Question

Sandersen Meat Processors has asked its lead process engineer to evaluate two different types of conveyors for the beef cutting line. Type A has an initial cost of $90196 and a life of 4 years. Type B has an initial cost of $ 192418 and a life expectancy of 5 years. The annual operating cost (AOC) for type A is expected to be $ 9459, while the AOC for type B is expected to be $7180. If the salvage values are $7320,6 and $5069 for type A and type B, respectively, Use present worth analysis to select the better method. Assume that i= 10% per year.

Answer 1

Let us write down the given data

Particulars	Type A	Type B
Initial cost	$90196	$192418
AOC	$9459	$7180
Salvage Value	$7320	$5069
Life	4 years	5 years
Interest (i)	10%

Solution

For present worth analysis we need to use repeated projects method or LCM method

Type A

Present worth of cost = 90196 + 90196(P/F,10,4) + 90196(P/F,10,8) + 90196(P/F,10,12) + 90196(P/F,10,16) + 9459(P/A,10,20) – [7320(P/F,10,4) + 7320 (P/F,10,8) + 7320 (P/F,10,12) + 7320 (P/F,10,16) + 7320 (P/F,10,20)]

Using DCIF tables

Present worth of cost = 90196 + 90196(0.6830) + 90196(0.4665) + 90196(0.3186) + 90196(0.2176) + 9459(8.514) – (7320(0.6830) + 7320 (0.4665) + 7320 (0.3186) + 7320 (0.2176) + 7320 (0.1486))

Present worth of cost = $309346.25

Type A

Present worth of cost = 192418 + 192418(P/F,10,5) + 192418(P/F,10,10) +192418(P/F,10,15) + 7180(P/A,10,20) – [5069(P/F,10,5) + 5069(P/F,10,10) + 5069(P/F,10,15) + 5069(P/F,10,20)

Using DCIF tables

Present worth of cost = 192418 + 192418(0.6209) + 192418(0.3855) +192418(0.2394) + 7180(8.514) – (5069(0.6209) + 5069(0.3855) + 5069(0.2394) + 5069(0.1486))

Present worth of cost = $486194.65

Since the Present worth of cost for Type A is less the same may be selected