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Lecture 6: Algorithm Approach to LP Soln

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?

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Lecture 6: Algorithm Approach to LP Soln

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  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: Resource Corn Soybeans Land 1 1 Cash 50 100 Storage 100 40

  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: • Identify the pivot column: the column with the most negative element in the objective row. • Identify the pivot cell in that column: the cell with the smallest RHS/column value. • Convert the pivot cell to a value of 1 by dividing the entire row by the coefficient in the pivot cell. • 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

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