Question

Machines K and L are mutually exclusive and have the following investment and operating costs. Note that machine K lasts for only 2 years. Year: 0 2 3 $10,000 $1,100 $1,200 1,100 a. Calculate the equivalent annual annuity of each investment by using a dis- b. Now suppose you have an existing machine. You can keep it going for 1 more 12,000 1,200 $1,300 count rate of 10%, which machine is the better buy? year only, but it will cost $2,500 in repairs and $1,800 in operating costs. Is it worth replacing now with either K or L?

Answer 1

Given data year 0 1 2 3 K $ 10,000 $ 1,100 $ 1,200 - L $ 12,000 $ 1,100 $ 1,200 $ 1,800 Machine K lost = 2 year step 1: To calculate the PV of each of machine outlay Note: All the amounts are outlays and since will be considered as negative To used PV formula PV formula = [C1]/(1 + r) + [C2/(1 + r^2)] + + [C n/(1 + r^2)] PV (K) = -$ 11,900 PV (L) = -$ 14,968.40 The negative sign implies that this is the lumpsum outlay if K/L is chosen Annuity payment = PV/factor factor = {1 - (L + r) - n/r} Annuity (k) = -$ 6.857.13 Annuity (L) = -$ 6.019.02 This shows that machine l is better to buy hence annual payment s lesser the PV outlay with keeping the present machine will be = (2500 + 1800/1.1) = -$ 3.909 This means that it�s cheaper to retain the present machine hence