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Homework answers / question archive / Claris Water Company makes and sells filters for water drinking fountains for the public

Claris Water Company makes and sells filters for water drinking fountains for the public

Economics

Claris Water Company makes and sells filters for water drinking fountains for the public. The filter sells for $50. Recently a make/buy analysis was done based on the need for new manufacturing equipment. The equipment first cost of $200,000 and $25,000 annual operation cost comprise the fixed cost, while Claris's variable cost is $20 per filter. The equipment has a 5-year life, no salvage value, and the MARR is 6% per year. The decision to make the filter was based on the breakeven point and the historical sales level of 5000 filters per year. a) Determine the breakeven point. b) An engineer at Claris learned that an outsourcing firm offered to make the filters for $30 each, but this offer was rejected by the president as entirely too expensive. Perform the breakeven analysis of the two options and determine if the "make" decision was correct. c) Develop and use the profit relations for both options to verify the preceding answers. d) Use a spreadsheet to verify the answers to parts (b) and (c) above by plotting the profit lines.

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Part a:

Filter sale price (A) $                50
Less: Variable cost per filter (B) $                20
Contribution per filter (C=A-B) $                30
   
Calculations of Fixed costs:  
Depreciation (200000 / 5) = D $        40,000
Annual operation cost (E ) $        25,000
Total Fixed Cost (F=D+E) $        65,000
Break-even Point in units per annum (F/C) 2166.666667
Break-even Point (rounded) 2167 units

Part b:

Break-even point of 'make' option = 2167 units (as found in part a. above)

Break-even point of 'buy' option = 0 units (as it would involved zero fixed cost)

Determination of better option = Simple break-even analysis suggests 'buy' option was better, but wait till we work out part c. as well.

Part c:

Analysis of 'making' 5000 units per year:  
Depreciation (200000 / 5) = D $            40,000
Annual operation cost (E ) $            25,000
Total Fixed Cost (F=D+E) $            65,000
Total Variable cost (G=5000 x $20) $        1,00,000
Total Cost of making (H=F+G) $        1,65,000
Sale value (I = 5000 x $50) $        2,50,000
Profit (J = I - H) $            85,000
   
Analysis of 'buying' 5000 units per year:  
Total Cost of buying (K = 5000 x $30) $        1,50,000
Sale value (I = 5000 x $50) $        2,50,000
Profit (L = I - K) $        1,00,000

Profit relation: 'buying' option fetches a profit of $100000 whereas 'making' option fetches only $85000. Our opinion in part b. above is confirmed here. 'Buy' option was more cost effective. Even if we carry out present value analysis @ 6% MARR, our point gets confirmed as below:

Net present value of the profit earned after 'making' 5000' units:

Year Fixed Cost Variable Cost @ $20 Total sale @ $50 Net cash flow PV @ 6% MARR
0 -200000     -200000 -200000
1 -25000 -100000 250000 125000 117925
2 -25000 -100000 250000 125000 111250
3 -25000 -100000 250000 125000 104952
4 -25000 -100000 250000 125000 99012
5 -25000 -100000 250000 125000 93407
          326545

Net present value of the profit earned after 'buying' 5000' units:

Year Cost@ $30 Sale@ $50 Net Cash Flow PV@ 6% MARR
1 -150000 250000 100000 94340
2 -150000 250000 100000 89000
3 -150000 250000 100000 83962
4 -150000 250000 100000 79209
5 -150000 250000 100000 74726
        421236

NPV of buying i.e. $ $421236 is more than NPV of making i.e. $326545. Hence even NPV analysis goes in favor of 'buying'.

Part d:

Year Profit from Making Profit from Buying
1 85000 100000
2 85000 100000
3 85000 100000
4 85000 100000
5 85000 100000

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