Question

Machine X has a first cost of $70,000 and an operating cost of $21,000 in year 1, increasing by $500 per year through year 5 with a salvage value of $13,000. Machine Y has a first cost of $62,000 and an operating cost of $21,000 in year 1, increasing by 3% per year through year 10 with a salvage value of $2000. If the interest rate is i =19% per year, evaluate which machine must you choose on the basis of:

(a) the present worth analysis,

(b) the conventional B/C analysis

(Show me all the steps)

Answer 1

The assets have unequal lives, therefore, Repeated projects method or LCM method is used with LCM = 10 years

Machine X

Present worth of cost = 70000 + 70000(P/F,19,5) + [21000 +500(A/G,19,5)](P/A,19,10) + 13000(P/F,19,5) + 13000(P/F,19,10)

Using DCIF tables

Present worth of cost =  70000 + 70000(0.4190) + [21000 +500(1.6566)](4.3389) + 13000(0.4190) + 13000(0.1756)

Present worth of cost =  70000 + 29330 + 94710.8+ 5447 + 2282.8

Present worth of cost = $201770.6

B/C ratio = [13000(P/F,19,5) + 13000(P/F,19,10)] / {70000 + 70000(P/F,19,5) + [21000 +500(A/G,19,5)](P/A,19,10)}

B/C ratio = (5447 + 2282.8) / (70000+29330+94710.8)

B/C ratio = 0.04

Machine Y

Present worth of cost = 62000 + [21000 +630(A/G,19,10)](P/A,19,10) + 2000(P/F,19,10)

Using DCIF tables

Present worth of cost = 62000 + [21000 +630(3.1331)](4.3389) + 2000(0.1756)

Present worth of cost = 62000 + 99681.25 + 351.2

Present worth of cost = $162032.45

B/C ratio = 2000(0.1756) / 62000 + [21000 +630(3.1331)](4.3389)

B/C ratio = (351.2) / (62000 + 99681.25)

B/C ratio = 0.0022

Based on both the analysis Machine Y is to be selected