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610Quiz34-1.docxENGR 610- Engineering Economics, Quiz-3&4, Apr.04 .2017
{Use this word file and same format exactly to place answers and Cash Flow diagrams only at designated places!} No more than Two pages would be accepted! Type all required; No hand writings.
Name: Instructor: Mutlu Ozer, F.2016
Problem-1
A multinational engineering consulting firm that wants to provide resort accommodations to certain clients is considering the purchase of a three-bedroom lodge in upper Montana that will cost $250,000. The property in that area is rapidly appreciating in value because people anxious to get away from urban developments are bidding up the prices. If the company spends an average of $500 per month for utilities and the investment increases at a rate of 1.5% per month, how long would it be before the company could sell the property for $100,000 more than it has invested in it?
$350,000
Answer: CFD:
Eq:
i=1.5%
Npw=0=-$250,000-$500(P/A,1.5%,n)+$350,000(P/F,1.5%,n)
0=-$250,000-$500(+$350,000
n=??
$500
$250,000
Problem-2
A chemical engineer is considering two styles of pipes for moving distillate from a refinery to the tank farm. A small pipeline will cost less to purchase (including valves and other appurtenances) but will have a high head loss and, therefore, a higher pumping cost. The small pipeline will cost $1.7 million installed and will have an operating cost of $14,000 per month. A larger-diameter pipeline will cost $2.1 million installed, but its operating cost will be only $8000 per month. Which pipe size is more economical at an interest rate of 1% per month on the basis of an annual worth analysis? Assume the salvage value is 12% of the first cost for each pipeline at the end of the 10-year project period.
Smallpipe:
$204,000
Answer: CFD:
i=1%
Eq
AWsmallpipe=-$1,700,000(A/P,1%,120)-$14000+$204000(A/F,1%,120)
$14,000
0 1 120 (Month)
=-$1,700,000(0.0143)-$14000+$204000(0.00435)
=-$37,422.6
$1,700,000
AWlargepipe=-$2,100,000(A/P,1%,120)-$8000+$252000(A/F,1%,120)
$252,000
=-$2,100,000(0.0143)-$8000+$252000(0.00435)
Largepipe:
=-$36,933.8
i=1%
Large pipe costs less money
$8,000
0 1 120 (Month)
Problem-3
$2,100,000
The cost of money is 12% per year.
CFDX
$10,000
i=12%
|
|
Machine X |
Machine Y |
|
Initial cost, $ |
–70,000 |
–50,000 |
|
Annual cost, $/year |
–10,000 |
–15,000 |
|
Salvage value, $ |
10,000 |
24,000 |
Life, years 6 3
0 1 6
$70,000
$10,000
Use present worth analysis to choose the best alternative? Answer:
i=12%
$24,000
PVX = -$106,044 Eq:PVx=-$70k-$10k(P/A,12%,6)+$10k(P/F,12%,6) CFDY:
=-$70k-$10k(4.111)+$10k(0.5066)=-$106,044
$15,000
0 1 3
PVY = -$68,946.8 Eq:PVY=-$50k-$15k(P/A,12%,3)+24k(P/F,12%,3)
$50,000
=-$50k-$15k(2.402)+24k(0.7118)=-$68,946.8
Problem-4
A $10,000 bond has an interest rate of 6% per year payable quarterly. The bond matures 20 years from now. At an interest rate of 8% per year, compounded quarterly, the present worth of the bond is closest to
Answer: CFD:
Eq:
$150
$10,000
8% compounded quarterly = 2%
PresentWorth= $150(P/A,2%,80)+$10,000(P/F,2%,80)
0 1Quarter 20x4=$80
= $150(39.745)+$10,000(0.2051)
= $8012.75.15<$10,000
$10,000
Problem-5
The cash flow associated with landscaping and maintaining a certain monument in Washington, D.C., is $150,000 now and $60,000 every 5 years forever. Determine its perpetual equivalent annual worth (in years 1 through infinity) at an interest rate of 10% per year.
Answer: CFD:
Eq: AnnualWorth=$60,000(A/F,10%,5)+$150,000(0.10)
0 1 5 10 infinity
=$60,000(0.1638)+$150,000(0.10)
AW=?? $60,000 $60,000
=$24,828
$150,000
Problem-6
The following cash flows has an interest rate of 12% per year.
|
|
Alternative X |
Alternative Y |
|
|
First cost, $ Annual cost, $/year Salvage value, $ |
–200,000 –60,000 20,000 |
–800,000 –10,000 150,000 |
|
|
Life, years |
5 |
∞ |
|
|
Use Annual worth analysis to choose the best alternative? |
|
||
|
AWX =-$112,332 |
|||
|
AWY =- $106,000 |
$20,000
i=12%
0 1 5
$60,000
$200,000
i=12% $150,000
0 1 (Infinity)
$10,000
$800,000
Eq: AWx=-$200,000(A/P,12%,5) - $60,000 + $20,000(A/F,12%,5)
=-$200,000(0.2774) - $60,000 + $20,000(0.1574)=-$112,332
Eq: AWy=-$800,000(0.12) - $10,000 + $150,000(A/F,12%,)
=-$800,000(0.12) - $10,000 + $0 = -$106,000
Problem-7
Barron Chemical uses a thermoplastic polymer to enhance the appearance of certain RV panels. The initial cost of one process was $130,000 with annual costs of
$49,000 and revenues of $78,000 in year 1, increasing by $1000 per year. A salvage value of $23,000 was realized when the process was discontinued after 8 years. What rate of return did the company make on the process?
Answer: CFD:
G=1,00000
Eq:
$78,000
NPV=-$130,000-$49,000(P/A,ROR,8)+$78,000(P/A,ROR,8)+$1,000(P/G,ROR,8)+23,000(P/F,ROR,8)=0
For ROR= 18%, NPV= $5209
ROR=??%
$23,000
For ROR= 20%, NPV=-$16,544.2
$49,000
0 1 8
$130,000
Problem-8
ASM International, an Australian steel company, claims that a savings of 30% of the cost of stainless steel threaded bar can be achieved by replacing machined threads with precision weld depositions. A U.S. manufacturer of rock bolts and grout-in fittings plans to purchase the equipment. A mechanical engineer with the company has prepared the following cash flow estimates. Determine the expected rate of return per year.
Answer : Eq: NVP=-$450k - $50k(P/A,i,5)+$10kP/G,i,5)+$10k(P/G,i,6)(F/P,i,1)+$80k(P/A,i,7)(P/F,i,5)=0
,
|
Quarter |
Cost, $ |
|
Savings, $ |
$10,000 G=$10,000 $80,0000
-1 0 1 G=10,000 5 6 12 $450,000 $50,000 i=??
|
|
0 |
|
–450,000 |
— |
|
|
1 |
|
–50,000 |
10,000 |
|
|
2 |
|
–40,000 |
20,000 |
|
|
3 |
|
–30,000 |
30,000 |
|
|
4 |
|
–20,000 |
40,000 |
|
|
5 |
|
–10,000 |
50,000 |
|
|
6–12 |
|
— |
80,000
|
|
Problem-9
A permanent endowment at the University of Alabama is to award scholarships to engineering students. The awards are to be made beginning 5 years after the $10 million lump-sum donation is made. If the interest from the endowment is to fund 100 students each year in the amount of $10,000 each, what annual rate of return must the endowment fund earn?
Answer: CFD:
Stds=(100x10,000/yr)
Eq: NPW=(-10*106)+
NPW=898571 , at 7% NPW=-812500 , at 8%
ROR=??
$10,000,000
0 5 (infinity)
, ROR=7.5%
Problem-10
Alternative R has a first cost of $100,000, annual M&O costs of $50,000, and a $20,000 salvage value after 5 years. Alternative S has a first cost of $175,000 and a
$40,000 salvage value after 5 years, but its annual M&O costs are not known. Determine the M&O costs for alternative S that would yield an incremental rate of return of 20% per year.
-x+50
Answer: CFD:
$20k
Eq:
The difference in NPV=
-$75,000+[-x+50][P/A,20%]+20,000(P/F,20%,5)=0
0 1 5
$75,000
Problem-11
The incremental cash flow between alternatives Z1 and Z2 is shown below (Z2 has the higher initial cost). Use an AW-based rate of return equation to determine the incremental rate of return and which alternative should be selected, if the MARR is 17% per year. Let k =1 to 10 from year 1 through 10.
CFD
$9,000
Incremental
G=-0.5k
Year Cash Flow, $(Z2 – Z1)
0 –40,000
1–10 9000 – 500k
ROR=??
0 1 10
Answer:
Eq: Annual Worth=
$40,000
