Question

1)

Consider the following investments:

Period                                                  Project Cash Flows

    n                                  A B C D
0                              $-8,700          $-7,400 $-8,100 $-9,100
1                              $2,600 $1,150    $2,400    $2,550
2                              $750     $2,250            $2,100 $3,000
3                             $3,000 $250              $1,500    $1,000
4                             $2,500 $400               $2,200    $2,500
    5 $2,200          $2,850            $2,000         $2,900
           
Compute the equivalent annual worth of each project at i=9%, and determine the acceptability of each project.

2)Y.I.W. Industries just purchased a new machine to speed up production on their assembly line. This machine costs $136,500. Because of the specialized function it performs, its useful life is expected to be six years. At the end of its useful life, its salvage value is estimated to be $17,300. What is the capital cost for this investment if the firm’s interest rate is 10%?

Answer 1

1. Compute the equivalent annual worth of each project at i=9%, and determine the acceptability of each project.

	A	B	C	D
0	$-8,700	$-7,400	$-8,100	$-9,100
1	$2,600	$1,150	$2,400	$2,550
2	$750	$2,250	$2,100	$3,000
3	$3,000	$250	$1,500	$1,000
4	$2,500	$400	$2,200	$2,500
5	$2,200	$2,850	$2,000	$2,900

Calculate the PW of each project and then calculate the AW

Project – A

PW = -8,700 + 2,600 (1+0.09) ^{– 1}+ 750 (1+0.09) ^{– 2}+ 3,000 (1+0.09) ^{– 3}+ 2,500 (1+0.09) ^{– 4}+ 2,200 (1+0.09) ^{– 5}

PW = -$165.96

AW = PW (A/P, 9%, 5)

AW = -$165.96 (0.2571) = -42.67

Project – B

PW = -7,400 + 1,150 (1+0.09) ^{– 1}+ 2,250 (1+0.09) ^{– 2}+ 250 (1+0.09) ^{– 3}+ 400 (1+0.09) ^{– 4}+ 2,850 (1+0.09) ^{– 5}

PW = -$2,122.45

AW = PW (A/P, 9%, 5)

AW = -$2,122.45 (0.2571) = -545.68

Project – C

PW = -8,100 + 2,400 (1+0.09) ^{– 1}+ 2,100 (1+0.09) ^{– 2}+ 1,500 (1+0.09) ^{– 3}+ 2,200 (1+0.09) ^{– 4}+ 2,000 (1+0.09) ^{– 5}

PW = $470.97

AW = PW (A/P, 9%, 5)

AW = $470.97 (0.2571) = 121

Project – D

PW = -9,100 + 2,550 (1+0.09) ^{– 1}+ 3,000 (1+0.09) ^{– 2}+ 1,000 (1+0.09) ^{– 3}+ 2,500 (1+0.09) ^{– 4}+ 2,900 (1+0.09) ^{– 5}

PW = $192.54

AW = PW (A/P, 9%, 5)

AW = $192.54 (0.2571) = 49.5

SELECT PROJECT –C, because it has positive and highest Annual Equivalent Cost.

2) Machine cost = 136,500

Salvage Value = 17,300

Life = 6 years

Interest rate = 10%

Calculate capital recovery cost

CRC = I (A/P, 10%, 6) – SV (A/F, 10%, 6)

CRC = 136,500 (0.2296) – 17,300 (0.1296) = 29,098.32

Alternatively, CRC = (I – SV) (A/P, 5%, 5) + SV*i

CRC = (136,500 – 17,300) (0.2296) + 17,300*0.10

CRC = 29,098.32