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Lecture 6: Algorithm Approach to LP Soln AGEC 352 Fall 2012 – Sep 12 R. Keeney 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… 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 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 Problem Setup (cont) The two crops the decision maker produces use limited resources at the following per acre rates: Resource Land Cash Storage Corn 1 50 100 Soybeans 1 100 40 Algebraic Form of Problem max 60 C 90 B s .t . land : C B 320 cash : 50 C 100 B 20 , 000 storage : 100 C 40 B 19 , 200 non neg : C 0 ; B 0 Problem Setup in Simplex Note the correspondence between algebraic form and rows/columns Initial Tableau C land cash stor obj B 1 50 100 -60 s1 1 100 40 -90 s2 1 0 0 0 s3 0 1 0 0 P 0 0 1 0 RHS 0 320 0 20000 0 19200 1 0 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 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. Step 1 B has the most negative ‘obj’ coefficient ◦ Most profitable activity Initial Tableau C land cash stor obj B 1 50 100 -60 s1 1 100 40 -90 s2 1 0 0 0 s3 0 1 0 0 P 0 0 1 0 RHS 0 320 0 20000 0 19200 1 0 Step 2 ID pivot cell: Divide RHS by elements in B column Most limiting resource identifcation Initial Tableau C land cash stor obj B 1 50 100 -60 s1 1 100 40 -90 s2 1 0 0 0 s3 0 1 0 0 P 0 0 1 0 RHS 0 320 0 20000 0 19200 1 0 320/1 20K/100 19200/40 Step 3 Convert pivot cell to value of 1 (*1/100) Initial Tableau C land cash stor obj B 1 50 100 -60 C new cash row s1 1 100 40 -90 B 0.5 s2 1 0 0 0 s1 1 s3 0 1 0 0 s2 0 P 0 0 1 0 s3 0.01 RHS 0 320 0 20000 0 19200 1 0 P 0 RHS 0 200 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) Example