In: Economics
Two equipments A and B have initial costs of $100,000 and $120,000 and expected to generate annual savings during the first year of $88,000 and $98,000 respectively. The value of these annual savings is expected to increase by 10% per year (over previous period). Assume service life of 2 years, operating hours per year of 4500, Use the NPW method to determining 5 savings/ hour for each equipment. Select the optimal equipment based on your results.
Answer: | |||||
Particulars | Equipment A | Equipment B | |||
Initial costs (A) | 100,000.00 | 120,000.00 | |||
PV of Annual savings (B) | 160,000.0 | 178,181.7 | |||
Year 1 | 80,000.01 | 89,090.82 | |||
88,000*PVIF(10%,1) | 98,000*PVIF(10%,1) | ||||
Year 2 | 79,999.97 | 89,090.88 | |||
96,800* PVIF(10%,2) | 1,07,800* PVIF(10%,2) | ||||
NPW (B-A) | 60,000.0 | 58,181.7 | |||
Operating hours per year | 4,500.00 | 4,500.00 | |||
Operating hours for 2 years | 9,000.00 | 9,000.00 | |||
Savings or NPW per operating hour | 6.67 | 6.46 | |||
(60,000/9,000) | (58,181.7/9,000) | ||||
NOTE: | Since, the discount rate is not given in the question, | ||||
MARR IS ASSUMED AT 10% PER ANNUM. | |||||
Based on the calculations made above, Equipment A should be selected as its savings/hour is | |||||
comparatively higher than equipment B. | |||||
Please please upvote. | |||||
Thanks in advance. | |||||