Linear Programming Prob.s 13 Must convert to standard form LP Problem!
Transforming LP to Std Form LP 1.If Max, then f(x) = - F(x) 2.If x is unrestricted, split into x + and x -, and substitute into f(x) and all g i (x) and r enumber all x i 3.If b i < 0, then multiply constraint by (-1) 4.If constraint is ≤, then add slack s i 5.If constraint is ≥, then subtract surplus s i 14
Std Form LP Problem 15 Matrix form All “ ≥0 ” i.e. non-neg. All “=“
Terms basic solutions - solutions created by setting (n-m) variables to zero basic feasible sol’ns - sol’ns @ vertices of feasibility polygon feasible solution - any solution in S polyhedron basic variables - dependent variables, not set to zero non-basic variables - independent variables, set to zero, i.e. not in basis. basis – identity columns of the coefficient matrix A 18
Method? 1.Set up LP prob in “tableau” 2.Select variable to leave basis 3.Select variable to enter basis (replace the one that is leaving) 4.Use Gauss-Jordan elimination to form identity sub-matrix, (i.e. new basis, identity columns) 5.Repeat steps 2-4 until opt sol’n is found! 19
Can we be efficient? Are we at the min? If not which non-basic variable should be brought into basis? Which basic variable should be removed to make room for the new one coming on? SIMPLEX METHOD! 20
Simplex Method – Part 1 of 2 Single Phase Simplex Method When the Standard form LP Problem has only ≤ inequalties…. i.e. only slack variables, we can solve using the Single-Phase Simplex Method! If surplus variables exist… we need the Two-Phase Simplex Method –with artificial variables… Sec 8.6 (after Spring Break) 21
Single-Phase Simplex Method 1. Set up LP prob in a SIMPLEX tableau add row for reduced cost, c j ’ and column for min-ratio, b/a label the rows (using letters) of each tableau 2. Check if optimum, all non-basic c’ ≥0? 3. Select variable to enter basis(from non-basic) Largest negative reduced cost coefficient/ pivot column 4. Select variable to leave basis Use min ratio column / pivot row 5. Use Gauss-Jordan elimination on rows to form new basis, i.e. identity columns 6.Repeat steps 2-5 until opt solution is found! 22
23 Figure 8.3 Graphical solution to the LP problem Example 8.7. Optimum solution along line C–D. z*=4. Ex 8.7 1 phase Simp Meth All constraints are “slack” type Therefore, can use single-phase Simplex Method
Step 1. Set up Simplex Tableau 24 Simplex Tableau rowbasicx1x2x3x4x5bb/a_pivot ax34310012 bx4210104 cx5120014 dc'-20000 Step 2. check if optimum? X1 and x2 are <0! Continue!
Step3 & 4 25 3. Select variable to enter basis(from non-basic) Largest negative reduced cost coefficient/ pivot column 4. Select variable to leave basis Use min ratio column / pivot row
Why use Min Ratio Rule? 26 We want to add x1 into basis, i.e. no longer is x1=0 How much of x1 can we add? Whoops!!!!
Step 5 Use Gauss-Jordan form new basis, i.e. identity columns 27 Step 6. Repeat steps 2-5. Step 2. Check if optimal? Since all c’ ≥ 0… We have found the optimal solution!
Use Excel to help with arithmetic? 28 See Excel spreadsheet on website simplexEx8_7.xls
Summary Need to transform into Std LP format Unrestricted, slack, surplus variables, min = - Max Opt solution is on a vertex Simplex Method moves efficiently from one feasible combination of basic variables to another. Use Single-Phase Simplex Method when only “slack” type constraints. Use Excel to assist w/arithmetic 29