ENGINEERING ECONOMY

1. ENGINEERING ECONOMY
Solved Problems
Text Book: Basics of Engineering Economy; Leland Blank and Anthony Tarquin; McGraw-Hill, 2008
Chapter 9:
9.23 AWX = -82,000(A/P,15%,2) – 30,000 + 42,000(A/F,15%,2)
= -82,000(0.61512) –30,000 + 42,000(0.46512)
= $-60,905
AWY = -97,000(A/P,15%,2) – 27,000 + 51,000(A/F,15%,2)
= -97,000(0.61512) –27,000 + 51,000(0.46512)
= $-62,946
Purchase robot X
9.5 AW1 = -10,000(A/P,10%,1) – 1000 + 7000(A/F,10%,1) = $-5000
AW2 = -10,000(A/P,10%,2) – 1000(P/F,10%,1)(A/P,10%,2)
+ (5000 – 1200)(A/F,10%,2) = $-4476
AW3 = -10,000(A/P,10%,3) – [1000(P/F,10%,1) +1200(P/F,10% 2)](A/P,10%,3)
+ (4200 – 1500)(A/F,10%,3) = $-3970
AW4 = -10,000(A/P,10%,4) – [1000(P/F,10%,1) +1200(P/F,10% 2)
+ 1500(P/F,10%,3)](A/P,10%,4) + (3000 – 2000)(A/F,10%,4) = $-3894
AW5 = -10,000(A/P,10%,5) – [1000(P/F,10%,1) + 1200(P/F,10% 2)
+ 1500(P/F,10%,3) + 2000(P/F,10%,4)](A/P,10%,5)
+ (2000 – 3000)(A/F,10%,5) = $-3961
Therefore, ESL is 4 years with AW = $-3894

2. Chapter 8:
Example: A company has two machine alternatives whose economic lives are 6 years. The
price and annual income of these machines are given in the following table. According to the
no return payback period, determine the alternative the company should invest.
Alternatives Cost (SR) Annual income (SR/yr)
Machine A 200 000 45 000
Machine B 300 000 60 000
Payback Period1= 200 000/45 000 = 4.4 years
Payback Period2= 300 000/60 000 = 5 years
According to the payback periods, first alternative should be preferred.
Example: A company wants to buy a production device for their new factory. They have
two alternatives, whose cash flows are given in the following table. According to these
cash flows,
determine the no return payback period of these alternatives.
Alternative A Alternative B
Cost 3 000 000 SR 3 500 000 SR
Annual income 1 200 000 SR first year,
decreasing by 300.000 SR
per year thereafter
100 000 SR for the first year,
increasing 300 000 SR per
year thereafter.
Useful life 4 years 8 years
Alternative A
Years 0 1 2 3 4
Cash Flow -3 000 000 1 200 000 900 000 600 000 300 000
Cumulative value -3 000 000 -1 800 000 -900 000 -300 000 0
PBA= 4 years

3. Alternative B
Years 0 1 2 3 4 5
Cash Flow -3 500 000 100 000 400 000 700 000 1 000 000 1 300 000
Cumulative value -3 500 000 -3 400 000 -3 000 000 -2 300 000 -1 300 000 0
PBB= 5 years
According to the payback periods, alternative A should be preferred.
8.2 (a) TC = FC + vQ
= 2.58 million + 395Q
(b) QBE = 2.58 million/(550-395)
= 16,645
(c) 1.2Q = 1.2(16,645) = 19,974
Profit = R – TC
= 19,974(550) – 2,580,000 - 395(19,974)
= $515,970
8.44 (a) Yes; cash flows sum to $139,100, which exceeds the $75,000 first cost.
(b) Solve PW = 0 relation for i*.
PW = -75,000 -10,500(P/F,i,1) + … + 105,000(P/F,i ,5) = 0
i* = 13.96% (IRR function)
(c) Calculate PW at 7% by year to determine when PW turns positive. Start with
N = 3 years.
N = 3: PW = -75,000 -10,500(P/F,7%,1) +18,600(P/F,7%,2) -2000(P/F,7%,3)
= -75,000 -10,500(0.9346) +18,600(0.8734) -2000(0.8163)
= $-70,201
N = 4: PW = -70,201 +28,000(P/F,7%,4)
= $-48,840
N = 5: PW = -48,840 +105,000(P/F,7%,5)
= $26,025
Investment is paid back plus 7% during year 5, in part due to large cash flow at
sale time.

4. Chapter 7:
CONVENTION B/C=
𝑩−𝑫
𝑰.𝑪.+𝑴𝑶−𝑺−𝑺𝒗
,
MODIFIED B/C=
𝑩−𝑫−𝑴𝑶+𝑺𝒗
𝑰.𝑪.−𝑺
, where I.C.: costs or initial cost, Sv: savings and S: salvage value.
7.11 Convert all cash flows to AW
B = 3,800,000(A/P,8%,20)
= 3,800,000(0.10185)
= $387,030
D = $65,000
C = 1,200,000(A/P,8%,20) + 300,000
= 1,200,000(0.10185) + 300,000
= $422,220
B/C = (387,030 – 65,000)/ 422,220
= 0.76< 1.0
7.16 Use annual worth, since most of the cash flows are in annual dollars.
(a) Conventional B/C ratio
B = 300,000(0.06) + 100,000
= 18,000 + 100,000
= $118,000
D = $40,000
C = 1,500,000(0.06) + 200,000(P/F,6%,3)(0.06)
= 90,000 + 200,000(0.8396)(0.06)
= $100,075
Sv = 70,000
B/C = (118,000 – 40,000)/(100,075 – 70,000)
= 2.59
(b) Modified B/C ratio = (B – D + Sv)/C
= (118,000 – 40,000 + 70,000)/100,075
= 1.48

5. 7.20 B = 1250(140) = $175,000 per year
C = 595,000(A/P,6%,5) + 56,000 + 19,000
= 595,000(0.23740) + 75,000
= $216,253
B/C = 0.81< 1.0 Therefore, the utility should send the samples out.
Final Exam: Chapters 1, 2, 4, 5, 7, 8, and 9. (50 Marks)
Chapter 9:
Read Sections 9.1-9.2, and 9.5.
Read and solve examples 9.4, 9.2.
Read and solve Problems: 9.5, 9.23.
HWK_CH. 9: PROPLEMS 9.6, 9.7, 9.8, 9.9, 9.22.
Chapter 8:
Read Sections 8.1, and 8.5.
Read and solve examples 8.1, 8.7.
Read and solve Problems: 8.2, 8.44.
HWK_CH. 8: PROPLEMS 8.12, 8.13, 8.45, 8.54 to 8.57.
Chapter 7:
Read Sections 7.1, and 7.2.
Read and solve example 7.2.
Read and solve Problems: 7.11, 7.16 AND 7.20.
HWK_CH. 7: PROPLEMS 7.10, 7.12, 7.14, 7.15, 7.18, 7.32 to 7.35.
Chapter 5:

6. Read Sections 5.1, 5.2 and 5.3.
Read and solve examples 5.1, 5.2, 5.3, 5.4 and 5.6.
HWK_CH. 5: PROPLEMS 5.2, 5.4, 5.7, 5.12, 5.13, 5.17, 5.20, 5.21 and 5.25 to 5.27.
Chapter 4:
Read Sections 4.1, 4.2, 4.3 and 4.5.
Read and solve examples 4.1, 4.3 and 4.7.
HWK_CH. 4: PROPLEMS 4.9, 4.15, 4.23, 4.28, 4.36, 4.42 and 4.45 to 4.48.
Chapter 2:
Read Sections 2.1 to 2.4.
Read and solve examples 2.1 to 2.11.
HWK_CH. 2: PROPLEMS 2.12, 2.35, 2.36, 2.40, 2.54, Find (P0)in Fig. 2.16a-b (page 44), 2.64, 2.74, 2.80
and 2.85.
Chapter 1:
Read Sections 1.1 to 1.7.
Read and solve examples 1.1 to 1.13.
HWK_CH. 1: PROPLEMS 1.19, 1.22, 1.28, 1.37, 1.41 to 1.45, and ) In the plan 5, example 1.6 (page 12):
Show that equal end-of-year payment is $1252.28.