Question

8.38 Four different machines were under consideration for materials flow improvement on a drug bottling line. An engineer performed the economic analysis to select the best machine, but some of his calcula- tions were removed from the report by a disgrun- tled employee. All machines are assumed to have a 10-year life. (a) Fill in the missing values in the comparison table. (b) Select the best machine at MARR = 18% per year, provided one must be chosen. 3 -72,000 ? -60,000 -98,000 - 70,000 -64,000 -61,000 -58,000 Machine Initial cost, S Annual cost, $ per year Annual savings, per year Overall ROR, 1*, % Machines compared Incremental investment, $ Incremental cash flow, $ per year ROR on increment, Ai .% +80,000 +80,000 18.6% ? 2 vs. 1 +80,000 23.1% 3 vs. 2 +82,000 20.8% 4 vs. 3 -16,000 ? -26,000 +6,000 35.7% +3,000 ? ? ?

I know how to solve some of these. My problem is when you have to solve for 'i' on the second machine via trial and error. How do you do this? Can you show how to solve machine two via trial and error? And for the ROR on Machine 3. How do you solve for i?

Answer 1

Showing trail and error method for finding i%

Let i% be the ROR of machine 2, then

-60000 + (80000 - 64000)*(P/A, i%,10) = 0

16000*(P/A, i%,10) = 60000

(P/A, i%,10) = 60000 / 16000 = 3.75

using trail and error method

When i = 20%, value of (P/A, i%,10) = 4.1925

When i = 22%, value of (P/A, i%,10) = 3.9232

When i = 23%, value of (P/A, i%,10) = 3.7993

When i = 24%, value of (P/A, i%,10) = 3.6818

using interpolation

i = 23% + (3.7993 - 3.75) / (3.7993 - 3.6818) *(24% - 23%)

i = 23% + 0.42%

i = 23.42%

Pls comment if you require further explanation