620-261
Solved Problems: Sensitivity Analysis
CHAPTER 8: SENSITIVITY ANALYSIS
1. Consider the LP problem whose first and final simplex tableaus are as follows:
B.V.
x5
x6
x7
Z
Eq. #
1
2
3
Z
x1
2
2
3
-8
x2
3
1
7
-5
x3
3
0
3
-1
x4
1
4
-2
-8
x5
1
0
0
0
x6
0
1
0
0
x7
0
0
1
0
b
8
10
12
0
B.V.
x1
x4
x7
Z
Eq. #
1
2
3
Z
x1
1
0
0
0
x2
11/6
-2/3
1/6
13/3
x3
2
-1
-5
7
x4
0
1
0
0
x5
2/3
-1/3
-8/3
8/3
x6
-1/6
1/3
7/6
4/3
x7
0
0
1
0
R.H.S
11/3
2/3
7/3
104/3
1.1
1.2
Determine the range of values of c2, c4, and b2 for which the final basis does not
change (one range at a time!).
What will be the new basis if you increase the value c4 just a bit above the upper
bound found in 2.2?
Solution:
1.1 Since the problem is in standard form, the optimal solution to the dual problem is
equal to the vector of reduced costs of the slack variables of the primal problem, namely
y*=(8/3,4/3,0). Hence, w*=y*b= 104/3.
1.2. If we change c2 to c2 + d, the reduced cost of x2 in the final tableau will be 13/3 - d.
The optimality condition is then 13/3 - d >= 0, hence d<=13/3. The range of c2 is thus c2
<= 5 + 13/3 =28/3.
If we change c4 to c4 + d, the reduced cost of x4 in the final tableau will be -d. Since x4
is a basic variable, we have to restore the canonical form of the tableau (by a pivot
operation). This is described by the following tables.
Before restoration of the canonical form:
B.V.
Eq. #
x1
x2
x3
x1
1
1
11/6
2
x4
2
0
-2/3
-1
x7
3
0
1/6
-5
Z
Z
0
13/3
7
x4
0
1
0
-d
x5
2/3
-1/3
-8/3
8/3
x6
-1/6
1/3
7/6
4/3
x7
0
0
1
0
R.H.S
11/3
2/3
7/3
104/3
After restoration of the canonical form:
B.V.
Eq. #
x1
x2
x1
1
1
11/6
x4
2
0
-2/3
x7
3
0
1/6
Z
Z
0
13/3-
x4
0
1
0
0
x5
2/3
-1/3
-8/3
8/3-
x6
-1/6
1/3
7/6
4/3+
x7
0
0
1
0
R.H.S
11/3
2/3
7/3
104/3 +
x3
2
-1
-5
7-d
Solved Problems: 8. 1
620-261
Solved Problems: Sensitivity Analysis
(2/3)d
(1/3)d
(1/3)d
(2/3)d
The optimality conditions are thus:
13/3 - (2/3)d >= 0, hence d <=13/2
7 - d >= 0, hence d<= 7
8/3 - (1/3)d >= 0, hence d <=8
4/3 + (1/3)d >= 0, hence d >= -4.
Therefore the range of appropriate values of d is [-4,13/2]. The range of c4 is thus
[4,29/2].
The modified final value of b is given by b" = B-1b', where b' is the modified value of the
initial RHS vector. Hence, in our case,
 2 / 3 1/ 6
b" 1/ 3 1/ 3

8 / 3 7/ 6
0 8 
010  d 
1

 12 

 2 / 3 1/ 6 0 8  2 / 3 1/ 6 0 0 
 1/ 3 1/ 3 0 10 1/ 3 1/ 3 0 d =

8/ 3 7 / 6 1 

12
 
8/ 3 7 / 6 1 

0 

11/ 3 (1/ 6)d  11/ 3  (1/ 6)d 
 2 / 3   (1/ 3)d   2 / 3  (1/ 3)d 

7 / 3 
 
 (7 / 6)d 
 
7/ 3  (7/ 6)d 

The feasibility conditions are therefore as follows:
11/3 - (1/6)d >= 0, hence d <= 22
2/3 + (1/3)d >= 0, hence d >= -2
7/3 + (7/6)d >= 0, hence d >= - 2.
Thus the feasible range of d is [-2, 22]. Therefore, the range of b2 is [8,32].
2.3. From the above analysis of changes to c4, we see that if we increase d above the
upper bound given above, the reduced cost of x2 will become negative (all other reduced
costs will remain non-negative). Hence, the new basic variable will be x2. The Ratio
Test for x2 requires x1 to leave the basis, hence the new basis will be and the new basis
will be (x2,x4,x7).
2 . Consider the LP problem whose first simplex tableau is as follows:
Solved Problems: 8. 2
620-261
Solved Problems: Sensitivity Analysis
B.V.
x5
x6
x7
Z
Eq. #
1
2
3
Z
x1
2
2
3
-8
x2
3
1
7
-5
x3
3
0
3
-1
x4
1
4
-2
-8
x5
1
0
0
0
x6
0
1
0
0
x7
0
0
1
0
b
8
10
12
0
x4
0
1
0
0
x5
2/3
-1/3
-8/3
8/3
x6
-1/6
1/3
7/6
4/3
x7
0
0
1
0
R.H.S
11/3
2/3
7/3
104/3
and whose final simplex tableau is as follows:
B.V.
x1
x4
x7
Z
2.1
2.2
Eq. #
1
2
3
Z
x1
1
0
0
0
x2
11/6
-2/3
1/6
13/3
x3
2
-1
-5
7
Determine the range of values of c1, c3, and b1 for which the final basis does not
change (one range at a time!).
What will be the new basis if you increase the value c3 just a bit above the upper
bound found in 2.1.
Solution.
2.1 Changes in C1: We add d to C1 and consider the following modified final
tableau:
B.V.
Eq. #
x1
x2
x3
x4
x5
x6
x7
R.H.S
x1
1
1
11/6
2
0
2/3
-1/6
0
11/3
x4
2
0
-2/3
-1
1
-1/3
1/3
0
2/3
x7
3
0
1/6
-5
0
-8/3
7/6
1
7/3
Z
Z
-d
13/3
7
0
8/3
4/3
0
104/3
We restore the canonical form of the tableau:
B.V.
x1
x4
x7
Z
Eq. #
1
2
3
Z
x1
1
0
0
x2
11/6
-2/3
1/6
13/3+11d/6
x3
2
-1
-5
7+2d
x4
0
1
0
0
x5
2/3
-1/3
-8/3
8/3+2d/3
x6
-1/6
1/3
7/6
4/3-d/6
x7
0
0
1
0
Hence the optimality conditions (opt=max) are (rj >=0):
26 + 11d >= 0 ; 7 + 2d >=0 ; 8 + 2d >= 0; 8 -d >= 0
d >= -26/11 ; d >= -7/2 ; d >= -4 ; d<= 8.
Hence d must be in the interval [-26/11, 8] and therefore c3 must be in the interval [8 –
26/11, 8+8] = [62/26,16].
Solved Problems: 8. 3
R.H.S
11/3
2/3
7/3
104/3
+11d/3
620-261
Solved Problems: Sensitivity Analysis
Changes in C3:
Since x3 is not in the basis, d<= reduced cost of x3, hence d<= 7. Hence the rage of c3
is c3<=1+7 =8.
Changes in b1. If we change b1 to b1 +d, the new final RHS would be
final new RHS
d 
= B b'  final old RHS + B 0 

0 

-1
-1
11/ 3
11/ 3 2 / 3  11/ 3  2d / 3
 2 / 3  dB1 1  2 / 3  d 1/ 3  2 / 3  d / 3 


7 / 3 

7 / 3 
 
8/ 3
 
7 / 3  8d / 3 

Thus, the non-negativity constraints are:
11 + 2d >= 0 ; 2 - d >= 0 ; 7 - 8d >=0.
Hence d >= -11/2 ; d <= 2 ; d <= 7/8, so that d must be in the interval [-11/2,7/8].
1.2 If c3 exceeds the upper bound implied above, then its reduced cost will become
negative, and the Greedy Rule will select it as the new basic variable. The Ratio
Test will then take x1 out of the basis. Thus the new basis will consist of
(x3,x4,x7).
3. Consider the LP problem whose first simplex tableau is as follows:
B.V.
x5
x6
x7
Z
Eq. #
1
2
3
Z
x1
2
2
23
-9
x2
1/3
1
7/3
-6
x3
1/3
0
4/3
-1
x4
1
4
-2
-9
x5
1
0
0
0
x6
0
1
0
0
x7
0
0
1
0
b
8
12
78
0
x4
-1
4
-10
14
x5
3
0
-4
3
x6
-1
1
-1
5
x7
0
0
1
0
R.H.S
12
12
34
84
and whose final simplex tableau is as follows:
B.V.
x3
x2
x7
Z
Eq. #
1
2
3
Z
x1
4
2
13
7
x2
0
1
0
0
x3
1
0
0
0
3.1 Determine the range of values of c1, c3, and b1 for which the final basis does not
change (one range at a time!).
3.2 What will be the new basis if you increase the value c1 just a bit above the upper
bound found in 3.1.
Solution:
Solved Problems: 8. 4
620-261
Solved Problems: Sensitivity Analysis
3.1 Determine the range of values of c1, c3, and b1 for which the final basis does not
change (one range at a time!).
Since x1 is not a basic variable, the optimality condition is 7->= 0, hence <=7. The
implication is therefore that for the current optimal basis will not change if c1 <= 9+7 =
16.
Since c3 is a basic variable, by changing c3 to c3 +  we face the following situation:
B.V.
x3
x2
x7
Z
Eq. #
1
2
3
Z
x1
4
2
13
7
x2
0
1
0
0
x3
1
0
0
-
x4
-1
4
-10
14
x5
3
0
-4
3
On restoring the canonical form of the x3 column we obtain:
B.V.
Eq. #
x1
x2
x3
x4
x5
x3
1
4
0
1
-1
3
x2
2
2
1
0
4
0
x7
3
13
0
0
-10
-4
Z
Z
0
0
7+4
14-
3+3
x6
-1
1
-1
5
x7
0
0
1
0
R.H.S
12
12
34
84
x6
-1
1
-1
5-
x7
0
0
1
0
R.H.S
12
12
34
84+12
Thus, the optimality condition (reduced costs are non-negative) yields:
7+4 >= 0 ; 14->=0 ; 3+3 >=0 ; 5- >=0
namely,
 >= -7/4
; 14 >=  ;  >= -1 ; 5 >= 
so that the critical range for d is [-1,5] and for c3 is [1-1, 1+5] =[0,6].
The impact of adding d to b1 is given by
RHS'  B 1 [b   e1 ]  B 1b  B1 e1  RHS   B 1 .1
12   3  12  3 
 12    0   12 

34
 
4
 
34  4  

Thus to insure non-negativity of the RHS, we need
12 + 3 >= 0
; 34 - 4 >= 0
Solved Problems: 8. 5
620-261
Solved Problems: Sensitivity Analysis
hence
 >= - 4 ;  <=17/2.
Consequently, the critical range for  is [-4,17/2] and for b1 is [12-4,12+17/2] =
[8,41/2].
3.2 What will be the new basis if you increase the value c1 just a bit above the upper
bound found in 1.1.
If x1 enters the basis, the ratio test will replace x7 by x1. Thus, the new basis will be
(x3,x2,x1).
4.( Consider the LP problem whose first and final simplex tableaus are as follows:
B.V.
x5
x6
x7
Z
Eq. #
1
2
3
Z
x1
2
2
3
-24
x2
3
1
7
-15
x3
3
0
3
-3
x4
1
4
-2
-24
x5
1
0
0
0
x6
0
1
0
0
x7
0
0
1
0
b
8
10
12
0
B.V.
x1
x4
x7
Z
Eq. #
1
2
3
Z
x1
1
0
0
0
x2
11/6
-2/3
1/6
13
x3
2
-1
-5
7
x4
0
1
0
0
x5
2/3
-1/3
-8/3
8
x6
-1/6
1/3
7/6
4
x7
0
0
1
0
R.H.S
11/3
2/3
7/3
104
4.1
Determine the optimal solution and the optimal value of the objective function of
the dual problem associated with this problem directly from the final tableau of
the primal problem. (2%).
4.2 Determine the range of values of c1, c3, and b1 for which the final basis does not
change (one range at a time!). (10%)
4.3 What will be the new basis if you increase the value c1 just a bit above the upper
bound found in 3.2? (3%)
4. Solution
4.1 The optimal solution to the dual problem is equal to the reduced cost of the slack
variables, hence y*=(8,4,0). Thus, the optimal value of the dual objective function is
equal to y*b=(8*8 + 4*10 + 0*12)=104=Z*.
4.2 Range of c1: Since x1 is in the basis we have to canonize the final tableau after
subtracting  from the reduced cost of x1: Thus
B.V.
x1
x4
x7
Eq. #
1
2
3
x1
1
0
0
x2
11/6
-2/3
1/6
x3
2
-1
-5
x4
0
1
0
Solved Problems: 8. 6
x5
2/3
-1/3
-8/3
x6
-1/6
1/3
7/6
x7
0
0
1
R.H.S
11/3
2/3
7/3
620-261
Z
Solved Problems: Sensitivity Analysis
Z
-
13
7
0
8
4
0
104
yields
B.V. Eq. #
x1
1
x4
2
x7
3
Z
Z
x1
1
0
0
0
x2
11/6
-2/3
1/6
13+(11/6) 
x3
2
-1
-5
7+2
x4
0
1
0
0
x5
2/3
-1/3
-8/3
8+(2/3) 
x6
-1/6
1/3
7/6
4-(1/6) 
x7 R.H.S
0 11/3
0
2/3
1
7/3
0
104
To ensure that all the reduced costs are non-negative, we require:
13+(11/6)  >= 0 ; 7+2 >= 0 ; 8+(2/3)  >= 0; 4-(1/6)  >= 0.
which in turn yields:
 >= -78/11 ;  >= -7/2 ; ;  >= - 12 ;  <= 24.
Hence the permissible range of  is [-7/2, 24] and consequently the permissible range
of c1 is [-55/2,0].
4.2 Range of c3: Since x3 is not in the basis, the final reduced cost of c3 will be 7-.
Hence, for the reduced cost of c3 to remain non-negative, we require 7- >=0, namely
 <=7. This means that the range of permissible values of c3 is <= 3+7=10.
Range of b1: The final RHS values are obtained by multiplying the original RHS
values by B-1. Thus, if we increase b1 by , the final RHS values will be
 2 / 3 1/ 6 08   
RHS'  B RHS  1/ 3 1/ 3 0 10 

8 / 3 7/ 6 1

 12 

1
11/ 3  2 / 3 1/ 6 0  
 2 / 3  1/ 3 1/ 3 0 0 

7 / 3 
 
8/ 3 7 / 6 1 

0 

11/ 3  (2 / 3)  (11/ 3)  (2/ 3) 
 2 / 3  (1/ 3)   (2/ 3)  (1/ 3) 

7 / 3 
 
(8/ 3) 
 
(7 / 3)  (8/ 3) 

For the final RHS values to be non-negative we require:
(11/3) + (2/3) >= 0 ; (2/3) - (1/3) >= 0 ; (7/3) -(8/3) >= 0
thus
 >= - 11/2 ;  <= 2 ;  <= 7/8.
Thus, the range of permissible values of d is [-11/2 , 7/8] and the range of permissible
values of b1 is [5/2, 57/8].
Solved Problems: 8. 7
620-261
Solved Problems: Sensitivity Analysis
4.3 If we increase c1 above the upper bound found in 4.2, then the reduced cost of x6
will become negative and therefore x6 will enter the basis. The ratio test indicates
that either x4 or x7 will leave the basis (there is a tie in the ratio test). Thus, the new
basis will be either (x1,x4,x6) or (x1,x6,x7).
Solved Problems: 8. 8