HW 4 Solution - #2

HW 4 Solution - #2 - 2 ½ =-$8.39 Average Annual Cost for Q...

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CSA 273 Solution to Homework Set #4 Problem #1 Calculations done on spreadsheet. IRR=16% Problem #2 Let Q = number of batteries purchased per order (a) Total Revenue per year = 8000 batteries/yr * $25/battery = $200,000 (b) Purchase Cost per year = 8000 batteries/yr * $20/battery = $160,000 Order cost per order = $100 + $ .10Q Number of orders per year = 8000/Q Order Cost per year = ($100 + $ .10Q)* 8000/Q = $800,000/Q + $800 Holding Cost per year = Annual holding cost per unit * Average inventory level = 1*Q/2 = Q/2 TC(Q) = Annual Purchase Cost + Annual Order Cost + Annual Holding Cost = $160,000 + $800,000/Q + $800 + Q/2 = $160,800 + $800,000/Q + Q/2 (c) Marginal Cost per year = TC'(Q) = -800,000/Q 2 + ½ Marginal Cost of 300 units = TC'(300)= -800,000/(300)
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Unformatted text preview: 2 + ½ = -$8.39 Average Annual Cost for Q =300 units = TC(300) = $160 ,800+$800,000/300 + $300/2 = $163,616.67 If we wish to find the average annual cost per battery, average annual unit cost = $163,616.67/8000 = $20.45 (d) TP(Q) = Total Revenue per year – TC(Q) = $200,000-($160,800 + $800,000/Q + Q/2) = $39,200 - $800,000/Q - Q/2 TP'(Q) = 800,000/Q 2- ½ Setting TP'(Q) = 0, we get 800,000/Q 2- ½ = 0 or Q = 1265 batteries as the positive root. Since TP''(Q) = -1,600,000/Q 3 < 0 for all positive Q, an order size of Q=1265 will maximize total profit. Problem #3 Calculations were done on spreadsheet. Within an interval length of .6, the optimal value to two decimal places is -1.00....
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This note was uploaded on 04/17/2008 for the course CSA 273 taught by Professor Patton during the Spring '08 term at Miami University.

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