Lecture 9 - Sensitivity

# Lecture 9 - Sensitivity - 540:311 DETERMINISTIC MODELS IN...

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540:311 DETERMINISTIC MODELS IN OPERATIONS RESEARCH Lecture 09: Chapter 5 Class Mee±ng: Mon Mar 7 th 10:20-11:40am Prof. W. Art Chaovalitwongse

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5.1 – A Graphical Approach to SensiDvity Analysis max z = 3X + 10Y Y = -3/10 X + z/3 slope = -0.3 4X + 10Y 40 Y -4/10 X + 40/4 slope = -0.4 X + Y 7 Y -1 X + 7 slope = -1 X,Y 0
2-D Graphical Representadon Y X Objective function: Z = 3X + 10Y ( slope = -10/3 ) Z = 4X + 10Y (slope = -2.5) Z = 3X+7.5Y Z = 3X + 10Y = 30 (0,3) (10,0) (0,1.5) (5,0) Z = 3X + 10Y = 15 Z = 3X + 10Y = 45 (0,4.5) (0,6) Z = 3X + 10Y = 60 Z = 4X + 10Y = 60 Z = 4X + 10Y = 40 (0,4) Z = 2.5X + 10Y (slope = -4) Z = 2.5X + 10Y = 25

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2-D Graphical Representadon Y X Z = 3X + 10Y ( slope = -10/3 ) Feasible Region X + Y 7 (slope = -1) 4X + 10Y 40 (slope = -2.5) X 3 (0,7) (7,0) (3,0) (3,2.8) (5,2) Z = 4X + 10Y ( slope = -2.5 ) Z = 5X + 10Y ( slope = -2 ) Z = 3X + 10Y = 30
5.1 – A Graphical Approach to SensiDvity Analysis SensiAvity analysis is concerned with how changes in an LP ` s parameters afect the opDmal soluDon. max z = 3x 1 + 2x 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:

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5.1 – A Graphical Approach to SensiDvity Analysis The optimal solution for this LP was z = 180, x 1 = 20, x 2 = 60 ( point B ) and it has x 1 , x 2 , and s 3 (the slack variable for the demand constraint. How would changes in the problem ` s objective function coefficients or right-hand side values change this optimal solution?
5.1 – A Graphical Approach to SensiDvity Analysis max z = 3x 1 + 2x 2 slope = -1.5 2 x 1 + x 2 100 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. 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

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5.1 – A Graphical Approach to SensiDvity Analysis Graphical analysis of the eFect of a change in an objecDve funcDon value for the GiapeNo LP shows: By inspection, we can see that making the slope of the isoprofit line more negative (tilt clockwise) than the finishing constraint (slope = -2) will cause the optimal point to switch from point B to point C . Likewise, making the slope of the isoprofit line less negative (tilt counter clockwise) than the carpentry constraint (slope = -1) will cause the optimal point to switch from point B to point A . Clearly, the slope of the isoprofit line must be between -2 and -1 for the current basis to remain optimal.
5.1 – A Graphical Approach to SensiDvity Analysis The values of the contribuDon to proFt for soldiers for which the current opDmal basis (x 1 ,x 2 ,s 3 ) will remain opDmal can be determined as follows: Let c 1 be the contribuDon (\$3 per soldier) to the proFt. ±or what values of c 1 does the current basis remain opDmal?

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Lecture 9 - Sensitivity - 540:311 DETERMINISTIC MODELS IN...

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