In: Accounting
A certain dealer owns a warehouse that can store a maximum of 600 units of a given commodity. He has on hand 100 units of the commodity and knows the buying and selling price for the next 5 time periods. They are:
PERIOD |
BUYING PRICE |
SELLING PRICE |
1 |
15 |
16 |
2 |
17 |
17 |
3 |
21 |
20 |
4 |
19 |
19 |
5 |
16 |
20 |
He is permitted to sell up to the amount that he has in his warehouse at the beginning of each period. If he buys in a given period, delivery will be made so that it is in storage at the beginning of the next period. Let (Xi, Yi) denote the amount he buys and sells in time-period i respectively. What is the optimal buying\selling strategy.
Formulate the linear programming problem.
Develop a spreadsheet and solve the problem.
See the table below for solution :
As per working given below situation 2 i.e Buying price 17 & Selling price 17 is the best resuting in profit.
Period |
Opening Stock |
Purchase price of Opening Stock |
Total Purchase Cost of opening stock |
Additional Purchase Qty |
Purchase Price of Additional purchase qty |
Total Purchase Cost of additional qty purchased |
Total Selling Price (Restricted to purchase cost of opening stock) |
Selling Price |
Sale Qty |
Profit (Selling price - Buying Price) * Qty sold |
Closing Stock in Qty |
Closing Stock in Value |
1 |
100 |
15 |
1500 |
500 |
15 |
7500 |
1488 |
16 |
93 |
(16-15)*93 = 93 |
507 |
7*15 = 105 500*15 = 7500 |
2 |
507 |
15 |
7605 |
93 |
17 |
1581 |
7599 |
17 |
447 |
(17-15)*500 = 894 |
153 |
60*15 = 900 93*17 = 1581 |
3 |
153 |
15 & 17 |
2481 |
447 |
21 |
9387 |
2480 |
20 |
124 |
(20-15)*60 + (20-17)*64 = 492 |
476 |
29*17 = 493 447*21 = 9387 |
4 |
476 |
17 & 21 |
9880 |
124 |
19 |
2356 |
9044 |
19 |
476 |
(19-17)*29 + (19-21)*457= -856 (loss) |
124 |
124*19 = 2356 |
5 |
124 |
19 |
2356 |
476 |
16 |
7616 |
2340 |
20 |
117 |
(20-19)*117 = 117 |
483 |
7*19 = 133 476*16 = 7616 |