Question

Engineering economy

5. Beckman Technologies, plans to purchase a precision laboratory equipment. Two manufacturers offered the estimates shown below: a. Draw the cash flow diagram for both vendors. (4 pts) b. Determine which vendor would be selected based on PW if the MARR is 10% per year (8 pts) C. This research lab has a standard practice of evaluating all options over a 3 year period. If a study period of 3 years is used and the salvage values are not expected to change, which vendor would be selected? (6 pts) Manufacturer A Manufacturer B First Cost, $ -15,000 -18,000 Annual Operating Costs (AOC), S/year -3,500 -3,100 Salvage Value S, $ 1,000 2,000 Life, Years 8

Answer 1

We have the following information

	Manufacturer A	Manufacturer B
First Cost ($)	– 15,000	– 18,000
Annual Operating Costs ($)	– 3,500	– 3,100
Salvage Value ($)	1,000	2,000
Life (n) in years	4	8
MARR (i)	10% or 0.1	10% or 0.1

We will be using Present Worth (PW) Method

Manufacturer A

PW(10%) = – First Cost – Annual Cost(P/A, i, n) + Salvage Value(P/F, i, n)

PW(10%) = – 15,000 – 3,500(P/A, 10%, 4) + 1,000(P/F, 10%, 4)

PW(10%) = – 15,000 – 3,500[((1+0.1)⁴ – 1)/0.1(1+0.1)⁴] + 1,000/(1 + 0.1)⁴

PW(10%) = – 15,000 – 11,094.53 + 683.01

PW(10%) of Manufacturer A = – ($25,411.52)

Manufacturer B

PW(10%) = – First Cost – Annual Cost(P/A, i, n) + Salvage Value(P/F, i, n)

PW(10%) = – 18,000 – 3,100(P/A, 10%, 8) + 2,000(P/F, 10%, 8)

PW(10%) = – 18,000 – 3,100[((1+0.1)⁸ – 1)/0.1(1+0.1)⁸] + 2,000/(1 + 0.1)⁸

PW(10%) = – 18,000 – 16,538.27 + 933.01

PW(10%) of Manufacturer B = – ($33,605.26)

Based on the PW, Manufacturer A should be selected as its PW is higher compared to the PW of Manufacturer B.

A Manufacturer 1000 1 2 . mt 14 3500 3500 3500 3500 15,000

Now it is given that

	Manufacturer A	Manufacturer B
First Cost ($)	– 15,000	– 18,000
Annual Operating Costs ($)	– 3,500	– 3,100
Salvage Value ($)	1,000	2,000
Life (n) in years	3	3
MARR (i)	10% or 0.1	10% or 0.1

Manufacturer A

PW(10%) = – First Cost – Annual Cost(P/A, i, n) + Salvage Value(P/F, i, n)

PW(10%) = – 15,000 – 3,500(P/A, 10%, 3) + 1,000(P/F, 10%, 3)

PW(10%) = – 15,000 – 3,500[((1+0.1)³ – 1)/0.1(1+0.1)³] + 1,000/(1 + 0.1)³

PW(10%) = – 15,000 – 8,703.98 + 751.31

PW(10%) of Manufacturer A = – ($22,952.67)

Manufacturer B

PW(10%) = – First Cost – Annual Cost(P/A, i, n) + Salvage Value(P/F, i, n)

PW(10%) = – 18,000 – 3,100(P/A, 10%, 3) + 2,000(P/F, 10%, 3)

PW(10%) = – 18,000 – 3,100[((1+0.1)³ – 1)/0.1(1+0.1)³] + 2,000/(1 + 0.1)³

PW(10%) = – 18,000 – 7,709.24 + 1,502.63

PW(10%) of Manufacturer B = – ($24,206.61)

Based on the PW, Manufacturer A should be selected as its PW is higher compared to the PW of Manufacturer B.