2.4 Postoptimality Analyses
Once an optimal solution has been obtained, the
natural question is “what (happens), if one (or more of
the given) parameter(s) change(s)?”
Two types of changes:
(1) Structural changes (addition or deletion of
variables or constraints)
(2) Parameter changes.
Structural changes:
Addition/deletion of a variable = addition/deletion of
an economic opportunity.
Example: diet problem. Add a variable = add a type of
food → the solution may get better (here: cheaper).
Delete a variable (here: delete some foodstuff): the
solution either stays the same (if the food was not used
anyway), or it gets more expensive (the food was used
& cannot be used anymore).
In general: z orig ≤ zaddvar for max problems, &
zorig ≤ zaddvar for min problems.
The deletion of variables is similar.
Here, z orig ≥ z delvar for maximization, &
zorig ≤ z delvar for minimization problems.
Addition of a constraint: The new problem is more
restricted. Hence, z addcon  z orig in maximization, &
z addcon  z orig in minimization problems.
Deletion of a constraint: zdelcon ≥ z orig holds for max,
while zdelcon ≤ z orig is valid for min problems.
Parameter changes: Consider the problem:
P: Max z = 5x1 + 6x2
s.t.
x1  2x2 ≥ 2
3x1 + 4x2 ≤ 12
x1, x2 ≥ 0.
cj: Objective function coefficients (here 5 & 6)
bi: RHS values (here 2, 12, 0, 0)
aij: LHS coefficients (technological coefficients).
Changes of the objective function coefficients
Max z = 3x1 + 2x2.
Conclusion: the angle of the gradient of the objective
function changes.
Example:
P: Max z = 1x1 + 2x2
s.t.
x2 ≤ 3
3x1 + 2x2 ≤ 11
x1  x2 ≤ 2
x1, x2 ≥ 0.
The original optimal solution is ( x1, x2 ) = (1⅔, 3).
Sensitivity analysis on c2. Increasing c2 tilts the
objective function in a counterclockwise direction.
Regardless of the magnitude of the increase, the
optimal solution stays at point B. However, the
objective function will increase, as x2 = 3, & as c2
increases (i.e., becomes more valuable), z increases.
Now decrease c2 → the gradient of the objective
function moves in a clockwise direction. For small
changes, the optimal solution stays at point B. As c2 =
⅔, B is still optimal, but so is C. As c2 decreases
further, point C = (x1, x2) = (3, 1) is now the unique
optimal solution.
For c2 below ⅔, point C is the unique optimum until c2
= –1. At this point, C & D are both optimal. For c2
below –1, point D remains optimal.
Summary:
Range of c2
Optimal
solution point
Optimal
coordinates
( x1 , x 2 )
Optimal
objective value
z
], 1[
1
]1,
⅔[
⅔
]⅔, +[
D
D&C
C
C&B
B
(2, 0)
(2, 0)
&
(3, 1)
(3, 1)
(3, 1) &
(1⅔, 3)
(1⅔, 3)
2
2
3 + c2
3⅔
1⅔ + 3c2
A similar analysis can be performed for c1.
Range of c1
Optimal
solution point
Optimal
coordinates
( x1 , x 2 )
Optimal
objective
value z
], 0[
A
(0, 3)
6
3
B&C
]3, [
C
(0, 3) & (1⅔, 3) (1⅔, 3)
(1⅔, 3)
&
(3, 1)
(3, 1)
0
A&B
6
]0, 3[
B
6+
1⅔c1
11
2 + 3c1
Now simultaneous changes.
100% rule: As long as the sum of the absolute values of
the increases or decreases of the objective function
coefficients is no more than 100%, the optimal
solution point remains optimal.
Our example: The optimal solution for the original
objective function Max z = 1x1 + 2x2 was ( x1 , x2 ) = (1⅔,
3).
This solution remains optimal as long as c1 (whose
original value is c1 = 1) does not increase by more than
c1 = 2 to the upper limit of the range at c1 = 3.
Similarly, the solution remains optimal as long as c1
does not decrease by more than c1 = 1 to the lower
end of the range at c1 = 0. Similarly, we obtain the
values c2 = + and c2 = 1⅓.
Hence, the solution remains optimal as long as
| c1 | | c2 | | c1 | | c2 |

 1 ≤ 1.
=
c1
c2
2
13
For instance, if c1 increases by ½ and c2 decreases by
½ ½
½, we obtain  1 = ¼ + ⅜ = 85 < 1, o the solution
2 13
remains optimal.
However, if c1 were to increase by ¾ and c2 were to
3/4 1
 1 = ⅜ + ¾ = 98 > 1, so that
decrease by 1, then
2 13
the optimal solution would
Changes of the right-hand side value:
Consider the constraint 2x1 + 3x2 ≤ 6.
Changes of the right-hand side value shift the
corresponding hyperplane in parallel fashion.
Example:
P: Max z = 1x1 + 2x2
s.t. x2 ≤ 3
3x1 + 2x2 ≤ 11
x1  x2 ≤ 2
x1, x2 ≥ 0.
(I)
(II)
(III)
The feasible set (shaded) has extreme points 0, A, B, C,
and D. Optimal solution: point B with coordinates
( x1 , x 2 ) = (1⅔, 3), its value of the objective function is
z = 7⅔.
Consider changes of the right-hand side value b2
Increasing b2 shifts the hyperplane in parallel fashion
to the right.
Presently, the constraint is essential (shaping the
feasible set), at b2 = 21 it is weakly redundant (primal
degeneracy), and for b2 > 21, it is strongly redundant.
On the other hand, reducing the value of b2 shifts the
hyperplane to the southwest. The feasible set shrinks,
it degenerates to a single point for b2 = 0 (primal
degeneracy), and if b2 decreases further, there is no
feasible solution.
Objective function: Original problem: Point B. As b2
increases, the optimal solution moves to points B1, B2,
and B3 = C. As b2 increases further, the optimal point
remains at B3 = C.
If b2 decreases from its original value, the
solution moves from B to B4 to B5 = A.
decreases of b2 force the optimal point down
A7 = 0. Any further decrease leaves no
solution.
optimal
Further
to A6 &
feasible
Note the difference to sensitivity analyses of objective
function coefficients: For objective function
coefficients, the solution (numerical values) does not
change within an interval. For changes of right-hand
side values, the basis does not change within an
interval, i.e., the same set of constraints determines the
optimal solution.
100% rule: In our example, we have obtained
intervals of [1, 5½], [6, 21], and [1⅓, [ for b1, b2, &
b3, whose original values were 3, 11, & 2, respectively.
Suppose now that b1 changes to 4½, b2 to 7, and b3
remains unchanged at 2. Δb1 = 1½, Δb2 = 4, and Δb3 =
0.
| b1 | | b2 | 1½ 4 7

   1, so that the basis will
=
b1
b2 2½ 5 5
change.
As
Same problem, different example: Let b1 = 2½ (Δb1 =
½), b2 = 13 (Δb2 = 2), & b3 = 1 (Δb3 = 1).
| b1 | | b2 | | b3 | ½ 2 1 3


Then
=     1,
b1
b2
b3 2 10 10 3 4
→ same basis, but the solution may (and most likely
will) change.
2.4.2 Economic Analysis of an Optimal
Solution
P: Max z =3x1  x2
s.t. x1  x2 = 2
2x1 + 3x2 ≤ 16
5x1 + x2 ≥ 15
x1, x2 ≥ 0.
SUMMARY OF RESULTS
VALUE OF THE OBJECTIVE FUNCTION 10.8000
DECISION
VARIABLE
VALUE AT
OPTIMUM
X1
X2
4.4000
2.4000
OPPORTUNITY
COST
0.0000
0.0000
SLACK/EXCESS
VARIABLE
CONSTR.
TYPE
OPT.
VALUE
CONSTRAINT 1:
CONSTRAINT 2:
CONSTRAINT 3:
EQ
LE
GE
0.0000
0.0000
9.4000
SHADOW
PRICE
2.2000
0.4000
0.0000
Part 1: Value of the objective function (here
z
= 10.8).
Part 2: Optimal values of decision variables (here x1 =
4.4, x2 = 2.4), & opportunity costs (= reduced costs):
indicates how far a variable is from being included in
the solution with a positive value. (Here: both
variables are included in the solution, so their
opportunity costs equal zero).
If a variable had a price of $5 & its opportunity costs
were $3. Then the lowest price for which we include
the variable in the solution is $8.
Part 3: Values of slack & excess variables. (Here: 0, 0,
9.4, meaning that the left-hand side values differ from
the right-hand side values by 0, 0, & 9.4, resp.)
Constraints with zero slack/excess are binding, i.e.,
they satisfy the relation as equation. They are
bottlenecks in the problem, as any change in them will
immediately change the solution.
Shadow prices indicate the change of the value of the
objective function given a unit change of the righthand side value of a constraint. It does not indicate
how the solution will change, however. (Here: If b1
increases by 1 from 2 to 3, the objective value
increases by 2.2. Similarly, if b2 increases by 1 from 16
to 17, z increases by 0.4. As the third constraint is not
satisfied as equation, a (small) change of b3 has no
effect on z .
The “Sensitivity” option generates additional printout:
SENSITIVITY ANALYSES
COEFFICIENTS OF THE OBJECTIVE FUNCTION
VARIABLE
LOWEST
ALLOWABLE
VALUE
ORIGINAL
VALUE
HIGHEST
ALLOWABLE
VALUE
X1
X2
1.0000
3.0000
3.0000
1.0000
INFINITY
INFINITY
RIGHT-HAND SIDE VALUES
CONSTRAINT
NUMBER
LOWEST
ORIGINAL
ALLOWABLE VALUE
VALUE
HIGHEST
ALLOWABLE
VALUE
CONSTR. 1
CONSTR. 2
CONSTR. 3
1.6154
8.1667
INF
8.0000
INF
24.4000
2.0000
16.0000
15.0000
Part 4: Coefficients of the Objective Function
Part 5: Right-Hand Side Values.
They indicate within what interval of parameters the
solution (basis) remains unchanged.
Example: At present, c1 = 3. The printout shows that
the solution remains unchanged as long as c1  [1, [.
Similarly, presently c2 = –1. The present solution
remains optimal as long as c2  [3, ∞[.
Miniature case study:
Products: hiking boots, “Walker,” “Hiker,” and
“Backpacker.”
Raw materials: NOwater (a lining that waterproofs
the boots), Fabrinsula (fabric insulation for warmth),
and Lugster (lug) soles. Currently, 10,000 sq. ft. of
NOwater are available at $20 per sq. ft. Similarly, the
manufacturer has access to up to 5,000 oz. of
Fabrinsula at $15 per ounce and up to 7,000 pairs of
Lugster soles at $10 per pair.
Products
Walker
NOwater
1/2
Fabrinsula
1/2
Lugster
1
Hiker
1
2/3
1
Backpacker
21/2
[sq. ft. per pair]
4/3
[oz. per pair]
1
[pair of soles]
Contracted sales: at least 3,000 pairs of “Walkers,” at
least 2,000 pairs of “Hikers,” and at least 1,000 pairs
of “Backpackers.”
Prices: $40, $65, and $110.
Note: The unit profits in the objective function below
are computed as price minus costs for the required
NOwater, Fabrinsula, and Lugster).
P: Max z = 12.5x1 + 25x2 + 30x3
s.t.
x2 + 2½x3 ≤ 10,000
(NOwater availability)
½x1 + ⅔x2 + 1⅓x3 ≤ 5,000
(Fabrinsula availability)
x1 + x2 +
x3≤ 7,000
(Lugster soles availability)
x1
≥ 3,000
(Walker requirement)
x2
≥ 2,000
(Hiker requirement)
x3 ≥ 1,000
½x1 +
(Backpacker requirement)
x2,
x3 ≥
0.
x 1,
SUMMARY OF RESULTS
VALUE
OF
143,749.60
DECISION
VARIABLE
THE
OBJECTIVE
VALUE AT
OPTIMUM
WALKER
HIKER
BACKPACKER
FUNCTION
OPPORTUNITY
COST
3,000.0000
2,750.0750
1,249.9249
0.0000
0.0000
0.0000
SLACK/
CONST.
EXCESS
TYPE
VARIABLE
OPTIMAL SHADOW
VALUE
PRICE
NOWATER
FABRINSULA
LUGSTER
WALKER
HIKER
BACKPACKER
2,625.1125
0.0000
0.0000
0.0000
750.0750
249.9250
LE
LE
LE
GE
GE
GE
0.0000
7.5008
19.9993
-11.2496
0.0000
0.0000
SENSITIVITY ANALYSES
COEFFICIENTS OF THE OBJECTIVE FUNCTION
VARIABLE
LOWEST
ALLOW.
VALUE
ORIG.
VALUE
HIGHEST
ALLOW.
VALUE
WALKER
HIKER
BACKPACKER
INF
16.00
25.00
12.50
25.00
30.00
23.75
30.00
50.00
RIGHT-HAND SIDE VALUES
CONSTR.
LOWEST
ALLOW.
VALUE
NOWATER
FABRINSULA
LUGSTER
WALKER
HIKER
BACKPACKER
7,374.89
4,833.40
6,625.00
2,001.60
INF
INF
ORIG.
VALUE
10,000
5,000
7,000
3,000
2,000
1,000
HIGHEST
ALLOW.
VALUE
INF
5,500.00
7,249.89
3,600.02
2,750.08
1,249.93
Q1: How many pairs of the boots should we
manufacture and what will be the associated profit?
A1: 3,000 pairs of Walkers,
2,750 pairs of Hikers, &
1,250 pairs of Backpackers.
The resulting profit is $ 143,750.
Q2: NOwater, Fabrinsula, and Lugster are critical
resources. How many of these do we use in the
suggested plan and how much is left over?
A2: 2,625 sq ft of NOwater are left over,
Fabrinsula and Lugster soles are completely used.
They are obvious bottlenecks in the process.
In other words, we are using 7,375 sq ft of NOwater,
the complete supply of 5,000 oz of Fabrinsula, and all
of the 7,000 pairs of Lugster soles.
Q3: Customers would be prepared to pay an
additional $10 for a pair of Hikers. Would such a
price hike change the optimal solution?
A3 A sensitivity analysis on the objective function
coefficient c2. The range within which the present
optimal solution remains optimal is [16, 30]. A price
increase by $10 leads to a price of $35 rather than the
original $25, which is not in the interval, thus the
solution will change.
Q4: We are presently making 3,000 Walkers, just
enough to satisfy one of our requirements. Would a
lower price lead to increased sales?
A4 Another sensitivity analysis on an objective function
coefficient. Given the present unit profit of $12.50 for
a pair of Walkers, the Sensitivyity Analyses part of the
printout indicates that as long as the unit profit
remains in the interval ]∞, 23.75], the optimal
solution will not change. Hence, it does not matter by
how much you decrease the price of Walkers, we will
not sell any more.
Q5: A salesman offered us an additional 100 oz of
Fabrinsula for $8.50 per ounce. Should we purchase
that? We used up all the Fabrinsula that we have.
What if I squeeze the man a bit and get it for $5?
Will we take it for that price? And what will happen
to our profit?
A5 A sensitivity analysis b2. The shadow price of
Fabrinsula is $7.50, so we should not purchase it for
more than that. The basis will not change if we buy up
to 500 ounces of Fabrinsula. If we can get it for $5
per ounce, we can buy up to 500 oz of it. For each
ounce that you get on top of what we have, our profit
will increase by $2.50. Beyond 500 extra ounces,
additional calculations are needed.
Q6: I just read an offer for Lugster Soles that we
could get for $19 a pair. Would you consider that?
A6 (to himself): The shadow price for Lugster Soles is
$20, and the Sensitivity Analyses tell me that this
holds for an increase of up to 250 pairs. Hence, if we
can get a pair for $19, get them. They are worth $20 to
us, so for each extra pair, we make $1 in addition to
our usual profit. We can get up to 250 pairs.
Q7: I just heard that 200 pairs of Lugster soles have
been damaged in our warehouse and can no longer
be used. What are we going to do? How much is that
going to cost me?
A7 The optimal basis remains unchanged as long as we
have between 6,625 and 7,250 pairs of soles. So 200
pairs less will not change the basis, but it will change
the solution. And since the shadow price for Lugster
Soles is $19, our profit will decrease by 200(19) =
$3,600.