Operational Research ppt Chapter 4

1. Linear Programming III:
Duality and Sensitivity Analysis
in Linear Programming
5/4/2018 4:12 PM 1

2. Dual means two. This implies that LPP exists in “pairs”, so
that corresponding to every LPP there is another LPP.
Meaning that we can face the LP models in other ways.
Primal model: the original model is known as primal LP
model.
Dual model: the one which can be obtained from the primal
is dual.
In simplex method, solving one of the LPP (primal or dual)
gives us the solution of both models.

3. Rules of converting from Primal to Dual or Vice versa:
Objective function: Max
Min
Coeff. Of objective function
RHS of Constraints
Coeff of LHS of the constraints
Sign of the constraint
>=
<=
Objective function: Min
Max
Right-hand-side of the constraints (bj)
Coeff. Of objective function
LHS of the constraints but in transpose
Sign of the constraint
<=
>=
Max Z= C1X1 + C2X2 + ..+ CnXn
St:
a11X1 + a12X2+ ..+ a1nXn <= b1
a21X1 + a22X2 + ..+ a2nXn <= b2
an1X1 + an2X2 +..+annXn <=bm
X1, X2, Xn>=O
Min Z= b1Y1 + b2Y2 + ..+ bnYn
St:
a11Y1 + a21Y2+ ..+ am1Ym <= C1
a12Y1 + a22Y2 + ..+ am2Ym <= C2
a1nY1 + a2nY2+…+amnYn <=Cn
Y1, Y2, Yn>=O

4. Advantages of dual
 To check the primal problem manually
 When there are more number of constraints than
variables in the primal, go to dual to reduce the
iterations

5. Example
Obtain the dual of the following problem:
Max. Z = 30X1 + 40X2
ST 60X1 + 120X2 <= 1200 Raw Material
8X1 + 5X2 <= 600 Machine Hour
3X1 + 4X2 <= 500 Man Hour
X1, X2>=0
If we let Y1, Y2, Y3 to be cost of unit resources raw
material, machine hour, and man hour, the problem can
be solved as follows:

6. Min Z= 1200Y1 + 600Y2 + 500Y3
ST 60Y1 + 8Y2 + 3Y3 >=30
120Y1 + 5Y2 4Y3 >= 40
Y1,Y2,Y3>=0
If you have n decision variables & m constraints in primal,
You will have m decision variables & n constraints in dual.
Example: Obtain the dual of the LP model:
Max Z= 40X1 + 50X2
ST X1 + 2X2 <= 60
4X1 + 3X2 <= 100
X1,X2>=0
Min Z= 60Y1 + 100Y2
ST Y1 + 4Y2 >= 40
2Y1 + 3Y2 >= 50
Y1,Y2>=0

7. Comparing the optimal solutions of primal & dual
Max Z= 160X1 + 200X2
ST 2X1 + 4X2 <= 40
18X1 + 18X2 <= 216
24X1 + 12X2 <= 240
X1,X2>=0
Cj 160 200 0 0 0
CB Basis X1 X2 S1 S2 S3 Soln
160 X2 0 1 ½ -1/8 0 8
200 X1 1 0 -1/2 1/9 0 4
0 S3 0 0 0 -2 1 48
Zj 160 200 20 20/3 0 Z= 2,240
Cj-Zj 0 0 -20 -20/3 0
The final optimal iteration (Optimal solution):

8. Comparing the optimal solutions of primal & dual
Dual of the above Primal:
Min Z= 40Y1 + 216Y2 + 240Y3
ST 2Y1 + 18Y2 + 24Y3 >= 160
4Y1 + 18Y2 + 12Y3 >= 200
Y1,Y2, Y3>=0
Cj 160 200 0 0 0
CB Basis Y1 Y2 Y3 S1 S2 Soln
216 Y2 0 1 2 -1/9 1/18 20/3
40 Y1 1 0 -6 ½ -1/2 20
Zj 40 216 192 -4 -8 Z= 2,240
Cj-Zj 0 0 48 4 8
The final optimal iteration (Optimal solution):

9. Primal:
1. Cj-Zj row values of slacks Si
2. Cj-Zj row values of decision variables Xi
3. Zx value (Objective function value)
Dual:
1. Values of the decision variables Yi
2. Values of the slack variables Si
3. Zy value

10. Sensitivity Analysis
After obtaining the optimal solution of LPP, we may ask
how the solution would be affected if the parameters Cj,
bi, or aij of the problem change.
This question is answered by sensitivity analysis (post-
optimality analysis).
It would be advantageous if we could obtain this
information from the simplex tableau containing the
final solution without carrying out the whole exercise
again.

11. Change in Cj: Coefficient of the objective function
Max Z= 160X1 + 200X2
ST 2X1 + 4X2 <= 40
18X1 + 18X2 <= 216
24X1 + 12X2 <= 240
X1,X2>=0
Cj 160 + ∆ 200 0 0 0
CB Basis X1 X2 S1 S2 S3 Soln
200 X2 0 1 ½ -1/18 0 8
160 + ∆ X1 1 0 -1/2 1/9 0 4
0 S3 0 0 0 -2 1 48
Zj 160 + ∆ 200 20 -∆/2 20/3+∆/9 0 Z= 2,240
Cj-Zj 0 0 -20 -20/3 0
New
Cj-Zj
0 0 -20 +∆/2 -20/3-∆/9 0
The final optimal iteration (Optimal solution):

12. To remain Optimal, all Cj-Zj<=0
Therefore,
-20 + ∆/2<= 0 and -20/3 - ∆/9 <= 0
∆/2 <= 20
∆ <= 40 (1)
-∆/9 <= 20/3
-∆ <= 180/3
-∆ <= 60
∆ >= -60 (2)Then:
C1 = 160 + ∆
C1 – 160 = ∆ <= 40
C1 -160 <= 40
C1 <= 160 + 40
C1 <= 200 (1)
C1 = 160 + ∆
C1 – 160 = ∆ >= -60
C1 -160 >= -60
C1 >= 160 - 60
C1 >= 100 (2)
100<= C1 <= 200

13. Cj 160 200+ ∆ 0 0 0
CB Basis X1 X2 S1 S2 S3 Soln
200+ ∆ X2 0 1 ½ -1/18 0 8
160 X1 1 0 -1/2 1/9 0 4
0 S3 0 0 0 -2 1 48
Zj 160 200+ ∆ 20 + ∆/2 20/3-∆/18 0 Z= 2,240
Cj-Zj 0 0 -20 -20/3 0
New
Cj-Zj
0 0 -20 -∆/2 -20/3+∆/18 0
Also for C2 = 200 + ∆
160<= C2 <= 320

14. Change in bi
Cj 160 200 0 0 0
CB Basis X1 X2 S1 S2 S3 Soln
160 X2 0 1 ½ -1/8 0 8
200 X1 1 0 -1/2 1/9 0 4
0 S3 0 0 6 -2 1 48
Zj 160 200 20 20/3 0 Z= 2,240
Cj-Zj 0 0 -20 -20/3 0
B1 = 40 + ∆ b2 = 216 b3 = 240
B1
B2
Bm
+ ∆ *
a1j
a2j
amj
>=
0
0
0
Current
Solution
Coeff. in the
corresponding
slack variables