everyone_ econ_eng

i already have the answer i just need some explintion for each question ( how do i get to the answer )

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eering

2 Economics Eng

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.25 per mile. Assuming a 40-year life for each road

00,000.

.

=Book value at beginning of year n – Salvage value,over, Remaining useful life at beginning of year n

n Dn

Dn Bn

0 $130,000 $130,000
1 $22,000

2 $22,000

3 $22,000

4 $22,000

$20,000

5 $22,000 $20,000 $0 $20,000

Sami
TECH-i	n	4	5
Vov 26, 2	0
Problem 8.5
Two different routes are under consideration for a new interstate highway:
Length of	Annual
Highway	First Cost	Upkeep Cost
The ” Long” route	22 miles	$21 million	$140,000
Transmountain shortcut	10 miles	$45 million	$165,000
For either route, the volume of traffic will be 400,000 cars per year. These cars
are assumed to operate at	$0
and an interest rate of 10%, determine which route should be selected.
User’s annual cost= 22miles*$0,25 per mile*400,000 cars=$2,200,000
Sponsor’s annual cost=$21,000,000(A/P,10%,40)+$140,000=$2,277,488
Short cut:
User’s annual cost= 10miles*$0,25*400,000 cars= $1,000,000
Sponsor’s annual cost=$45,000,000(A/P,10%,40)+$165,000=$4,766,674
Problem 9.1
Identify which of the following expenditures is considered as a capital expenditure that
must be capitalized (depreciated):
(a) Purchase land to build a warehouse at $	3
(b)Purchased a copy machine at $15,000.
( c) Installed a conveyor syatem at a cost of $55,000 to automate some part of production processes.
(d)Painted the office building, both interior and exerior, at a cost of					$22,000
(e)Repaved the parking lot at a cost of $25,000.
(f) Installed a purified water fountain in the employee lounge at a cost of $3,000.
(g) Purchased a spare part for a stamping machine at a cost of $3,800.
(h) Paid $12,000 to lease a dump truck for six months.
(i) Purchased a patent on an energy-saving device over five years at a cost of $30,000.
a,b,e,f, and h (amortization,rather than depreciation)
Problem 9.9
Consider the following data on an asset:
Cost of the asset, I			$130,000
Useful life, N	5 years
Salvage value, S				$20,000
Compute the depreciation allowance and the resulting book values using the following method:
		Dn
SL	DDB
	Bn
$108,000	$52,000	$78,000
$86,000	$31,200	$46,800
$64,000	$18,720	$28,080
$42,000	$693
(a) Depreciation rate=1/5 for SL
(b) Deprciation rate= 2/5 for DDB

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,000.The remaining $22,000 will be financed by the dwaler.The dealer computes your monthlypayment

N

0

)

1

2 $23,000 $32,000 ($18,000)

3

($25,000) ($18,000)

el.

PW(I)=-$25,000+$12,000(P/F,I,1)+$23,000(P/F,I,2)+$34,000(P/F,I,3)=0

N

0

1

2

3 x

Sami

TECH 452 Economics Engineering

N

0

Problem 7.4

Assume that you are going to buy a new car for $25,000.You will be able to make a down payment of

$

3

to be $547.47 for48 months of financing. What is the dealer’s annual rate of return on this car loan?

$22,000=$547.47(P/A,I,48)

i=0.75%

r=0.75*12=9%

i=(1+i)^-1

i=(1+0.0075)^12 – 1=9.38% per year

Problem 7.11

Consider four projects with the following sequences of cash flows

Net Cash Flow

Project A

ProjectB

ProjectC

ProjectD

($25,000)

(

$23,000

$43,233

($56,600)

$12,000

$32,000

($18,000)

($2,500)

($6,459)

$34,000

$88,345

a) Identify all the simple investments.

Simple investment: Project A,D project C Simple borrowing.

b) Identify all the nonsimple investments

Non- simple investment: Project B

c) Comare the i* for each project using E

x

Project A:

i^=59.32%

ProjectB:

PW(I)=$23,000+$32,000(P/A,I,2)- $25,000(P/F,I,3)=0

I^=82.72%

Project C:

PW(I)=$43,233-$18,000(P/A,I,3)=0

I^=12% A=borrowing rate of return

Project D:

PW(I)= – $56,500 – $2,500(P/F,I,1) – $6,459(P/F,I,2)+$88,345(P/F,I,3)=0

I^=11.37%

d) Which projet has no rate of return?

The answer could be project C on the grounds that its cash flowrepresent a loan not an investment.

Problem 7.20

Consider the following project’s cash flows:

net cash flow

($3,000)

$800

$900

Assume that the project’s IRR is 10%

(A) find the value of X

PW(I)=P/F,I,N)

PW(10%)= – $3,000+ $800(P/F,10%,1) + $9,000(P/F,10%,2)+ X (P/F,10%,3)=0

X= $2,035

(B) Is this porject acceptable at MARR=8%?

Since IRR=8%, the project is acceptable.

&A Page &P

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0,000 from a local bank at an interest rate of 9% over five years.

0

1

2 $0 $1,800

$1,800

3

$1,800

Sami Almalki
TECH 452 – Engineering Economics
5-Nov-12
Homework # 4
Problem 6.1
An engineering design firm needs to borrow $	3	0
What is the required annual equal payment to retire the lona in five years?
AE(9%)=$300,000(A/P,9%,5)= $77,127.74
Problem 6.9
Cosider the following sets of investment projects:
Period	Project Cash Flow
($4,300)	-3,500	-5,500	-3,800
	$0	$1,500	$3,000				$1,800
$2,000
$5,500	$2,100	$1,000
Cumpute the equivalent annual worth of each project at i=13% determine the acceptability of each project.
AE(13%)a= -$4,300(A/P,13%,3)+$5,000(A/F,13%,3)= -$206.8 Not accept
AE(13%)b= – $3,500(A/P,13%,3)+$1,500+$300((A/G,13%,3)=$293. Accept
AE(13%)c= – $5,500(A/P,13%,3)+$ 3,000 – $1,000(A/G,13%,3)= – $247.95 Not accept
AE(13%)d= – $3,800(A/P,13%,3)+$1,800=$ 190.7 Accept
Problem6.15
Susan is considring buying a 2011 Smart for Two costing $21,635 and finds that the retaining values of the vehical over
next five years are at follow :
Prercent of the total value retained after 36 months:28%.
If her interst rate is 6% compounded annually, what is the ownership cost of the vehicle over three years? Five years?
CR(6%)3 years= ($21,635 – $6,057.80)(A/P,6%,3)+(0.06)($6,057.80)=$6,191.05
CR(6%)5 years= ($21,635 – $3,677.95)(A/P,6%,5)+(0.06)($3,677.95)=$4,483.68
Problem6.16
Nelson Electronics, Inc., just purchased a soldering machineto be used in its assembly cell for flexible disk drives.
This machine costs $248,000. because of the specialized function it performs, its useful life is estimated to be five
years. At the end of the time, its salvage value is estimated to be $43,000. What is the capital cost for the investment in the firm’s interest rate is 18%?
CR(18%)=($248,000 – $43,000)(A/P,18%,5)+$43,000(0.18)=$73,299

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