 $43.2 $0 $0
5 1.2 $20 $25.2 $0 $0 $40 $48 $0 $0
6 1.00 $20 $20 $0 $0 $50 $50 $180 $180
Total $68.95 $58.29 $102.1 $178.32

Table (6)

The alternative G has the maximum future worth.

Thus, the alternative G should be selected.

Conclusion:

The alternative G should be selected.

To determine

(d)

The benefit cost ratio for alternate G.

Expert Solution
Check Mark

Answer to Problem 55P

The benefit cost ratio for alternate G is 0.96.

Explanation of Solution

Given:

The interest rate is 10%.

Concept used:

Write the expression for present worth of benefits.

PWB=G[( 1+i)nin1i2( 1+i)n] ...... (I)

Here rate of interest is i, time period is n and uniform gradient is G.

Write the expression for benefit cost ratio.

BC ratio=PWBPWC ...... (II)

Here, the present worth of costs is PWC.

Calculations:

Calculate the present worth of benefits.

Substitute $10 for G, 10% for i and 6 for n in Equation (I).

PWB=$10[ ( 1+0.1) 60.1×61 0.1 2 ( 1+0.1) 6]=$10[1.771.60.01×1.77]=$10(9.6)=$96

Calculate the benefit cost ratio.

Substitute $96 for PWB, and $100 for PWC in Equation (I).

BC ratio=$96$100=0.96

Thus, the benefit cost ratio for alternate G is 0.96.

Conclusion:

The benefit cost ratio for alternate G is 0.96.

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Engineering Economic Analysis

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