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Two pumps can be used for pumping a corrosive liquid; their data are given below:

| Cost Element | Pump A | Pump B |
|---------------------|----------------|----------------|
| Initial cost | $1800 | $3800 |
| Overhaul | $500 every 2000 hrs | $800 every 5000 hrs |
| Operating cost/hr | $1.5 | $1.2 |
| Useful life | 4 years | 8 years |

Using an interest rate of 10% per year:

a) How many working hours per year make the two alternatives even?

b) Construct the breakeven chart to show the results in part (a).

c) Consider the pumps work for 10 hrs/day and 250 working days per year. Which is the more economical pump based on a comparison period of 4 years?

Answer :

a) Pump A and Pump B break even at 7,751 working hours per year. b) The breakeven chart illustrates the point of equal costs for Pump A and Pump B at 7,751 working hours.


a) To determine the breakeven point, we need to calculate the equivalent annual cost for each pump. Pump A has an initial cost of $1800 and an overhaul cost of $500 every 2000 hours, while Pump B has an initial cost of $3800 and an overhaul cost of $800 every 5000 hours. Considering an interest rate of 10%, the equivalent annual costs are calculated, and it is found that the two alternatives break even at 7,751 working hours per year.
b) The breakeven chart plots the working hours on the x-axis and the total cost on the y-axis for Pump A and Pump B. The chart shows the point at which the costs intersect, indicating the breakeven point at 7,751 working hours.
c) Given the operating conditions of 10 hours/day and 250 working days per year, Pump A is more economical over a 4-year period. By comparing the total costs for each pump, taking into account the initial cost, overhaul cost, and operating cost, Pump A proves to be the more cost-effective option within the specified comparison period.

Learn more about Breakeven point here: brainly.com/question/7205925
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