Payback Period¶
Formula:¶
Some Practice Questions for Payback Period with Answers.
Question
Question:¶
A Firm is considering purchase two machinery, P and Q, given the following cash flows and an initial investment of ₹200,000 for each.Calculate it by payback period?
Cash Flows:
Year | Machine P | Machine Q |
---|---|---|
1 | 60,000 | 20,000 |
2 | 80,000 | 60,000 |
3 | 100,000 | 80,000 |
4 | 60,000 | 120,000 |
5 | 40,000 | 80,000 |
Answer:¶
Steps for Calculating Payback Period:¶
- Calculate cumulative cash flow for each year until it equals or exceeds the initial investment.
- Identify the year in which the cumulative cash flow first exceeds the initial investment.
- Calculate the fraction of the year needed to pay back the remainder of the investment.
Steps for Machine P:¶
Formula:
Calculation:
-
Year 1: 60,000
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Year 2: 60,000 + 80,000 = 140,000
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Year 3: 140,000 + 100,000 = 240,000
At the start of Year 3, 60,000 is needed to recover the full investment.
Steps for Machine Q:¶
Calculation:
-
Year 1: 20,000
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Year 2: 20,000 + 60,000 = 80,000
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Year 3: 80,000 + 80,000 = 160,000
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Year 4: 160,000 + 120,000 = 280,000
At the start of Year 4, 40,000 is needed to recover the full investment.
Conclusion:¶
-
The payback period for Machine P is approximately 2.6 years.
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The payback period for Machine Q is approximately 3.33 years.
so Machine P is better than Machine Q which has less payback period for the same investment.
Question
Question:¶
Neeraj Books Ltd. is considering the purchase of a new machine. There are two alternative models X and Y. Prepare a statement of profitability showing the payback period from the following information:
Particulars | Machine X | Machine Y |
---|---|---|
Cost of machine | 1,80,000 | 3,00,000 |
Estimated life in Years | 10 yrs | 15 yrs |
Estimated savings in scrap per year | 12,000 | 15,000 |
Additional cost of supervision per year | 14,400 | 19,200 |
Additional cost of maintenance per year | 8,400 | 13,200 |
Cost of indirect material per year | 7,200 | 9,600 |
Estimated savings in wages: | ||
Workers not required | 150 | 200 |
Wages per worker per year | 720 | 720 |
Assume tax at 50% of profit.
Which model would you recommend?
Answer:¶
Particulars | Machine X | Machine Y |
---|---|---|
Estimated Savings in scrap | 12,000 | 18,000 |
Estimated Savings in wages (720 * Workers) | 1,08,000 | 1,44,000 |
Total savings (A) | 1,20,000 | 1,52,000 |
Estimated cost of Additional materials | 7,200 | 9,600 |
Cost of supervision | 14,400 | 19,200 |
Additional cost of Maintenance | 8,400 | 13,200 |
Total additional cost (B) | 30,000 | 42,000 |
Annual Cash inflow (A-B) | 90,000 | 1,20,000 |
Less: Depreciation (Cost/Life) | 18,000 | 20,000 |
Profit before tax | 72,000 | 100,000 |
Less: Tax (50% of profit) | 36,000 | 50,000 |
Profit after tax | 36,000 | 50,000 |
Add: Depreciation | 18,000 | 20,000 |
Profit after tax and depreciation (Annual Cash inflow) | 54,000 | 70,000 |
Payback period (Original Investment/Annual Cash inflow) | 180,000/54,000 | 300,000/70,000 |
Payback period | 3.33 years | 4.28 years |
Considering the profit after tax, Machine Y seems more profitable in the long run even though it has a higher initial cost.
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