Lecture_14_Linear programming: sensitivity analysis

(A) Change Affecting optimality:
The current solution will be not optimal only if the coefficients of the objective equation violate the optimality condition.
(1) Changes in the objective function:

1- If the changes in the objective function involve the coefficients of a current basic variable, determine the new dual values and then use them to recompute the new Z- equation coefficients.
2- If the changes involve non basic variables only, use the current dual values (directly from the current tableau) and recomputed the Z-equation coefficients of the involved non basic variables only. No other changes will occur in the tableau.
Example(3): Suppose that the objective function in example (1) is changed from Z=3x1 +2x2 to Z=5x1 +4x2


the primal model:

Max Z=5x1+4x2
s.t x1+2x2?6
2x1+2x2?8
-x1+x2?1
x2?2
x1,x2 ?0


The dual model:
Min W=6y1+8y2+y3+2y4
s.t y1+2y2 –y3?5
2y1+y2+y3+y4?4
y1 ,y2 ,y3 ,y4?0






1


The changes involve both x1 and x2 which happen to be basic in the current solution. Thus we must determine the new dual values. Notice that the order of the basic variables in the current tableau is(x2,x1,S3,S4)









(y1 ,y2 ,y3,y4)=(4,5,0,0)* =(1,2,0,0)





The next step is to recompute the Z-equation coefficients by taking the difference between the left and right sides of the dual constraints.
x1 – coefficient =y1+2y2–y3-5
=1(1) +2(2)-0-5=0
x2 – coefficient =2y1+y2+y3+y4-4
=2(1) +2+0+0-4=0
s1 – coefficient =y1- 0 =1-0 =1
s2 – coefficient =y2-0 =2-0 =2
s3 – coefficient =y3-0 =0-0 =0
s4 – coefficient =y4-0 =0-0 =0
??Z-Cj >= 0, the changes indicated in the objective function will not change the optimum variables or their values. The only change is the value of Z=5*(10/3)+4*(4/3)=22 ,where(x1=10/3,x2=4/3 in the primal optimal tableau in example (1) ).