Sections 4.5 and 4.6 Notes

Transcription

1 Sections 4.5 and 4.6 Notes

2 Transportation Problem: Delivering Bread A supermarket chain gets bread deliveries from a bakery chain that does its baking in different places. Each supermarket store needs a certain number of loaves/day. The supplier bakes enough bread to exactly meet the demands. How many loaves of bread should be shipped for each locale to each of the stores to stay within the demands and to minimize the cost?

3 Transportation Problem: Reshuffling Rental Cars After a long holiday weekend, a car rental company will have extra cars in some cities and too few cars in other cities It is faced with the problem of reshuffling the cars at minimal cost.

4 Transportation Problem Definition A group of suppliers must meet the needs of users of these supplies. There is a cost for shipping from a particular supplier to a particular user (demander). The transportation problem involves minimizing the total shipping cost of meeting the required demands from the supplies available.

5 Delivering Bread Example There are three stores who require 3 dozen, 7 dozen, and 1 dozen loaves of bread, respectively. There are three bakeries who can supply 8 dozen, 1 dozen, and 2 dozen loaves, respectively. On the next slide is a table which is a visual representation of this problem.

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7 The roman numerals (rows) will stand for the suppliers, in this case, the bakeries. The numbers (columns) will stand for the demanders, in this case, the stores. The numbers at the bottom are the demands, in this case, how many loaves of bread the stores need. The numbers on the right side are the supplies available, in this case, how many loaves of bread the bakeries can bake. These two sets of numbers are called the rim conditions.

8 The numbers in the upper-right-hand corner of each cell are the aggregate costs. These are the costs for getting the needed supplies from the supplier to the demander.

9 This table is called a Tableau. A tableau is a table showing costs and rim conditions for transportation problem.

10 A possible solution for the bread transportation problem The circled number 6 in row I, column 2, means we plan to ship six loaves of bread to store 2 from bakery I. These circled numbers must add up to the rim conditions across the rows and also down the column = 11

11 Once we have a possible solution we can figure out the cost of shipping these loaves of bread from the bakeries to the stores. For example, the cost of shipping 2 loaves of bread from Bakery I to Store 1 is 2(8) or $16. The total shipping cost is 2(8) + 6(9) + 1(15) + 1(3) + 1(5) = $93. We do not know if this is the optimal solution but there is a way to help find the optimal solution. It is called the Northwest Corner Rule.

12 Northwest Corner Rule (NCR) An easier approach that is a simple rule since it is based on the geometry of the table and does not look at the costs (initially).

13 Northwest Corner Rule Algorithm 1. Locate the top far-left-hand cell and ship this cell along with the costs. Ship via this cell with the smaller of the two rim cells (call the value s ) and put a circle around the entry in the tableau. 2. Cross out the row or column that had rim value s and reduce the other rim value for this cell by s. 3. When a single cell remains, there will be a tie for the rim conditions of both the row and column involved, and this amount is entered into the cell and circled.

14 1 2 3 I 8 II 1 III

15 Now that you have gotten the number 3 from the bottom of the 1 st column, you will no longer use that column.

16 2 3 I 85 3 = 5 II 1 III 2 7 1

17 Now that you have used the number 5 on the right side of 1 st row, you will no longer use this row.

18 2 3 II 1 III = 2 1

19 Now that you have used the number 1 on the right side of the row, you will no longer use this row.

20 2 3 III = 1 1

21 Now that you have used the number 1 on the bottom of the column, you will no longer use this column.

22 3 III 12 1 = 1 1 Now let s put the completed tableau together.

23 1 2 3 I 8 II 1 III

24 Total Cost = 3(8) + 5(9) + 1(1) + 1(3) + 1(5) = 78