# The first machine was \$240,000 three years ago. The operating and maintenance cost of a costs of this machine

The first machine was \$240,000 three years ago. The operating and maintenance cost of a costs of this machine along the past three years have been as follows.

Investment of \$240,000 is made at t = 0. The life of the machine is 3+6 = 9 years.

Economic life is the period which minimizes the annual equivalent (AE) costs.

AE(n=1) = (240,000-60,000)*(A/P,8%,1)+0.08*60,000+34984 = 234,184

AE (n=2) = (240,000-60,000)*(A/P,8%,2)+0.08*60,000+[34,984*(P/F,8%,1)+44,984*(P/F,8%,2)](A/P,8%,2)

= 145,530.15

Similarly using the above formula the AE for all other years are computed as shown below:

 Time Investment Operating cost Salvage value A/P P/F AE 0 240,000.00 1 34,984.00 60,000.00 1.0800 0.9259 234,184.00 2 44,984.00 60,000.00 0.5608 0.8573 145,530.15 3 54,984.00 60,000.00 0.3880 0.7938 119,117.46 4 65,000.00 60,000.00 0.3019 0.7350 108,172.89 5 75,000.00 60,000.00 0.2505 0.6806 103,336.55 6 85,000.00 60,000.00 0.2163 0.6302 101,491.31 7 95,000.00 60,000.00 0.1921 0.5835 101,301.76 8 105,000.00 60,000.00 0.1740 0.5403 102,100.72 9 115,000.00 60,000.00 0.1601 0.5002 103,518.07

As we can see from the above table that AE is minimum at 101,301.76 and this happens when t = 7.

Thus the economic life of machine is 7 years (from the time of purchase i.e from t=0.) Current point is end of 3 years so economic life is 4 years from the current point of end of 3 years.

Asked on May 20, 2017 in