Depletion Engineering Economy

Question

An engineer sold his machine for $17,000$, after using it for 5 years. He bought a new machine worth $75,000$ with an expected life of 12 years, and a salvage value of $2,000$. The operating cost is $5,500$ per year. The old machine which he bought for $50,000$ when new will be useful for 10 years with a junk value of $1,000$, but because of appropriate preventive maintenance it will be useful for another 5 years with an annual operating cost of twice the new one. If the money is worth 12%, was the engineer justified in selling the old machine? Use straight-line depreciation.

Here is my Solution

For Old

Life is 10 years so using straight line where D= (Cost - Salvage value) / Life

D= $\frac{50,000-1000}{10}$=$4900$ and since after 10 years if the machine was given a maintenance it will last for another $5$ years which is twice the cost of operating cost of the first one so $(2)(5500)$$5$years = $55,000$ a total of $59,900$

for the new one using again the formula I get $6083.33$ and add the operating cost which is $5500(12)$years = $66,000$

I'm not really sure if I'm getting somewhere here, also what is the $12$% for?

Answer 1

For the new machine, consider the following:

  PERIOD  COST                           UPDATED COST
   0     -75000   `=-75000/(1,12)^0        -75000,00
   1      -5500   `= -5500/(1,12)^1         -4910,71
   2      -5500   `= -5500/(1,12)^2         -4384,57
   3      -5500   `= -5500/(1,12)^3         -3914,79
   4      -5500   `= -5500/(1,12)^4         -3495,35
   5      -5500   `= -5500/(1,12)^5         -3120,85
   6      -5500   `= -5500/(1,12)^6         -2786,47
   7      -5500   `= -5500/(1,12)^7         -2487,92
   8      -5500   `= -5500/(1,12)^8         -2221,36
   9      -5500   `= -5500/(1,12)^9         -1983,36
   10     -3500   `= -3500/(1,12)^10        -1126,91
                                          -105432,28

So, after 10 years his updated costs at time zero are -105432,28 monetary units.

Now, do the same for the other machine and compare values.