Lecture 17 - Final Review 1

# Lecture 17 - Final Review 1 - 540: OPERATIONSRESEARCH...

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540:311 DETERMINISTIC MODELS IN OPERATIONS RESEARCH Lecture 17: Final Review 1 Prof. W. Art Chaovalitwongse
Topics Ch. 4: Simplex Method Ch. 5: [email protected] Analysis Graphical [email protected] Analysis LINDO [email protected] Analysis (including 100% rule) Ch. 6: Matrix [email protected] Simplex Method [email protected] Analysis Duality Ch. 8: Network Models Shortest Path Problem Maximum Flow Problem Minimum Spanning Tree Problem Min Cost Network Flow Ch. 9: Integer Programming Modeling Branch and Bound
5.1 – A Graphical Approach to [email protected] Analysis SensiDvity analysis is concerned with how changes in an LP s parameters affect the [email protected] [email protected] max z = 3 x 1 + 2 x 2 2 x 1 + x 2 100 (finishing constraint) x 1 + x 2 80 (carpentry constraint) x 1 40 (demand constraint) x 1 ,x 2 0 (sign restriction) Where: x 1 = number of soldiers produced each week x 2 = number of trains produced each week. Reconsider the Giapetto problem from Chapter 3 shown to the right:
5.1 – Change in Obj Fcn Coeﬃcient max z = 3x 1 + 2x 2 slope = -1.5 2 x 1 + x 2 100 (finishing constraint) x 1 + x 2 80 (carpentry constraint) x 1 40 (demand constraint) x 1 ,x 2 0 (sign restriction) x 1 = number of soldiers produced each week x 2 = number of trains produced each week. X1 X2 10 20 40 50 60 80 20 40 60 80 100 finishing constraint Slope = -2 carpentry constraint Slope = -1 demand constraint Feasible Region A D Isoprofit line z = 120 Slope = -3/2 C B The optimal solution for this LP was z = 180, x 1 = 20, x 2 = 60 ( point B ) z = 2x 1 + 2x 2 slope = -1 z = 1.5x 1 + 2x 2 slope = -0.67 z = 1.5x 1 + 2x 2 z = 4x 1 + 2x 2 slope = -2 z = 4x 1 + 2x 2 z = 5x 1 + 2x 2 slope = -2.5
5.1 – Change in RHS Determine whether a change in the rhs of a constraint will make the current basis no longer optimal . For example, let b 1 = number of available finishing hours. The current optimal solution (point B) is where the carpentry and finishing constraints are binding. If the value of b 1 is changed, then as long as where the carpentry and finishing constraints are binding , the optimal solution will still occur where the carpentry and finishing constraints intersect . !" !# 10 20 40 50 60 80 20 40 60 80 100 finishing constraint Slope = -2 carpentry constraint Slope = -1 demand constraint Feasible Region A D Isoprofit line z = 120 Slope = -3/2 C B
5.1 – A Graphical Approach to [email protected] Analysis max z = 3x 1 + 2x 2 slope = -1.5 2 x 1 + x 2 100 (finishing constraint) x 1 + x 2 80 (carpentry constraint) x 1 40 (demand constraint) x 1 ,x 2 0 (sign restriction) x 1 = number of soldiers produced each week x 2 = number of trains produced each week. X1 X2 10 20 40 50 60 80 20 40 60 80 100 finishing constraint Slope = -2 carpentry constraint Slope = -1 demand constraint Feasible Region A D Isoprofit line z = 120 Slope = -3/2 C B The optimal solution for this LP was z = 180, x 1 = 20, x 2 = 60 ( point B ) 2 x 1 + x 2 80 2 x 1 + x 2 40 2 x 1 + x 2 120 2 x 1 + x 2 80 2 x 1 + x 2 40 2 x 1 + x 2 120 2 x 1 + x 2 100 2 x 1 + x 2 140 2 x 1 + x 2 140
5.1 – Shadow Price To determine how a change in a constraint s rhs changes the

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