DUALITY AND SENSITIVITY ANALYSIS
Understanding Duality
No learning of Linear Programming is complete unless we learn the concept of ‘Duality” in
linear programming. It is impossible to separate the linear programming problem and its dual.
Several algorithms to solve special linear programming problems are centered around the
concept of duality.
Even though the emphasis of this book is to solve LP problems and understand to interpret
the solutions, we deviate from the main focus of applications to understand duality. We will
however retain the methodology of solving LPs to understand the concepts and applications.
We explain duality using an example.
In Illustration 1.1, we formulated and solved the LP problem to maximize the revenue for the
bakery. The problem is to
Maximize 32X1 + 25X2
Subject to
5 X 1  4 X 2  59
4 X 1  3 X 2  46
X1 , X 2  0 .
The optimum solution obtained using the solver is X1 = 7, X2 = 6 with Z = 374
We consider another LP given below:
Illustration 1.18
Minimize 59Y1 + 46Y2
Subject to
5Y1  4Y2  32
4Y1  3Y2  25
Y1 , Y2  0 .
We solve this LP using the solver. The final output is given in Table 1.16
Table 1.16 – Solution for Illustration 1.18
y1
4
y2
3
374
32
>=
32
25
4
3
>=
>=
>=
25
0
0
The optimum solution is Y1 = 4, Y2 = 3 with objective function value W = 374.
Now considering the maximization problem (called P1) and the minimization problem (called
P2) we make the following observations:
1. P1 and P2 had the same value of the objective function at the optimum
2. The objective function coefficient coefficients of P1 are the RHS values of P2 and
vice versa.
3. The number of variables in P1 and the number of constraints in P2 are equal and vice
versa.
The obvious question is “Is there a relationship between P1 and P2?” The relationship is
established using the discussion that follows:
1. Consider P1 (the maximization problem). If there were no constraints the objective
function value is ∞. Let us try to get upper estimates for the value of Z.
2. We multiply the second constraint by 9 to get 36X1 + 27X2 ≤ 414. Since X1 and X2
are ≥ 0, 32X1 + 25X2 ≤ 36X1 + 27X2 ≤ 414. Therefore Z* ≤ 414.
3. We multiply the first constraint by 7 to get 35X1 + 28X2 ≤ 413. Since X1 and X2 are ≥
0, 32X1 + 25X2 ≤ 35X1 + 28X2 ≤ 413. Therefore Z* ≤ 413 and this is a better upper
estimate for Z*
4. We add the two constraints to get 9X1 + 7X2 ≤ 105. This inequality holds because X1
and X2 are ≥ 0, We multiply this constraint by 4 to get 36X1 + 28X2 ≤ 420. Since X1
and X2 are ≥ 0, 32X1 + 25X2 ≤ 36X1 + 28X2 ≤ 420. Therefore 420 is an upper
estimate of Z* but we ignore this because our current best estimate is 413.
5. We multiply the constraint 4X1 + 3X2 ≤ 46 by 25/3 to get 33.33X1 + 25X2 ≤ 383.33.
Based on the above discussions, 383.33 is a better upper estimate for Z*.
6. We multiply the constraint 5X1 + 4X2 ≤ 59 by 32/5 to get 32X1 + 25.6X2 ≤ 377.6.
Based on the above discussions, 377.6 is a better upper estimate for Z*.
7. We multiply the constraint 9X1 + 7X2 ≤ 105 by 25/7 to get 32.14X1 + 25X2 ≤ 375.
Now 375 is a better upper estimate for Z*. Here we have added the two constraints
and multiplied by 25/7. This is the same as multiplying the constraints by 25/7 and
adding them.
8. We can multiply the first constraint by a and the second by b and add them. If on
addition the coefficients of X1 and X2 are ≥ 59 and 46 respectively, the RHS value
(which is 59a + 46b) is an upper estimate of Z. We should have a, b ≥ 0, otherwise the
sign of the inequality will change and we cannot add the inequalities.
9. If we want the best estimate of Z* (as small an upper estimate as possible) we need to
define a and b such that 59a + 46b is as small as possible.
10. We therefore define a and b to minimize 59a + 46b is minimized. This leads us to
another LP problem (where the variables are defines as Y1 and Y2 instead of a and b)
which is P2.
Minimize 59Y1 + 46Y2
Subject to
5Y1  4Y2  32
4Y1  3Y2  25
Y1 , Y2  0 .
Therefore problem P2 is born out of P1. The problem P2 is called the dual of the given
problem P1 (called the primal).
1. Do the primal and dual have the same value of the objective function at the optimum?
Yes. If both have optimum solutions their objective function value is the same. This result is
called the main duality theorem. If the primal and dual have optimal solutions, the objective
function value at the optimum for both is the same.
2. Can we read the solution of the dual from the solution of the primal?
Yes. When the LP is written into the solver, we click options and click linear model and non
negative.
After we solve the LP, the solver dialogue box shows option for three reports. We can click
on them and press OK. The reports are available as active sheets. If we go to the sensitivity
report, we see a table under “constraints”. This part is shown in Table 1.17
Table 1.17- Sensitivity analysis report
Constraints
Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$B$6 x1
59
4
59
2.333333333
1.5
$B$7 x1
46
3
46
1.2
1.75
$B$8 x1
7
0
0
7
1E+30
$B$9 x1
6
0
0
6
1E+30
The shadow prices against the two constraints are 4 and 3 respectively. These are the values
of Y1 and Y2 at the optimum. (We will see the relationship between the shadow prices and the
dual solution later).
3. Does the algebraic method show the dual solution?
YES. The algebraic method terminated for our LP problem with the equations
X1 = 7 + 3X3 – 4X4
X2 = 6 - 4X3 + 5X4 and
Z = 374 - 4X3 – 3X4.
We terminated by saying that the revenue function cannot be maximized further by
increasing X3 or X4 (by increasing the non utilization or by keeping some resources idle). At
the optimum the solution is X1 = 7 and X2 =6. The coefficients of X3 and X4 in the objective
function at the optimum (in the algebraic method) represent the dual solution (with a negative
sign). The dual solution is Y1 = 4 and Y2 =3 with W = 374 (It is customary to use W to
represent the objective function of the dual).
4. Since the primal and dual have the same value of the objective function, how do we
define the dual variables?
The optimum solution to the dual is Y1 = 4 and Y2 =3 with W = 59 x 4 + 46 x 3 = 374. The
values Y1 = 4 and Y2 =3 represent the worth of the 59 units of the first resource and 46 units
of the second resource at the optimum. The dual solution represents the worth of the
resources at the optimum.
5. What is shadow price and why is it related to the resources (primal constraints)?
We have already seen that the dual solution represents the worth of one unit of each resource
at the optimum. Let us explain it further using an example. Here we increase the first resource
of the primal by 1 to 60.
Illustration 1.19
Solve the LP - Maximize 32X1 + 25X2
Subject to
5 X 1  4 X 2  60
4 X 1  3 X 2  46
X1 , X 2  0 .
The optimum solution to the primal (using solver) is given in Table 1.18
Table 1.18 – Solver Solution
x1
4
obj fn
x2
10
378
60
<=
60
46
4
10
<=
>=
>=
46
0
0
The optimum solution is X1 = 4, X2 = 10 with Z = 378.
We continue to produce both the items but with different quantities. The objective function
value has increased by 4 to 378. This shows that if we can increase the RHS of the first
resource by 1, the profit increases by 4 (as long as we continue to produce the same products
as before). The worth of the additional resource at the optimum is the extent to which the
revenue goes up on increase of the resource. This is called the shadow price of the resource
at the optimum or the worth of an additional amount of the resource at the optimum or the
marginal value of the resource at the optimum. This is represented by the optimum solution
of the dual.
We restore the RHS of the first constraint to 59 and decrease the second resource by 1 to 45.
Illustration 1.20
Solve the LP - Maximize 32X1 + 25X2
Subject to
5 X 1  4 X 2  59
4 X 1  3 X 2  45
X1 , X 2  0 .
The optimum solution to the primal (using solver) is given in Table 1.19
Table 1.19 – Optimum solution from solver
x1
x2
3
11
obj fn
371
59
45
3
11
<=
<=
>=
>=
59
45
0
0
The optimum solution is X1 = 3, X2 = 11 with Z = 371.
We continue to produce both the items but with different quantities. The objective function
value has decreased by 3 to 371. This shows that if we decrease the RHS of the first resource
by 1, the profit decreases by 3 (as long as we continue to produce the same products as
before). The worth of the additional resource at the optimum is the extent to which the
revenue decreases on decrease of the resource. This is called the shadow price of the
resource at the optimum. This is represented by the optimum solution of the dual.
6. How do I write the dual for a given primal?
The relationship between the primal and the dual are given in Table 1.20
Table 1.20 – Primal dual relationships
Primal
Dual
Maximization
Minimization
Number of variables
Number of constraints
Number of constraints
Number of variables
Objective function values
RHS values
RHS values
Objective function values
Coefficient
matrix
for Transpose of the constraint coefficient
constraints
matrix
≥ variable
≥ constraint
≤ variable
≤ constraint
Unrestricted variable
Constraint is an equation
≥ constraint
≤ variable
≤ constraint
≥ variable
equation
Unrestricted variable
Illustration 1.21
Consider the LP
Minimize 59Y1 + 46Y2 + 37Y3
Subject to
5Y1  4Y2  3Y3  32
4Y1  3Y2  8Y3  26
2Y1  5Y2  9Y3  60
3Y1  2Y2  3Y3  15
Y1  0, Y2unrestricted , Y3  0 .
Write the dual to this LP and solve the dual?
Using the information in Table 1.20 (treating the minimization problem as primal) we get the
dual as
Maximize 32X1 +26X2 + 60X3+15X4
Subject to
5 X1  4 X 2  2 X 3  3 X 4  59
4 X1  3 X 2  5 X 3  2 X 4  46
3 X1  8 X 2  9 X 3  3 X 4  37
X1unrestricted , X 2  0, X 3  0, X 4  0
We solve the primal and the dual optimally using the solver. Since we have a ≤ variable and
an unrestricted variable, we go to options and leave out the non negativity restriction there.
We write Y1≥ 0 and Y2 ≤ 0 as explicit constraints. We don’t have a restriction on Y2 and we
do not restrict it. There is no constraint for Y2 unrestricted. Table 1.21 gives the solver output.
Table 1.21 – Solver output
y1
y2
8
-2
obj
y3
0
380
32
26
6
20
8
0
=
>=
<=
>=
>=
<=
32
26
60
15
0
0
The optimum solution is Y1 = 8, Y2 = -2 and Y3 = 0 with Z = 380
We now solve the dual. The solution to the dual is given in Table 1.22
Table 1.22 – Solution to the dual
x1
x2
x3
7
6
0
obj fn
x4
0
380
59
46
69
6
0
0
<=
=
>=
>=
<=
>=
59
46
37
0
0
0
The solution is X1 = 7, X2= 6 with W = 380.
7. It was mentioned that the dual value for a resource is the increase in profit for unit
increase in the resource. If we increase the resource by a large value will the objective
function value increase at the rate of the dual?
Illustration 1.22
Maximize 32X1 + 25X2
Subject to
5 X 1  4 X 2  59
4 X 1  3 X 2  45
X1 , X 2  0
We know that the dual solution is Y1 =4, Y2 = 3. We have already seen that if the RHS of the
first constraint becomes 60, the solution to the primal is X1 =4, X2 = 10 with Z = 378.
When RHS = 62, the solution is X1 = 0, X2 = 15.333 with Z = 383.33. We observe that the
objective function has not increased by 12 when the RHS of the first constraint is increased
by 3.
When RHS = 61.3, the solution is X1 = 0.1, X2 = 15.2with Z = 383.2 We observe that the
objective function has increased by 9.2 when the RHS of the first constraint is increased by
3.2. The linear increase continues because we continue to produce X1 and X2.
At RHS = 61.333, we produce X1 = 0, X2 = 15.333 with Z = 383.33. The limiting value up to
which the linearity holds is 2.333. This is given in Table 6.2 as the allowable increase and
allowable decrease for both the resources.
As we keep increasing (decreasing) the value of the RHS (resource) the combination of
products made changes. At that point, the shadow price changes and the proportional increase
(decrease) in the objective function also changes.
8. Is the first dual variable related to the first constraint or the primal (resource)?
Yes. We have as many dual variables as the number of constraints in the primal. For every
primal constraint there is a dual variable. The dual variable represents the value of the
corresponsing resource at the optimum.
9. Can a dual variable take zero value?
Yes. We explain this through an example
Illustration 1.23
Maximize 32X1 + 25X2
Subject to
5 X 1  4 X 2  59
4 X 1  3 X 2  46
3 X 1  2 X 2  30
X1 , X 2  0 .
The optimum solution to the primal is given in Table 1.23
Table 1.23 – Optimum solution from the solver
X1
X2
1
13.5
369.5
59
44.5
30
<=
<=
<=
59
46
30
The solution is given by X1 = 1, X2 = 13.5 and Z = 369.5. We store the sensitivity result for
this solution. Table 1.24 shows the sensitivity results.
Table 1.24 – Sensitivity results
Microsoft Excel 12.0 Sensitivity Report
Worksheet: [Book1]Sheet1
Report Created: 24-05-2010 22:19:38
Adjustable Cells
Final
Cell Name Value
$B$2 X1
1
$C$2 X2 13.5
Reduced
Cost
0
0
Objective Allowable Allowable
Coefficient Increase Decrease
32
5.5
0.75
25
0.6
3.666666667
Shadow
Price
5.5
0
Constraint Allowable Allowable
R.H. Side Increase Decrease
59
3
1E+30
46
1E+30
1.5
Constraints
Final
Cell Name Value
$B$6 X1
59
$B$7 X1 44.5
$B$8 X1
30
1.499999999
30
3
1E+30
From the shadow prices, we find that the dual solution is Y1 = 5.5, Y2 = 0 and Y3 = 1.5 with
W = 59 x 5.5 + 30 x 1.5 = 369.5. We observe that the objective function value of the primal
and dual at the optimum is 369.5.
Here we observe that Y2 = 0 indicating that the value of the second dual variable is zero.
From Table 6.8 we observe that the primal optimal solution X1 = 1, X2 = 13.5 completely
utilizes resource 1 and 3 (in the primal) but uses only 44.5 out of the available 46 units of the
second resource. We also observe that Y2 (the dual variable corresponding to this resource) is
zero.
At the optimum, when the primal resource is utilized fully, the dual variable has a positive
value. When a resource is not utilized fully the corresponding dual variable is zero.
The dual variable is the marginal value of the resource at the optimum. It also represents the
price at which an additional resource can be bought at the optimum. It is obvious that when a
resource is not utilized fully, increasing it will not increase the revenue. Therefore the shadow
price or marginal value of such a resource is zero.
Illustration 1.24
Consider a small factory that makes two types of furniture – tables and chairs. They use two
resources wood and steel. The amount of steel used per unit is 4 kg and 5 kg for a table and a
chair. The amount of wood required per table and chair is 6 and 5 kg. We have 60 kg of steel
and 80 kg of wood. The selling price of table and chair is Rs 80 and 70 respectively. Find the
number of tables and chairs to be made to maximize revenues.
Let X1 and X2 be the number of tables and chairs produced. The corresponding LP problem is
to Maximize 80X1 + 70X2
Subject to
4X1 + 5X2 ≤ 60
6X1 + 5X2 ≤ 80
X1, X2 ≥ 0
The optimum solution to the LP is X1 = 10, X2 = 4 with revenue = 1080.
Worth of the resource at the optimum
Consider the optimum solution X1 = 10, X2 = 4. We observe that we have consumed 60 kg of
steel and 80 kg of wood to make 10 tables and 4 chairs. In the example, at the optimum we
have consumed all the resources that we had, converted them into produces that have fetched
us the revenue. It is therefore not incorrect to consider that the worth if the resources put
together is the total revenue that we have earned by converting them into products.
Let us try to compute the worth of the resources at the optimum using the example
Illustration 1.25
Consider the data given in Illustration 1.24. Use the optimum solution X1 = 10 and X2 = 4 (if
required) and try to formulate an LP problem to determine the value (or worth) of the
resources at the optimum?
Let Y1 and Y2 be the worth of 1 kg of steel and wood at the optimum. The worth of a table is
4Y1 + 6Y2 and the worth of a chair is 5Y1 + 5Y2. Now 4Y1 + 6Y2 should be ≥ 80 and 5Y1 +
5Y2 ≥ 70 (The worth of a table should not be less than the selling price of 80; otherwise we
would not convert the raw material into products and sell them). If the value of the unit
resource represents the cost price of the resource (at the optimum), we would minimize the
cost price which is to minimize 60Y1 + 80Y2. The corresponding LP that determines the
value of the resources at the optimum is to
Minimize 60Y1 + 80Y2
Subject to
4Y1 + 6Y2 ≥ 80
5Y1 + 5Y2 ≥ 70
Y1, Y2 ≥ 0
The optimum solution to this LP is Y1 = 2, Y2 = 12 with total worth = 1080.
It is to be noted that expectedly, the total revenue is equal to the total worth of the resource at
the optimum.
Shadow price and the marginal value of the resources
When we computed the value of the resource, we assumed that Y1 and Y2 represented the
cost price or market price of a unit of the resource. This price is the shadow price of the
resource and represents the price of unit resource at the optimum.
In our example, we were able to convert 60 kg of steel and 80 kg of wood to 10 tables and 4
chairs having total revenue (worth) of Rs 1080. Let us solve the problem assuming that we
have 61 kg of steel and 80 kg of wood.
Illustration 1.26
Solve the problem in Illustration 1.24 considering 61 kg of steel and interpret the solution?
What happens if we wish to have 70 kg and 90 kg of steel?
The corresponding LP would be Maximize 80X1 + 70X2
Subject to
4X1 + 5X2 ≤ 61
6X1 + 5X2 ≤ 80
X1, X2 ≥ 0
The optimum solution is given by X1 = 9.5, X2 = 4.6 with revenue = 1082. We observe that a
unit increase of steel results in an increase of revenue of Rs 2. If we were to buy the extra
resource (at the optimum) from the market, its worth is Rs 2 because it can fetch us additional
revenue of Rs 2. This is the same as Y1 = 2 in example 2.
The value of Y1 is called the marginal value of resource 1 at the optimum which is the
additional worth of a unit of the resource.
The reader would observe that the optimum solution now has X1 = 9.5 and X2 = 4.6. The
reader can ask a question as to whether we can make and sell a fractional number of tables
and chairs. The answer is that we are solving LP problem that allows fractional values for the
variables. In fact the proportionality assumption makes us believe that 0.5 tables is worth Rs
40 just as half a kg of the first resource is worth Re 1 at the optimum!
When we solve for 70 kg of steel, the optimum solution is given by X1 = 5, X2 = 10 with
revenue = 1100. The additional 10 kg of steel fetched us additional revenue of Rs 20 at the
rate of Rs 2/kg for the additional 10kgs.
We solve for 70 kg of steel and 90 kg of wood. The optimum solution is given by X 1 = 10
and X2 = 6 with revenue = 1220. The additional revenue is 1220 – 1080 = 140. The extra 10
kg of steel is worth 20 and the extra 10 kg of wood is worth 120 (at the rate of 12 per kg for
10 kg – note that Y2 =10, indicating that the worth of 1 kg of wood at the optimum is Rs 10).
We observe that the additional cost of the resource is equal to the additional revenue.
We solve for 90 kg of steel and 80 kg of wood. The optimum solution is given by X 2 = 16
with revenue = 1120. If we assume that the marginal value of steel is Rs 2, the additional cost
is Rs 60 while the additional revenue is only Rs 40. Here, the additional cost does not get
reflected as additional revenue. We observe that the proportionality works till we increase the
steel quantity to 80 after which we do not produce tables. The proportionality will hold only
till the same set of products is made (at different quantities). When the set of products made
changes, the extra cost and the extra revenue are not equal.
Illustration 1.27
Consider the data in Illustration 1.24. In addition it is observed that it takes 8 hours to make a
table and 5 hours to make a chair. The company works 5 days a week and 16 hours a day.
Assume that the products have to be made in a week.
We have a third constraint which relates the time required to the time available. The problem
is to
Maximize 80X1 + 70X2
Subject to
4X1 + 5X2 ≤ 60
6X1 + 5X2 ≤ 80
8X1 + 5X2 ≤ 80
X1, X2 ≥ 0
The optimum solution to the LP is given by X1 = 5, X2 = 8 with revenue = 960. It is observed
that 60 units of steel is consumed, 70 units of wood is consumed and 80 hours of time is
consumed. It is observed that the second resource is not fully consumed.
The sensitivity report from the excel solver shows that the shadow prices are 8, 0 and 6 for
the three resources. The worth of the resources are 60 x 8 + 80 x 0 + 80 X 6 = 960 which is
equal to the revenue.
It is observed that the shadow price of resource 2 is zero. This is because this resource is not
fully consumed. Obviously the marginal value (cost) of this resource is zero because we need
not pay extra to buy a unit of this resource since it is available with us.
Shadow prices are positive for resources that are scarce (fully consumed). Shadow price is
zero when a resource is not fully consumed.
Does it mean that the wood used has no value? The answer is No. It is acceptable to say that
the marginal value (or the implied cost of procuring an extra kg) is zero since the resource is
available with us. When the shadow price is zero we should not assume that the worth is zero.
The 70 kg of wood consumed has a value which is not visible at the optimum. Only the
marginal value at the optimum has to be interpreted as zero. The total revenue of 960 is
distributed between the value of the other two resources at the optimum
Sensitivity Analysis
Sensitivity analysis deals with the sensitivity of the solution to an LP to changes in problem
parameters. The parameters that can change are:
1. Change in Right hand side (RHS) values in the constraints
2. Change in objective function values
3. Changes in coefficients of constraints.
In addition two more aspects are included in sensitivity analysis. These are
1. Adding a variable (product)
2. Adding a constraint.
In the days when LPs were solved by hand or when solvers were not available freely, it was
necessary to compute the effect of the above changes from the optimal solution. Sensitivity
analysis methods essentially did that and obtained the optimum solution after the changes
from the given optimum. So long as the changes were not on the coefficients of the
constraints corresponding to the decision variables in the optimum solution, it is possible to
carry out sensitivity analysis from the present optimum solution.
Since we are using the excel solver, we do not emphasize much on sensitivity analysis
because it is possible to solve the problem from the beginning using the solver for every
change made. We need to make changes in the template and modify the constraints or the
objective in the solver.
Illustration 1.28
Maximize 32X1 + 25X2
Subject to
5 X 1  4 X 2  59
4 X 1  3 X 2  46
X1 , X 2  0 .
Table 1.25 shows the results of modifying some of the parameters.
Table 1.25 – Results after modifying some parameters
No.
Parameter
Original
Changed value
value
1.
Objective coeff 1
32
33.333
2.
Objective coeff 1
32
33.5
3.
Objective coeff 1
32
31.25
4.
Objective coeff 1
32
31
5.
Objective coeff 2
25
25.6
6.
Objective coeff 2
25
26
7.
Objective coeff 2
25
24
8.
9.
RHS 1
RHS1
59
59
60
61.333
10.
11.
12.
RHS1
RHS2
RHS2
59
46
46
57.5
47
47.2
13.
RHS2
46
44.25
From Table 1.25 we observe the following:
Solution
X1 = 7, X2 = 6 with Z = 383.33
X1 = 11.5 X2 = 0 with Z =
385.25
X1 = 7, X2 = 6 with Z = 368.75
X1 = 0, X2 = 14.75 with Z =
368.75
X1 = 0, X2 = 14.75 with Z =
368.75
X1 = 7, X2 = 6 with Z = 377.6
X1 = 0, X2 = 14.75with Z =
377.6
X1 = 0, X2 = 14.75with Z =
383.5
X1 = 7, X2 = 6 with Z = 368
X1 = 11.5, X2 = 0 with Z = 368
X1 = 4, X2 = 10 with Z = 378
X1 =0, X2 = 15.333 with Z =
383.333
X1 = 11.5, X2 = 0 with Z = 368
X1 = 11, X2 = 1 with Z = 377
X1 = 11.8, X2 = 0 with Z =
377.6
X1 = 0, X2 = 14.75 with Z =
368.75
1. When the objective function coefficient of the first variable (C1) increases up to
33.33, we produce X1 and X2. When it increases to 33.5 we produce only X1. The
limiting value for this coefficient is an increase of 1.5 at/after which the products
made changes.
2. The lower limiting value for C1 is a reduction of 0.75 at/after which we make only X2.
At the limiting value we can have two (or more) solutions. We have shown two
solutions under item 3. In fact at all limiting values we have alternate optima.
3. When the objective function coefficient of the first variable (C2) can be increased up
to 25.6, we produce X1 and X2. When it increases to 25.6 we produce only X2. The
limiting value for this coefficient is an increase of 0.6 at/after which the products
made changes. At the limiting value we can have two (or more) solutions. We have
shown two solutions under item 5. In fact at all limiting values we have alternate
optima.
4. The lower limiting value for C2 is a reduction of 1 at/after which we make only X1. At
the limiting value we can have two (or more) solutions. We have shown two solutions
under item 7. In fact at all limiting values we have alternate optima.
5. We observe that when the RHS of the first constraint goes up to 61.33 we start
producing only X2 and if it reduces to 57.5 we make only X1. The limiting values are
2.333 and 1.5 respectively.
6. We observe that when the RHS of the second constraint goes up to 47.2 we start
producing only X1 and if it reduces to 44.25 we make only X2. The limiting values are
1.2 and 1.75 respectively.
The observations in Table 1.25 are based on solving the LP problem at least ten times using
the solver. It is not a difficult task but one of the reports that can be obtained from the
solution to the original problem (sensitivity analysis report) gives all the above information.
After we solve the problem, we get a message “keep solver solution” and we can retain the
solver solution. On the right hand side we have the following three options:
1. Answer
2. Sensitivity
3. Limits
We can click on the “Sensitivity” and the solver creates a sheet called Sensitivity report 1.
The sheet looks as shown in Table 1.26
Table 1.26 – Sensitivity Report
Microsoft Excel 12.0 Sensitivity Report
Worksheet: [Book1]Sheet1
Report Created: 5/31/2010 3:09:56 PM
Adjustable Cells
Final
Cell Name
Value
$B$2 x1 6.999999998
$C$2 x2 6.000000002
Reduced
Cost
0
0
Objective Allowable Allowable
Coefficient Increase
Decrease
32
1.333333334 0.749999999
25
0.599999999
1
Constraints
Cell Name
$B$6 x1
$B$7 x1
Final
Value
59
46
Shadow Constraint Allowable Allowable
Price
R.H. Side Increase
Decrease
4.000000003
59
2.333333332
1.5
2.999999996
46
1.2
1.749999999
The allowable increase and decrease for the coefficients in the objective function and for the
constraints can be read from Table 1.26. The shadow prices indicate the optimum solution to
the dual.
Thus the sensitivity analysis report provides certain basic information about the solution to
the dual. It also tells us the extent to which the solution is sensitive to changes in coefficients
in the objective function and to RHS values. The limiting values where there is a change in
the products made is known when the original problem is solved.
Illustration 1.29
Consider the data given in Illustration 1.24. The sensitivity analysis report given when the
problem is solved using Excel solver is reproduced in Table 1.27 Interpret the solver report to
explain the sensitivity of the LP parameters.
Table 1.27 – Solver Report
Variable
Values
X1
10
X2
4
Constraint
Value
1
60
2
80
Obj fn
80
70
Shadow Price
2
12
Increase
4
30
Increase
20
10
Decrease
24
3.333
Decrease
6.6666
20
The table shows that the optimum solution to the given problem is X1 = 10, X2 = 4. The
shadow prices for the two resources are 2 and 12 respectively.
When the objective function coefficient for X1 can be increased to 84 and decreased to 56, we
will continue to produce both items. When this value goes out of this range, we will not
produce tables. Similarly when the revenue from chair is within 100 and 66.666, we will
produce chairs. When the revenue goes beyond these values, we will produce a different
combination of items.
When the availability of resource 1 is between 80 and 53.333, the revenue increase (or
decrease) will be at the rate of Rs 2 per kg. Similarly, till the availability of the second
resource is between 90 and 60, the increase or decrease in revenue will be Rs 12 for unit
increase or decrease of the resource.
What happens when the objective function coefficients are 50 and 50 and the availability of
resources is 60 and 50? The optimum solution to the LP is X2 = 10 and revenue = 500. The
problem becomes a new LP which is quite unrelated to the old LP that we are solving. It only
shows that some linear relationships among the variables are not valid at the new values.