1. Lecture 6: Algorithm Approach to LP Soln AGEC 352 Fall 2012 – Sep 12 R. Keeney

2. Linear Programming Corner Point Identification ◦ Solution must occur at a corner point ◦ Solve for all corners and find the best solution What if there were many (thousands) of corner points? ◦ Want a way to intelligently identify candidate corner points and check when we have found the best Simplex Algorithm does this…

3. Assigned Reading 5 page handout posted on the class website ◦ Spreadsheet that goes with the handout Lecture today will point out the most important items from that handout

4. Problem Setup Let: C = corn production (measured in acres) B = soybean production (measured in acres) The decision maker has the following limited resources: 320 acres of land 20,000 dollars in cash 19,200 bushels of storage The decision maker wants to maximize profits and estimates the following per acre net returns: C = $60 per acre B = $90 per acre

5. Problem Setup (cont) The two crops the decision maker produces use limited resources at the following per acre rates: ResourceCornSoybeans Land11 Cash50100 Storage10040

6. Algebraic Form of Problem

7. Problem Setup in Simplex Note the correspondence between algebraic form and rows/columns

8. Simplex Procedure: Perform some algebra that is consistent with equation manipulation ◦ Multiply by a constant ◦ Add/subtract a value from both sides of an equation Goal: Each activity column to have one cell with a 1 and the rest of its cells with 0 Result: A solution to the LP can be read from the manipulated tableau

9. Simplex Steps The simplex conversion steps are as follows: 1)Identify the pivot column: the column with the most negative element in the objective row. 2)Identify the pivot cell in that column: the cell with the smallest RHS/column value. 3)Convert the pivot cell to a value of 1 by dividing the entire row by the coefficient in the pivot cell. 4)Convert all other elements of the pivot column to 0 by adding a multiple of the pivot row to that row.

10. Step 1 B has the most negative ‘obj’ coefficient ◦ Most profitable activity

11. Step 2 ID pivot cell: Divide RHS by elements in B column Most limiting resource identifcation 320/1 20K/100 19200/40

12. Step 3 Convert pivot cell to value of 1 (*1/100)

13. Step 4 Convert other elements of pivot col to 0, by multiplying the new cash row and adding to the other rows Multiplying factor for land row ◦ -1  (-1*1 + 1 = 0) Multiplying factor for stor row ◦ -40  (-40*1 + 40 = 0)

14. Example