Chapter 4 Duality & sensitivity analysis hand out last .pdf

4/28/2020 Duality and Sensitivity Analysis
Compiled by Tsegay Berhe [ MSc in production engineering & Management ]
MEKELLE UNVERISTY

COMPILED BY TSEGAY BERHE [ MSC IN PRODUCTION ENGINEERING & MANAGEMENT ] 1
Contents
4. Duality and Sensitivity Analysis .............................................................................................. 2
4.1. Primal-dual relationship;................................................................................................. 2
4.2. Rules for Constructing the Dual Problem........................................................................ 3
4.3. Economic interpretation of duality .................................................................................. 4
4.4. Simple way of solving dual problems [optimal Dual solution]........................................ 6
4.5. Post-optimal [Simplex method sensitivity Analysis] ........................................................ 8

4. Duality and Sensitivity Analysis
The dual problem is defined systematically from the primal (or original) LP model. The two
problems are closely related, in the sense that the optimal solution of one problem automatically
provides the optimal solution to the other. As such, it may be advantageous computationally in
some cases to determine the primal solution by solving the dual.
The term 'Duality' implies that every linear programming problem, whether of maximization or
minimization, is associated with another linear programming problem based on the same data
which is called dual.
The primal problem is dealing with determining the number of units of the products, time etc.
While the dual problem deals with determining the unit worth (price) of the resource.
When taking the dual of a given LP, we refer to the given LP as the primal. If the primal is a max
problem, then the dual will be a min problem, and vice versa. For convenience, we define the
variables for the max problem to be Z, X1, X2, ..., Xn and the variables for the min problem to be
W, Y1, Y2, . . ., YM.
To find the dual to a max problem in which all the variables are required to be nonnegative and
all the constraints are ≤ constraints (called normal max problem) the problem may be written as:
4.1. Primal-dual relationship;
Primal Dual problem
𝑂𝑝𝑡𝑖𝑚𝑖𝑧𝑒(𝑀𝑎𝑥 𝑜𝑟 𝑀𝑖𝑛), 𝑍 = ∑ 𝐶𝑖 ∗ 𝑋𝑖
𝑛
𝑖
𝑂𝑝𝑡𝑖𝑚𝑖𝑧𝑒(𝑀𝑖𝑛𝑖 𝑜𝑟 𝑀𝑎𝑥), 𝑊 = ∑ 𝑏𝑖 ∗ 𝑦𝑖
𝑛
𝑖
𝑆. 𝑡
{
𝑎11𝑋1 + 𝑎12𝑋2 + . . . + 𝑎1𝑛𝑋𝑛 (≤, =, ≥)𝑏1
𝑎21𝑋1 + 𝑎22𝑋2 + . . . + 𝑎2𝑛𝑋𝑛 (≤, =, ≥)𝑏2
𝑎31𝑋1 + 𝑎32𝑋2 + . . . + 𝑎3𝑛𝑋𝑛 (≤, =, ≥)𝑏3
.
.
.
.
𝑎𝑚1𝑋1 + 𝑎𝑚2𝑋2 + . . . + 𝑎𝑚𝑛𝑋𝑛 (≤, =, ≥)𝑏𝑚
𝑆. 𝑡
{
𝑎11𝑌1 + 𝑎21𝑌2 + . . . + 𝑎𝑚1𝑌𝑚 (≥, =, ≤)𝐶1
𝑎12𝑌1 + 𝑎22𝑌2 + . . . + 𝑎𝑚𝑛𝑌𝑚 (≥, =, ≤)𝐶2
𝑎13𝑌1 + 𝑎23𝑋2 + . . . + 𝑎𝑚𝑛𝑋𝑛 (≥, =, ≤)𝐶3
.
.
.
.
𝑎1𝑛𝑌1 + 𝑎2𝑛𝑌2 + . . . + 𝑎𝑚𝑛𝑌𝑛 (≥, =, ≤)𝐶𝑚
The following is a summary of how the dual is constructed from the (equation form) primal:
I. A dual variable is assigned to each primal (equation) constraint and a dual constraint is
assigned to each primal variable.
II. The right-hand sides of the primal constraints provide the coefficients of the dual objective
function.
Table 4.1

Rules for constructing the dual problem
Primal problem objective
Dual problem
objective Constraint type Variable sign
Maximization Minimization ≥ Unrestricted
Minimization Maximization ≤ Unrestricted
✓ All primal constraints are equations with nonnegative right-hand sides, and all the
variables are nonnegative.
✓ A convenient way to remember the constraint type (≤ or ≥) in the dual is that if the dual
objective is a “pointing-down” minimization, then all the constraints are “pointing-up”
(≥) inequalities. The opposite applies when the dual objective is maximization.
III. The dual constraint corresponding to a primal variable is constructed by transposing the
primal variable column into a row with;
a. the primal objective coefficient becoming the dual right-hand side and
b. the remaining constraint coefficients comprising the dual left-hand side
coefficients.
IV. The sense of optimization, direction of inequalities, and the signs of the variables in the
dual are governed by the rules in Table 4.1
4.2. Rules for Constructing the Dual Problem
Maximization Problem Minimization Problem
constraints Variables
≥ ≤
≤ ≥
= Unrestricted
Variables constraints
≥ ≥
≤ ≤
Unrestricted =
Primal Dual
Objective is minimization Objective is maximization & vice versa
≥ type constraints ≤ type constraints
Number of columns Number of rows
Number of rows Number of columns
Number of decision variables Number of constraints
Number of constraints Number of decision variables
Coefficient of objective function RHS value
RHS values Coefficient of objective function

Example
Finding the Dual of a Normal Max /Min
Primal Dual
1. 𝑀𝑎𝑥 𝑍 = 60𝑋1 + 30𝑋2 + 20𝑋3
𝑆. 𝑡 {
8𝑋1 + 6𝑋2 + 𝑋3 ≤ 48
4𝑋1 + 2𝑋2 + 1.5𝑋3 ≤ 20
2𝑋1 + 1.5𝑋2 + 0.5𝑋3 ≤ 8
𝑋𝑖 ≥ 0
𝑀𝑖𝑛 𝑊 = 48𝑌1 + 20𝑌2 + 8𝑌3
𝑆. 𝑡 {
8𝑌1 + 4𝑌2 + 2𝑌3 ≥ 60
6𝑌1 + 2𝑌2 + 1.5𝑌3 ≥ 30
𝑌1 + 1.5𝑌2 + 0.5𝑌3 ≥ 20
𝑌𝑖 ≥ 0
2. 𝑀𝑎𝑥 𝑍 = 5𝑋1 + 6𝑋2
𝑆. 𝑡 {
3𝑋1 + 2𝑋2 ≤ 120
4𝑋1 + 6𝑋2 ≤ 260
𝑋1, 𝑋2 ≥ 0
𝑀𝑖𝑛 𝑤 = 120𝑌1 + 260𝑌2
𝑆. 𝑡 {
3𝑌1 + 4𝑌2 ≥ 4
2𝑌1 + 6𝑌2 ≥ 6
𝑌1, 𝑌2 ≥ 0
3. 𝑀𝑎𝑥 𝑍 = 𝑋1 − 2𝑋2+3𝑋3
𝑆. 𝑡 {
−2𝑋1 + 𝑋2 + 3𝑋3 = 2
2𝑋1 + 3𝑋2 + 4𝑋2 = 1
𝑋𝑖 ≥ 0
𝑀𝑖𝑛 𝑤 = 2𝑌1 + 𝑌2
𝑆. 𝑡 {
−2𝑌1 + 2𝑌2 ≥ 1
𝑌1 + 3𝑌2 ≥ −2
3𝑌1 + 4𝑌2 ≥ 3
𝑌1, 𝑌2 𝑎𝑟𝑒 𝑢𝑛𝑟𝑒𝑠𝑡𝑟𝑖𝑐𝑡𝑒𝑑 𝑖𝑛 𝑠𝑖𝑔𝑛
4. 𝑀𝑖𝑛 𝑤 = 𝑋1 − 3𝑋2 − 2𝑋3
𝑆. 𝑡 {
3𝑋1 − 𝑋2 + 2𝑋3 ≤ 7
2𝑋1 − 4𝑋2 ≥ 12
−4𝑋1 + 3𝑋2 + 8𝑋3 = 10
𝑋1, 𝑋2 ≥ 0 𝑎𝑛𝑑 𝑋3 = 𝑢𝑛𝑟𝑒𝑠𝑡𝑟𝑖𝑐𝑡𝑒𝑑 𝑖𝑛 𝑠𝑖𝑔𝑛
𝑀𝑎𝑥 𝑍 = 7𝑌1 + 12𝑌2 + 10𝑌3
𝑆. 𝑡 {
3𝑌1 + 2𝑌2 − 4𝑌3 ≥ 2
−𝑌1 − 4𝑌2 + 3𝑌3 ≤ 1
2𝑌1 + 8𝑌3 = 2
𝑌𝑖 ≥ 0
4.3. Economic interpretation of duality
Example: A Dakota work shop want to produce desk, table, and chair with the available resource of:
Timber, finishing hours and carpenter hours as revised in the table below. The selling price and
available resources are also revised in the table. Formulate this problem as Primal and Dual Problem?
[ Amare Matebu Kassa (Dr.-Ing)]
Resource Desk Table Chair Availability
Timber 8 board ft 6 board ft 1 board ft 48 boards fit
Finishing 4 hours 2 hours 1.5hours 20 hours
Carpentry 2hours 1.5hours 0.5 hours 8 hours
Selling price $60 $30 $20

Interpreting the Dual of the Dakota (Max) Problem;
The primal;
𝑀𝑎𝑥 𝑍 = 60𝑋1 + 30𝑋2 + 20𝑋3
8𝑋1 + 6𝑋2 + 𝑋3 ≤ 48 (Timber constriants)
4𝑋1 + 2𝑋2 + 1.5𝑋3 ≤ 20(Finishing constriants)
2𝑋1 + 1.5𝑋2 + 0.5𝑋3 ≤ 8 (Carpentry constriants)
𝑋1, 𝑋2, 𝑋3 ≥ 0
The dual;
Min w = 48Y1 + 20Y2 + 8Y3
8Y1 + 6Y2 + 2Y3 ≥ 60 (Desk constriants)
6Y1 + 2Y2 + 1.5Y3 ≥ 30(Table constriants)
Y1 + 1.5Y2 + 0.5Y3 ≥ 8 (Chair constriants)
Y1, Y2, Y3 ≥ 0
The first dual constraint is associated with desks, the second with tables, and the third with chairs.
Decision variable y1 is associated with Timber, y2 with finishing hours, and y3 with carpentry
hours. Suppose an entrepreneur wants to purchase all of Dakota’s resources. The entrepreneur
must determine the price he or she is willing to pay for a unit of each of Dakota’s resources.
To determine these prices, we define:
• y1 = price paid for 1 boards ft of lumber
• y2 = price paid for 1 finishing hour
• y3 = price paid for 1 carpentry hour
The resource prices y1, y2, and y3 should be determined by solving the Dakota dual.
The total price that should be paid for these resources is 48 y1 + 20y2 + 8y3. Since the cost of
purchasing the resources is to minimized:
Min w = 48y1 + 20y2 + 8y3 is the objective function for Dakota dual.
In setting resource prices, the prices must be high enough to induce Dakota to sell.
For example, the entrepreneur must offer Dakota at least $60 for a combination of resources
that includes 8 board feet of timber, 4 finishing hours, and 2 carpentry hours because Dakota
could, if it wished, use the resources to produce a desk that could be sold for $60. Since the
entrepreneur is offering 8y1 + 4y2 + 2y3 for the resources used to produce a desk, he or she must
choose y1, y2, and y3 to satisfy: 8y1 + 4y2 + 2y3 ≥ 60. Similar reasoning shows that at least $30
must be paid for the resources used to produce a table.
Thus y1, y2, and y3 must satisfy: 6y1 + 2y2 + 1.5y3 ≥ 30
Likewise, at least $20 must be paid for the combination of resources used to produce one chair.
Thus y1, y2, and y3 must satisfy: y1 + 1.5y2 + 0.5y3 ≥ 20. The solution to the Dakota dual yields
prices for timber, finishing hours, and carpentry hours.

4.4. Simple way of solving dual problems [optimal Dual solution]
The primal and dual solutions are closely related, in the sense that the optimal solution of either
problem directly yields the optimal solution to the other, as is explained subsequently. Thus, in an
LP model in which the number of variables is considerably smaller than the number of constraints,
computational savings may be realized by solving the dual because the amount of computations
associated with determining the inverse matrix primarily increases with the number of constraints.
Notice that the rule addresses only the amount of computations in each iteration but says nothing
about the total number of iterations needed to solve each problem.
This section provides two methods for determining the dual values.
Method 1.
(
𝑂𝑝𝑡𝑖𝑚𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑑𝑢𝑎𝑙
𝑣𝑎𝑟𝑎𝑖𝑏𝑙𝑒 𝑌𝑖
) = (
𝑂𝑝𝑡𝑖𝑚𝑎𝑙 𝑝𝑟𝑖𝑚𝑎𝑙 𝑍 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑋𝑖
+
𝑜𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑋𝑖
)
Example 1
𝑀𝑎𝑥 𝑍 = 2𝑋1 + 2𝑋2 + 5𝑋3 + 4𝑋4
S. t
𝑋1 + 3𝑋2 + 4𝑋3 + 3𝑋4 ≤ 10
4𝑋1 + 2𝑋2 + 6𝑋3 + 8𝑋4 ≤ 25
(
𝑣𝑎𝑟𝑎𝑖𝑏𝑙𝑒 𝑌1
) = (
4
5
+ 0 ) =
4
5
(
) =
3
10
+ 0 =
3
10
Method 2
(
𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓
𝑑𝑢𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
) = (
𝑅𝑜𝑤 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓
𝑜𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠
𝑜𝑓 𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑝𝑟𝑖𝑚𝑎𝑙 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
) ∗ (
𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑝𝑟𝑖𝑚𝑎𝑙
𝑖𝑛𝑣𝑒𝑟𝑠𝑒
)
Cj 2 2 5 4 0 0
C.B.V B.V X1 X2 X3 X4 S1 S2 solution
5 X3 0 1 1 2/5 2/5 - 1/10 1.5
2 X1 1 -1 0 1.4 - 3/5 2/5 4
Zj 2 3 5 24/5 4/5 3/10
Zj-Cj 0 1 0 4/5 4/5 3/10
Optimal table B
-1

(𝑌1 𝑌2) = (5 2) ∗ (
2
5
−
1
10
−
3
5
2
5
) = (5 ∗
2
5
−
3
5
∗ 2 5 ∗ (−
1
10
) +
2
5
∗ 2) = (
4
5
3
10
)
𝑀𝑎𝑥 𝑍 = 𝑀𝑖𝑛 𝑤
2𝑋1 + 2𝑋2 + 5𝑋3 + 4𝑋4 = 10𝑌1 + 25𝑌2
2 ∗ 4 + 2 ∗ 0 + 5 ∗ 1.5 + 4 ∗ 0 = 10 ∗ 0.9 + 25 ∗ 0.3 = 𝟏𝟓. 𝟓
Example 2
Maximize
22X1 + 6X2 + 2X3
Subject to:
10X1 + 2X2 + X3 ≤ 100
7X1 + 3X2 + 2X3 ≤ 72
2X1 + 4X2 + X3 ≤ 80
X1, X2, X3 ≥ 0
Max Z=22X1 + 6X2 + 2X3+0S1+0S2+0S3
Subject to:
10X1 + 2X2 + X3 + S1= 100
7X1 + 3X2 + 2X3+ S2= 72
2X1 + 4X2 + X3+ S3 = 80
X1, X2, X3, S1, S2, S3 ≥ 0
Optimal table
Cj 22 6 2 0 0 0
CBV
Basic
Variable
X1 X2 X3 S1 S2 S3
Basic
solution
Min
Ratio
22 X1 1 0 -0.06 0.19 -0.13 0 9.75
6 X2 0 1 0.81 -0.44 0.63 0 1.25
0 S3 0 0 -2.13 1.38 -2.25 1 55.5
Zj 22 6 3.5 1.5 1 0 222
Cj-Zj 0 0 -1.5 -1.5 -1 0
(
) = (1.54 + 0 ) = 1.5

(
) = 1 + 0 = 1
(
) = 0 + 0 = 0
Method 2;
(
𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓
𝑑𝑢𝑎𝑙 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
) = (
𝑅𝑜𝑤 𝑣𝑒𝑐𝑡𝑜𝑟 𝑜𝑓
𝑜𝑟𝑔𝑖𝑛𝑎𝑙 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠
𝑜𝑓 𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑝𝑟𝑖𝑚𝑎𝑙 𝑏𝑎𝑠𝑖𝑐 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
) ∗ (
𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑝𝑟𝑖𝑚𝑎𝑙
𝑖𝑛𝑣𝑒𝑟𝑠𝑒
)
(𝑌1 𝑌2 𝑌3) = (𝑋1 𝑋2 𝑆3) ∗ (
0.19 −0.13 0
−0.44 0.63 0
1.38 −2.25 1
)
(𝑌1 𝑌2 𝑌3) = (22 6 0) ∗ (
0.19 −0.13 0
−0.44 0.63 0
1.38 −2.25 1
)
(𝑌1 𝑌2 𝑌3) = (1.5 1 0)
Max z=Min W
222=100*1.5+72*1+0*80
222=222
4.5. Post-optimal [Simplex method sensitivity Analysis]
While solving a linear programming problem for optimal solution, we assume that:
a. Technology is fixed,
b. Fixed prices,
c. Fixed levels of resources or requirements,
d. The coefficients of variables in structural constraints (i.e. time required by a product
on a particular resource) are fixed,
e. profit contribution of the product will not vary during the planning period.
The condition in the real world however, might be different from those that are assumed by the
model. It is, therefore, desirable to determine how sensitive the optimal solution is to different
types of changes in the problem data and parameters.
Why we use sensitivity analysis?
(a) Sensitivity analysis allow us to determine how "sensitive" the optimal solution is to
changes in data values.
(b) Sensitivity analysis is important to the manager who must operate in a dynamic
environment with imprecise estimates of the coefficients.
(c) Sensitivity analysis is used to determine how the optimal solution is affected by changes,
within specified ranges, in:
i. the objective function coefficients (cj ), which include:
❖ Coefficients of basic variables.

❖ Coefficients of non-basic variables.
ii. the right-hand side (RHS) values (bi ), (i.e. resource or requirement levels).
iii. Change in the consumption rate (Technological coefficients)
The above changes may result in one of the following three cases
Case I. The optimal solution remains unchanged, that is the basic variables and
their values remain essentially unchanged.
Case II. The basic variables remain the same but their values are changed.
(d)
Case III. The basic solution changes completely.
𝑂𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛(𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑜𝑟 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑎𝑡𝑖𝑜𝑛), 𝑍 = 𝐶1𝑋1+𝐶2𝑋2+ . . . . . . . . +𝐶𝑛𝑋𝑛
𝑆. 𝑡
{
𝑎11𝑋1 + 𝑎12𝑋2 + . . . + 𝑎1𝑛𝑋𝑛 (≤, =, ≥)𝑏1
𝑎21𝑋1 + 𝑎22𝑋2 + . . . + 𝑎2𝑛𝑋𝑛 (≤, =, ≥)𝑏2
𝑎31𝑋1 + 𝑎32𝑋2 + . . . + 𝑎3𝑛𝑋𝑛 (≤, =, ≥)𝑏3
.
.
.
𝑎𝑚1𝑋1 + 𝑎𝑚2𝑋2 + . . . + 𝑎𝑚𝑛𝑋𝑛 (≤, =, ≥)𝑏𝑚
𝑋1, 𝑋2, , , 𝑋𝑛 ≥ 0
➢ Sensitivity of the optimal solution to the changes in the available resources, (i.e. the right
hand side RHS of the constraints bij)
➢ Sensitivity of the optimal solution to the changes in the unit profit or unit cost, (i.e. the
coefficient of the objective function Cij)
➢ Change in the consumption rate (Technological coefficients)
The right hand side of the constraint denotes present level of availability of resources (or
requirement in minimization problems). When this is increased or decreased, it will have effect
on the objective function and it may also change the basic variable in the optimal solution.
Example 1
𝑀𝑎𝑥 𝑍 = 2𝑋1 + 2𝑋2 + 5𝑋3 + 4𝑋4
S. t
𝑋1 + 3𝑋2 + 4𝑋3 + 3𝑋4 ≤ 10, 𝑀𝑎𝑛 − ℎ𝑜𝑢𝑟𝑠 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡𝑠
4𝑋1 + 2𝑋2 + 6𝑋3 + 8𝑋4 ≤ 25, 𝑀𝑎𝑐ℎ𝑖𝑛𝑒 ℎ𝑜𝑢𝑟𝑠
𝑋ij ≥ 0

Optimal table
Cj 2 2 5 4 0 0
5 X3 0 1 1 2/5 2/5 - 1/10 3/2
2 X1 1 -1 0 7/5 - 3/5 2/5 4
Zj 2 3 5 24/5 4/5 3/10
Zj-Cj 0 1 0 4/5 4/5 3/10
N.B. From this optimal table
➢ {X1, X3} are Basic variables (B.V) because there are in the solution
➢ {X2, X3} are Non Basic variables (N.B.V) because there are not in the solution
Solution X1 = 4; X2 = 0; X3 = 1.5; S1 = 0; S2 = 0; ) Z = 15.5
➢ Man-hours are completely utilized hence S1 = 0.
Machine hours are completely utilized, hence S2 = 0
I. The shadow price of the man-hours resource is $4/5. Hence it means to say that as we go
on increasing one hour of man-hour resource, the objective function will go on increasing
by $4/5 per hour.
II. Similarly, the shadow price per unit of machine hour is $3/10. Similar reasoning can be
given, that is every unit increase in machine hour resource will increase the objective
function by $3/10.
If the management want to increase the capacity of both man-hours and machine-hours, which
one should receive priority?
• The answer is man-hours, since it is shadow price is greater than the shadow price of
machine-hours.
If the management considers to increase man-hours by 10 hours i.e., from 10 hours to 20 hours
and machine hours by 20 hours i.e., 25 hours to 45 hours will the optimal solution remain
unchanged?
Use example 1 for more illustration
1. Change in the coefficient of objective function (Ci)
Case 1;Change in the coefficient of objective Non basic variable(N.B.V)
Δ Coeff of
Objective.
Function
Case 2;Change in the coefficient of objective basic variable(B.V)
Case 1; Change in the coefficient of objective Non basic variable (N.B.V)
a. Change in the coefficient of objective of X2 [C2]

2+Δ2
C2
2-Δ2
If the coefficient of X2 is changed then only Z2-C2 will change be changed and the other Zj-Cj along
the column are still constant. In addition, in order to do the sensitivity analysis, the current optimal
table should be optimal. So the optimal table is still optimal if Z2-C2 ≥0.
Case 1: Then if C2 =2+Δ2 [Maximum Increment]
then
𝑍2 − 𝐶2 ≥ 0
𝟑 − [𝟐 + 𝛥2] ≥ 0
𝟏 − 𝛥2 ≥ 0
𝜟𝟐 = 𝟏
𝑖𝑓 𝛥2 ≥ 1, 𝑡ℎ𝑒𝑛 𝑍2 − 𝐶2 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒, 𝑡ℎ𝑖𝑠 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑒𝑠 𝑡ℎ𝑒 𝑓𝑖𝑛𝑎𝑙 𝑡𝑎𝑏𝑙𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑜𝑝𝑡𝑖𝑚𝑎𝑙
𝐶2 = 𝟐 + 𝛥2 = 2 + 1 = 3 , 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑖𝑛𝑐𝑟𝑒𝑚𝑒𝑛𝑡
Case 2: Then if C2 =2-Δ2
𝑍2 − 𝐶2 ≥ 0
𝟑 − [𝟐 − 𝛥2] ≥ 0
𝟏 + 𝛥2 ≥ 0
𝛥2 ≥ −1,
𝐼𝑛 𝑜𝑟𝑑𝑒𝑟 𝑡𝑜 𝑑𝑜 𝑡ℎ𝑒 𝑠𝑒𝑛𝑠𝑡𝑖𝑣𝑖𝑡𝑦 𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠 𝑡ℎ𝑒 𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑡𝑎𝑏𝑙𝑒 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑟𝑒𝑚𝑎𝑖𝑛 𝑜𝑝𝑡𝑖𝑚𝑎𝑙, 𝑡ℎ𝑒𝑛
𝛥2 = ∞
Then
𝑪𝟐 = 𝟐 − ∞ = −∞
Then the range of optimality for the coefficient of non-basic variable X2 which is C2.
−∞ ≤ 𝑪𝟐 ≤ 𝟑
b. Change in the coefficient of objective of X4 [C4]
4+Δ4
C4
4-Δ4
Case 1: Then if C2 =4+Δ4 [Maximum Increment]

then
𝑍4 − 𝐶4 ≥ 0
𝟐𝟒
𝟓
− [𝟒 + 𝛥4] ≥ 0
𝟒
𝟓
− 𝛥4 ≥ 0
𝜟𝟒 =
𝟒
𝟓
𝑖𝑓 𝛥24 ≥
4
5
, 𝑡ℎ𝑒𝑛 𝑍4 − 𝐶4 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒, 𝑡ℎ𝑖𝑠 𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑒𝑠 𝑡ℎ𝑒 𝑓𝑖𝑛𝑎𝑙 𝑡𝑎𝑏𝑙𝑒 𝑖𝑠 𝑛𝑜𝑡 𝑜𝑝𝑡𝑖𝑚𝑎𝑙
𝑪𝟒 = 𝟒 +
𝟒
𝟓
=
𝟐𝟒
𝟓
, 𝒕𝒉𝒊𝒔 𝒊𝒔 𝒕𝒉𝒆 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒊𝒏𝒄𝒓𝒆𝒎𝒆𝒏𝒕
Case 2: Then if C4 =4-Δ4
𝑍4 − 𝐶4 ≥ 0
24
5
− [4 − 𝛥2] ≥ 0
4
5
+ 𝛥4 ≥ 0
𝜟𝟐 ≥ −
𝟒
𝟓
,
𝐼𝑛 𝑜𝑟𝑑𝑒𝑟 𝑡𝑜 𝑑𝑜 𝑡ℎ𝑒 𝑠𝑒𝑛𝑠𝑡𝑖𝑣𝑖𝑡𝑦 𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠 𝑡ℎ𝑒 𝑜𝑝𝑡𝑖𝑚𝑎𝑙 𝑡𝑎𝑏𝑙𝑒 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑟𝑒𝑚𝑎𝑖𝑛 𝑜𝑝𝑡𝑖𝑚𝑎𝑙, 𝑡ℎ𝑒𝑛
𝛥4 = ∞
Then
𝑪𝟒 = 𝟒 − ∞ = −∞
Then the range of optimality for the coefficient of non-basic variable X4 which is C4;
−∞ ≤ 𝑪𝟒 ≤
𝟐𝟒
𝟓
Case 2; Change in the coefficient of objective basic variable (B.V)
From the above optimal table, the basic variables are X1, X3, because these variables are within
the solution with the value of 4,3/2 respectively.
a. Change in the coefficient of X1,
2+Δ1
C1
2-Δ1
N.B. if the coefficient of the basic variable is changed, then the whole Zj-Cj value will be also
changed. Though a new value of Zj-Cj should be determined using the current optimal table.

Case (a.1) when the coefficient of X1 which is C1 is changed to 2+Δ1
Then the new values of Zj-Cj respective to each variable along the column are;
𝑍1 = (5 ∗ 0) + (2 + 𝛥1) ∗ 1 = 2 + 𝛥1
𝑍2 = 5 − (2 + 𝛥1) = 3 − 𝛥1
𝑍3 = 5 − (2 + 𝛥1) ∗ 0 = 5
𝑍4 = 2 +
7
5
∗ (2 + 𝛥1) =
24
5
+
7
5
𝛥1
𝑍5 = 2 −
3
5
(2 + 𝛥1) =
4
5
−
3
5
𝛥1
𝑍6 = −
1
2
+
2
5
(2 + 𝛥1) =
3
10
+
2
5
𝛥1
The determine Zj-Cj
𝑍1 − 𝐶1 ≥ 0
[𝟐 − 𝛥1] − [𝟐 − 𝛥1]
𝟎 = 0
𝛥1 = ∞
𝑍2 − 𝐶2 ≥ 0
[3 − 𝛥1] − 2 ≥ 0
1 − 𝛥1 ≥ 0
𝜟𝟏 = 𝟏
𝑍3 − 𝐶3 ≥ 0
5 − 5 ≥ 0
0 ≥ 0
𝛥1 = ∞
𝑍4 − 𝐶4 ≥ 0
[
24
5
+
7
5
𝛥1] − 4 ≥ 0
4
5
+
7
5
𝛥1 ≥ 0
𝛥1 = ∞
𝑍5 − 𝐶5 ≥ 0

[
4
5
−
3
5
𝛥1] − 0 ≥ 0
4
5
−
3
5
𝛥1 ≥ 0
𝛥1 = 4/3
𝑍6 − 𝐶6 ≥ 0
[
3
10
+
2
5
𝛥1] − 0 ≥ 0
3
10
+
2
5
𝛥1 ≥ 0
𝛥1 = ∞
Then the next step is selecting the value of Δ1
max 𝑖𝑛𝑐𝑟𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝛥1 = 𝑀𝑖𝑛 {∞, 1, ∞, ∞,
4
3
, ∞} = 1
𝛥1 = 1
Then 𝐶1 = 2 + 𝛥1 = 2 + 1 = 3
Case (a.2) when the coefficient of X1 which is C1 is changed to 2-Δ1
Performing the same analysis as case (a.1) then;
𝑍1 − 𝐶1 ≥ 0
5 − 5 ≥ 0
0 ≥ 0
𝛥1 = ∞
𝑍2 − 𝐶2 ≥ 0
[5 − (−1 ∗ (2 − 𝛥1)] − 2 ≥ 0
1 + 𝛥1 ≥ 0
𝛥1 = ∞
𝑍3 − 𝐶3 ≥ 0
5 − 5 ≥ 0
𝜟𝟏 = ∞
𝑍4 − 𝐶4 ≥ 0
[[
10
5
+
7
5
(2 − 𝛥1)] − 4 ≥ 0
4
5
−
7
5
𝛥1 ≥ 0
𝜟𝟏 =
𝟒
𝟕
𝑍5 − 𝐶5 ≥ 0
[[
10
5
−
3
5
(2 − 𝛥1)] − 0 ≥ 0

4
5
+
3
5
𝛥1 ≥ 0
𝛥1 = ∞
𝑍6 − 𝐶6 ≥ 0
[[
−5
10
+
2
5
(2 − 𝛥1)] − 0 ≥ 0
3
10
−
2
5
𝛥1 ≥ 0
𝜟𝟏 =
𝟑
𝟒
Then for determining the maximum decrement
𝑴𝒂𝒙 𝒅𝒆𝒄𝒓𝒆𝒂𝒎𝒆𝒏𝒕𝜟𝟏 = 𝑴𝒊𝒏{∞, ∞, ∞,
𝟒
𝟕
, ∞,
𝟑
𝟒
}
𝜟𝟏 =
𝟒
𝟕
𝑪𝟏 = 𝟐 −
𝟒
𝟕
= 𝟏𝟎/𝟕
Therefor the range of optimality for C1 is;
𝟏𝟎
𝟕
≤ 𝑪𝟏 ≤ 𝟑
2. Change in the RHS of constraints
✓ Let the initial RHS is a column matrix represented by” b”
✓ Let B is m by m matrix of optimal basic variable in the initial table (according their order)
✓ B-1
is the inverse matrix of B in which B* B-1
=I
✓ In the optimal simplex table B-1
is the matrix of slack and surplus variables coefficients.
Then the simplex iteration has the following important formula.
𝑩𝒂𝒔𝒊𝒄 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏[𝑩. 𝒔](𝒐𝒑𝒕𝒊𝒎𝒂𝒍) = 𝑩−𝟏
∗ 𝒃
𝑩𝒂𝒔𝒊𝒄 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏[𝑩. 𝒔](𝒐𝒑𝒕𝒊𝒎𝒂𝒍) = (
𝟐
𝟓
−
𝟏
𝟏𝟎
−
𝟑
𝟓
𝟐
𝟓
) ∗ [
𝟏𝟎
𝟐𝟓
] ≥ 0
Cj 2 2 5 4 0 0
5 X3 0 1 1 2/5 2/5 - 1/10 1.5
2 X1 1 -1 0 1.4 - 3/5 2/5 4
Zj 2 3 5 24/5 4/5 3/10
Zj-Cj 0 1 0 4/5 4/5 3/10
Optimal table B
-1

𝑋3 =
2
5
∗ 10 −
1
10
∗ 25 =
3
2
= 𝟏. 𝟓
𝑋1 = −
3
5
∗ 10 +
2
5
∗ 25 = 𝟒
Case
Case 1;Change in the R.H.S of constraints 1
Change in
RHS
Case 2;Change in the R.H.S of constraints 2
A. Change in the R.H.S of constraints 1
Let the RHS constraint one be changed by 𝜟𝟏,Then ;
✓ 𝐶𝑎𝑠𝑒 𝐴. 1; 𝒃𝟏 = 𝟏𝟎 + 𝜟𝟏
✓ 𝐶𝑎𝑠𝑒 𝐴. 2; 𝒃𝟏 = 𝟏𝟎 − 𝜟𝟏
𝐶𝑎𝑠𝑒 𝐴. 1; 𝒃𝟏 = 𝟏𝟎 + 𝜟𝟏,then
𝐁−𝟏
∗ 𝐛 = (
𝟐
𝟓
−
𝟏
𝟏𝟎
−
𝟑
𝟓
𝟐
𝟓
) ∗ [
𝟏𝟎 + 𝚫𝟏
𝟐𝟓
] ≥ 𝟎
2
5
∗ (10 + 𝛥1) −
1
10
∗ (25) ≥ 0
15
10
+
2
5
𝛥1 ≥ 0
𝜟𝟏 = ∞
−3
5
∗ (10 + 𝛥1) +
2
5
∗ (25) ≥ 0
20
5
−
3
5
𝛥1 ≥ 0
𝜟𝟏 =
𝟐𝟎
𝟓
𝜟𝟏 = 𝒎𝒊𝒏 {∞,
𝟐𝟎
𝟑
} =
𝟐𝟎
𝟑
𝒃𝟏 = 𝟏𝟎 + 𝜟𝟏 = 𝟏𝟎 +
𝟐𝟎
𝟑
= 𝟓𝟎/𝟑

𝐶𝑎𝑠𝑒 𝐴. 2; 𝒃𝟏 = 𝟏𝟎 − 𝜟𝟏
𝐁−𝟏
∗ 𝐛 = (
𝟐
𝟓
−
𝟏
𝟏𝟎
−
𝟑
𝟓
𝟐
𝟓
) ∗ [
𝟏𝟎 − 𝚫𝟏
𝟐𝟓
] ≥ 𝟎
2
5
∗ (10 − 𝛥1) −
1
10
∗ (25) ≥ 0
15
10
−
2
5
𝛥1 ≥ 0
𝜟𝟏 =
𝟏𝟓
𝟒
−3
5
∗ (10 − 𝛥1) +
2
5
∗ (25) ≥ 0
20
5
+
3
5
𝛥1 ≥ 0
𝜟𝟏 = ∞
𝛥1 = 𝑚𝑖𝑛 {∞,
15
4
} =
15
4
𝑏1 = 10 − 𝛥1 = 10 −
15
4
=
25
4
𝟐𝟓
𝟒
≤ 𝒃𝟏 ≤
𝟓𝟎
𝟑
, 𝒓𝒂𝒏𝒈𝒆 𝒐𝒇 𝒇𝒆𝒂𝒔𝒊𝒃𝒍𝒊𝒕𝒚 𝒇𝒐𝒓 𝒄𝒐𝒏𝒔𝒕𝒓𝒂𝒊𝒏𝒕 𝒐𝒏𝒆
N.B. then we can increase constraint 1 up to 50/3 and we can decrease up to 25/4
B. Change in the R.H.S of constraints 2
Let the RHS constraint one be changed by 𝜟𝟐,Then ;
✓ 𝐶𝑎𝑠𝑒 𝐵. 1; 𝒃𝟐 = 𝟐𝟓 + 𝜟𝟐
✓ 𝐶𝑎𝑠𝑒 𝐵. 2; 𝒃𝟐 = 𝟐𝟓 − 𝜟𝟐
𝐶𝑎𝑠𝑒 𝐴. 2; 𝒃𝟏 = 𝟐𝟓 + 𝜟𝟐

𝐁−𝟏
∗ 𝐛 = (
𝟐
𝟓
−
𝟏
𝟏𝟎
−
𝟑
𝟓
𝟐
𝟓
) ∗ [
𝟏𝟎
𝟐𝟓 + 𝚫𝟐
] ≥ 𝟎
2
5
∗ (10) −
1
10
∗ (25 + 𝜟𝟐) ≥ 0
15
10
−
1
10
𝛥2 ≥ 0
𝜟𝟐 = 𝟏𝟓
−3
5
∗ (10) −
2
5
∗ (25 + 𝜟𝟐) ≥ 0
−30
5
+
50
5
+
2
5
𝛥2 ≥ 0
𝜟𝟐 = ∞
𝜟𝟐 = 𝒎𝒂𝒙 𝒊𝒏𝒄𝒓𝒆𝒂𝒎𝒆𝒏𝒕 = 𝒎𝒊𝒏{𝟏𝟓, ∞} = 𝟏𝟓, 𝒕𝒉𝒆𝒏
𝒃𝟐 = 𝟐𝟓 + 𝟏𝟓 = 𝟒𝟎
𝐶𝑎𝑠𝑒 𝐵. 2; 𝒃𝟐 = 𝟐𝟓 − 𝜟𝟐
𝐁−𝟏
∗ 𝐛 = (
𝟐
𝟓
−
𝟏
𝟏𝟎
−
𝟑
𝟓
𝟐
𝟓
) ∗ [
𝟏𝟎
𝟐𝟓 − 𝚫𝟐
] ≥ 𝟎
2
5
∗ (10) −
1
10
∗ (25 − 𝜟𝟐) ≥ 0
15
10
+
1
10
𝛥2 ≥ 0
𝜟𝟐 = ∞
−3
5
∗ (10) −
2
5
∗ (25 − 𝜟𝟐) ≥ 0

20
5
−
2
5
(𝜟𝟐) ≥ 0
𝜟𝟐=10
𝜟𝟐 = 𝒎𝒊𝒏{∞, 𝟏𝟎} = 𝟏𝟎
𝒃𝟐 = 𝟐𝟓 − 𝟏𝟎 = 𝟏𝟓 then range of feasibility for constraint two;
𝟏𝟓 ≤ 𝒃𝟐 ≤ 𝟒𝟎
3. Change in the technological coeffiecnt [Consumption rate]
✓ Let Xi is initial column matrix of variables
✓ Xj is the optimal column matrix of variables
𝑿𝒋
⃗⃗⃗⃗ = 𝐁−𝟏
∗ 𝑿𝒊
𝑋1 = [
𝟏
𝟒
]
𝑋1 = [
1 + 𝛥1
4
] = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑖𝑛𝑐𝑟𝑒𝑎𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑜𝑒𝑓𝑓. 𝑜𝑓 𝑋1 𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡 1
𝑋1 = [
1 − 𝛥1
4
] = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑐𝑟𝑒𝑎𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑜𝑒𝑓𝑓. 𝑜𝑓 𝑋1 𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡 1
𝑋1 = [
1
4 + 𝛥1
] = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑖𝑛𝑐𝑟𝑒𝑎𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑜𝑒𝑓𝑓. 𝑜𝑓 𝑋1 𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡 2
𝑋1 = [
1
4 − 𝛥1
] = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑐𝑟𝑒𝑎𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑜𝑒𝑓𝑓. 𝑜𝑓 𝑋1 𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑟𝑎𝑖𝑛𝑡 2
Let coefficient of X1 in the first constraint changed by Δ1; Then
𝐶𝑎𝑠𝑒1; 𝑎11𝑐ℎ𝑎𝑛𝑔𝑒𝑑 𝑡𝑜 𝑎11 + 𝛥1 = 1 + 𝛥1
𝐶𝑎𝑠𝑒2; 𝑎11𝑐ℎ𝑎𝑛𝑔𝑒𝑑 𝑡𝑜 𝑎11 − 𝛥1 = 1 − 𝛥1
𝑪𝒂𝒔𝒆𝟏; 𝒂𝟏𝟏𝒄𝒉𝒂𝒏𝒈𝒆𝒅 𝒕𝒐 𝒂𝟏𝟏 + 𝜟𝟏 = 𝟏 + 𝜟𝟏
𝑿𝒋
⃗⃗⃗⃗ = 𝐁−𝟏
∗ 𝑿𝒊

𝑿𝟏
⃗⃗⃗⃗⃗ = (
𝟐
𝟓
−
𝟏
𝟏𝟎
−
𝟑
𝟓
𝟐
𝟓
) ∗ [
𝟏 + 𝜟𝟏
𝟒
]
2
5
+
2
5
𝛥1 −
4
10
=
𝟐
𝟓
𝜟𝟏
−
3
5
−
3
5
𝛥1 +
2
5
∗ 4 = 𝟏 −
𝟑
𝟓
𝜟𝟏
𝑋1
⃗⃗⃗⃗ = (
2
5
−
1
10
−
3
5
2
5
) ∗ [
1 + 𝛥1
4
] = (
𝟐
𝟓
𝜟𝟏
𝟏 −
𝟑
𝟓
𝜟𝟏
)
If 𝑿𝟏
⃗⃗⃗⃗⃗ will changed then Zj-Cj will also change
Z1-C1 ≥ 0
5 ∗
2
5
𝛥1 + 2 ∗ (1 −
3
5
𝛥1) − 2 ≥ 0
4
5
𝛥1 + 2 − 2 ≥ 0
𝛥1 = ∞
𝑎11 + 𝛥1 = 1 + ∞ = ∞
𝑪𝒂𝒔𝒆𝟐;𝒂𝟏𝟏𝒄𝒉𝒂𝒏𝒈𝒆𝒅 𝒕𝒐 𝒂𝟏𝟏 − 𝜟𝟏 = 𝟏 − 𝜟𝟏
𝑋1
⃗⃗⃗⃗ = (
2
5
−
1
10
−
3
5
2
5
) ∗ [
1 − 𝛥1
4
]
2
5
−
2
5
𝛥1 −
4
10
= −
𝟐
𝟓
𝜟𝟏
−
3
5
+
3
5
𝛥1 +
2
5
∗ 4 = 𝟏 +
𝟑
𝟓
𝜟𝟏
❖ 𝑋1
⃗⃗⃗⃗ = (
2
5
−
1
10
−
3
5
2
5
) ∗ [
1 + 𝛥1
4
] = (
−
𝟐
𝟓
𝜟𝟏
𝟏 +
𝟑
𝟓
𝜟𝟏
)
If 𝑿𝟏
⃗⃗⃗⃗⃗ will changed then Zj-Cj will also change

Z1-C1 ≥ 0
5 ∗ (−
2
5
𝛥1) + 2 ∗ (1 +
3
5
𝛥1) − 2 ≥ 0
−
4
5
𝛥1 + 2 − 2 ≥ 0
−
4
5
𝛥1+≥ 0
𝛥1 = 0
𝒂𝟏𝟏 = 𝟏 − 𝛥1 = 1
𝟏 ≤ 𝒂𝟏𝟏 ≤ ∞, 𝒄𝒐𝒏𝒔𝒖𝒑𝒕𝒊𝒐𝒏 𝒓𝒂𝒏𝒈𝒆 𝒐𝒇 𝑿𝟏 𝒐𝒏 𝒄𝒐𝒏𝒔𝒕𝒓𝒂𝒊𝒏𝒕 𝟏
Let coefficient of X2 in the first constraint is changed by Δ2;
𝐶𝑎𝑠𝑒1; 𝑎12𝑐ℎ𝑎𝑛𝑔𝑒𝑑 𝑡𝑜 𝑎12 + 𝛥2 = 3 + 𝛥2
𝐶𝑎𝑠𝑒2; 𝑎12𝑐ℎ𝑎𝑛𝑔𝑒𝑑 𝑡𝑜 𝑎12 − 𝛥2 = 3 − 𝛥2
𝑪𝒂𝒔𝒆𝟏; 𝒂𝟏𝟐𝒄𝒉𝒂𝒏𝒈𝒆𝒅 𝒕𝒐 𝒂𝟏𝟐 + 𝜟𝟏 = 𝟑 + 𝜟𝟐
𝑋2
⃗⃗⃗⃗ = (
2
5
−
1
10
−
3
5
2
5
) ∗ [
3 + 𝛥2
2
] = (
1 +
2
5
𝜟𝟐
−1 −
3
5
𝜟𝟐
)
Z2-C2 ≥ 0
5 ∗ (1 +
2
5
𝛥2) + 2 ∗ (−1 −
3
5
𝛥2) − 2 ≥ 0
3 +
4
5
𝛥2 ≥ 0
𝛥2 = ∞
𝑎12 + 𝛥2 = 3 + ∞ = ∞
𝑪𝒂𝒔𝒆𝟐; 𝒂𝟏𝟐𝒄𝒉𝒂𝒏𝒈𝒆𝒅 𝒕𝒐 𝒂𝟏𝟐 − 𝜟𝟐 = 𝟑 − 𝜟𝟐
𝑋2
⃗⃗⃗⃗ = (
2
5
−
1
10
−
3
5
2
5
) ∗ [
3 − 𝛥2
2
] = (
1 −
2
5
𝜟𝟐
−1 +
3
5
𝜟𝟐
)
Z2-C2 ≥ 0
5 ∗ (1 −
2
5
𝛥2) + 2 ∗ (−1 +
3
5
𝛥2) − 2 ≥ 0
1 −
4
5
𝛥2 ≥ 0

𝛥2 =
5
4
𝑎12 − 𝛥2 = 3 −
5
4
=
7
4
𝟕
𝟒
≤ 𝒂𝟏𝟐 ≤ ∞, 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑋2𝑜𝑛 𝑐𝑜𝑛𝑠𝑡𝑟𝑖𝑎𝑛𝑡 1

Exercise
𝑀𝑎𝑥 𝑍 = 12𝑋1 + 3𝑋2 + 𝑋3
S. t
10𝑋1 + 2𝑋2 + 𝑋3+≤ 100
7𝑋1 + 3𝑋2 + 2𝑋3 ≤ 77
2𝑋1 + 4𝑋2 + 𝑋3 ≤ 80
𝑋1, 𝑋2, 𝑋3 ≥ 0
Optimal table
Cj 12 3 1 0 0 0
C.B. V B. V X1 X2 X3 S1 S2 S3 SOLUTION
12 X1 1 0 - 1/16 3/16 - 1/8 0 73/8
3 X2 0 1 13/16 - 7/16 5/8 0 35/8
0 S3 0 0 -2 11/8 -9/4 1 177/4
Zj 12 3 27/16 15/16 3/8 0
Zj-Cj 0 0 18/16 15/16 3/8 0
i. Determine the dual values
ii. Determine the range of optimality of C1, C2 and C3 (change in the objective function
coefficient)
iii. Determine the range of feasibility b1 (change in the RHS constraints)
iv. Determine the range of optimality of the consumption rate (a11)

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