Solutions Manual For Fundamentals of Engineering Economics, 4e Chan Park (All Chapters, 100% Original Verified, A+ Grade)

Solutions Manual For Fundamentals of Engineering Economics, 4e
Chan Park (All Chapters, 100% Original Verified, A+ Grade)
Chapter 1 Engineering Economic Decisions
1.1) Not provided

For The Wall Street Journal, go to the Front page to find the section on ―What’s
News.‖ This is a section on a brief summary on major headlines of the day’s
news. Quickly browse through the news summary to see if there is any news
related to business investment. The best places to find the major business news on
investment are sections on ―BUSINESS‖, ―MARKETS‖ or ―TECH.‖

1.2)
Not provided

Some of the well-known business publications are:

 Daily Newspapers:

o The Wall Street Journal
o The New York Times (Business Section)
o Financial Times

 Weekly or Monthly Magazines:

o BusinessWeek
o Forbes
o Money
o Smart Money
o Fortune

,Chapter 2: Time Value of Money
2.1) I  (iP)N  (0.06)($2, 000)(5)  $600

2.2)
 Simple interest:

$20,000  $10,000(1  0.075N )
(1  0.075N )  2
1
N  13.33 14 years
0.075

 Compound interest:

$20,000  $10,000(1  0.07)N
(1  0.07)N  2
N  10.24  11years


2.3)

 Simple interest:

I  iPN  (0.10)($10, 000)(5)  $5, 000

 Compound interest:

I  P[(1 i)N 1]  $10, 000(1.6105 1)  $6,105


2.4)

 Option 1: Compound interest with 8.5%:

F  $4, 500(1 0.085)5  $4, 500(1.5037)  $6, 766.65

 Option 2: Simple interest with 9%:

$4, 500(1 0.09  5)  $5, 000(1.45)  $6, 525

 Option 1 is still better.

,2.5)
 Compound interest:

F  $1, 000(1 0.065)5
 $1, 370.09

 Simple interest:

F  $1, 000(1 0.068(5))
 $1, 340
The compound interest option is better.

2.6)

End of Year Principal Interest Remaining
Repayment payment Balance
0 $15,000.00
1 $4,620.50 $1,200.00 $10,379.50
2 $4,990.14 $830.36 $5,389.36
3 $5,389.35 $431.15 $0


2.7)

P  $22, 000(P / F , 5%, 5)  $22, 000(0.7835)  $17, 237.58

2.8) F  $30, 000(F / P, 9%, 3)  $30, 000(1.295)  $38,850.87

2.9)
F  $100(F / P,10%,10)  $200(F / P,10%,8)  $688

2.10) F  $250, 000(F / P, 6%,10)  $447, 712

2.11) P  $300, 000(P / F ,8%,10)  $138, 958

2.12) i  10.5% , two-year discount rate is (1 0.105)2  1.221 (or 22.1%)

2.13)


(a) F  $5, 000(F / P, 7%, 5)  $7, 013

, (b) F  $7, 250(F / P, 9%,15)  $26, 408

(c) F  $9, 000(F / P, 6%, 33)  $61, 565

(d) F  $12, 000(F / P, 5.5%,8)  $18, 416


2.14)
(a) P  $25, 500(P / F ,12%,8)  $10, 299

(b) P  $58, 000(P / F , 4%,12)  $36, 227

(c) P  $25, 000(P / F , 6%, 9)  $14, 797

(d) P  $35, 000(P / F , 9%, 4)  $24, 795

2.15) (a) P  $12, 000(P / F ,13%, 4)  $7, 360

(b) F  $30, 000(F / P,13%, 5)  $55, 273

2.16)
F  3P  P(1 0.08)N
log 3  N log(1.08)
N  14.27  15 years

2.17)
F  2P  P(1 0.06)N
log 2  N log 1.06
N  11.896 years (or 12 years)

2.18)
F  2P  P(1 0.06)N
 log 2  N log(1.06)
N  11.90 years 12 years

 Rule of 72:  12 years




2.19) F  $1(1.08)394  $14, 755, 694, 730, 611