A city hospital has an emergency power supply driven by a gasoline engine. The e

Question

A city hospital has an emergency power supply driven by a gasoline engine. The engine is 5 years old and requires a lot of maintenance to keep it in reliable condition. The engineer has made a study of replacement engines and has concluded that the best choice is between a K-400 gasoline engine or an E-450 diesel engine. The data for the decision are as shown.

K-400 E-450

First cost $12,000 $24,000

Estimated economic life 4 years 8 years

Estimated salvage value $1,000 $2,000

Annual fuel and maintenance $3,700 $2,800

Explanation / Answer

Here we need to find the equated uniform annual cost for both machines

For K-400, EUAC = 12000(A/P,10%,4)+3700-1000(A/F,10%,4)
=12000*[((0.1*(1+0.1)^4)/(((1+0.1)^4)-1))]+3700-1000*[0.1/(((1+0.1)^4)-1)]
=12000*0.3154+3700-1000*(0.2154)
=7484.8-215.4=7269.4

For E-450
EUAC = 24000(A/P,10%,8)+2800-2000(A/F,10%,8)
=24000*[((0.1*(1+0.1)^8)/(((1+0.1)^8)-1))]+2800-2000*[0.1/(((1+0.1)^8)-1)]
=24000*0.1874+2800-2000*(0.08744)
=7297.66-174.88 = 7122.72

Since the EUAC for E-450 is lesser than K-400 purchasing E-450 is a better option

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