A project engineer is considering a new labelling style that can be produced by two alternative machines

NOTE: The problem comprises of two assets having unequal lives, in order to compare the same (if mutually exclusive) we need to use ether LCM method / Repeated projects method or study period method. Since, nowhere both the aspects of mutually exclusiveness and use of any method is mentioned the NPV and BCR are computed directly for the respective useful lives. Maintenance cost is also assumed to be annual.

Machine X

NPV = Present worth of benefits - Present worth of costs

NPV = 28500(P/A,12,10) + 16000(P/F,12,10) – [125000 + 4000(P/A,12,10)]

Using DCIF Tables

NPV = 28500(5.650) + 16000(0.3220) – (125000 + 4000(5.650))

NPV = RM 18577

BCR = Present worth of benefits / Present worth of costs

BCR = [28500(P/A,12,10) + 16000(P/F,12,10)] / [125000 + 4000(P/A,12,10)]

Using DCIF Tables

BCR = 28500(5.650) + 16000(0.3220) / (125000 + 4000(5.650))

BCR = 1.13

Machine Y

NPV = Present worth of benefits - Present worth of costs

NPV = 26500(P/A,12,7) + 13000(P/F,12,7) – [115000 + 2500(P/A,12,7)]

Using DCIF Tables

NPV = 26500(4.564) + 13000(0.4523) – [115000 + 2500(4.564)]

NPV = RM 415.9

BCR = Present worth of benefits / Present worth of costs

BCR = [26500(P/A,12,7) + 13000(P/F,12,7)] – [115000 + 2500(P/A,12,7)]

Using DCIF Tables

BCR = [26500(4.564) + 13000(0.4523)] / [115000 + 2500(4.564)]

BCR = 1.0

Conclusion: Based on the NPV approach Machine X can be chosen over Machine Y since it has higher net present value