Aerotron Electronics is considering purchasing a water filtration system to assist in circuit board manufacturing. The system costs $32,000. It has an expected life of 7 years at which time its salvage value will be $5,000. Operating and maintenance expenses are estimated to be $2,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $13,000 per year for water purification. If the system is purchased, no water purification from Bay City will be needed. Aerotron Electronics must borrow 1/2 of the purchase price, but they cannot start repaying the loan for 2 years. The bank has agreed to 3 equal annual payments, with the 1st payment due at the end of year 2. The loan interest rate is 8.5 % compounded annually. Aerotron electronics’ MARR is 12.5 % compounded annually.

Required:

a. What is the present worth of this investment? (Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is £10.)
b. What is the decision rule for judging the attractiveness of investments based on present worth?
c. Should Aerotron Electronics buy the water filtration system?

Explanation:

a) Present worth of the system:

First step :

Calculation of bank installment:

We are given:

·nitial costs = $32,000

Borrow amount = ½ of purchase price

Payment of borrow amount = EOY 2 to EOY 4 (3 Equal installments)

Bank loan interest = 8.5% = 8.5/100 = 0.085

Assume the installment amount is F. They will be paid at end of year 2 to end of year 4. Their present value must be equal to borrow amount.

Present value of cost to be incurred in future can be calculated by below formula:

F P (1 + i)

F= Future cost

i = Rate of interest

n = time (in years)

Therefore,

F F $32,000 2 F (1 +0.085)2 (1 +0.085) 3 (1 +0.085) + +

.: 2.35394F = 16,000 or F = 16,000 2.35394 - $6,797.11

Step 2: Present worth of the system:

Given Data:

Initial costs = $32,000

Expected life = 7 years

Salvage value = $5,000

O&M Costs = $2,000 per year

MARR = 12.5%

= 12.5/100

= 0.125

Present value of uniform recurring payments is given by below formula:

P=A (1 + i) - 1 i(1+i)n 72

Where,

P = Present Value

A = Recurring payments per annum

i = rate of interest

n = time (in years)

Hence present value of O&M costs,

P1 = -2,000 x (1 + 0.125) 7-1 0.125 x (1 + 0.125) 7 -$8,984.60

Present worth of the system calculated in below table:

Description

F ($)

MARR (i)

per year

n (years)

P ($)

Initial investment (1/2 of purchase price)

-16,000.00

0.125

0

-16,000.00

Bank installment EOY2

-6,797.11

0.125

2

-5,370.56

Bank installment EOY3

-6,797.11

0.125

3

-4,773.83

Bank installment EOY4

-6,797.11

0.125

4

-4,243.40

Salvage value

5,000.00

0.125

7

2,192.31

O&M Costs

-8,984.60

0.125

0

-8,984.60

The current worth of the new system

-37,180.07

Part b) Decision rule of judgment:

Assuming current value of costs is lower than current value of benefit, an alternative is known to be economic to use based on current worth analysis.

Part c) Decision for the water filtration system:

Given Data:

· Annual savings from filtration system (A) = $13,000 per year

· Expected life (n) = 7 years

· MARR = 12.5%

= 12.5/100

= 0.125

Present value of benefits = 13,000 X (1 + 0.125)7 - 1 0.125 x (1 + 0.125) 7 $58,399.91

Since current value of costs is less than current value of benefit, this is an worthwhile system and has to be be purchased.

answer:

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