Operations Research

Lecture(1)

BY: DR. HEMAT MOUSTAFA

Definition of the Operations Research (O.R.)

Operations research is the application of scientific

methods, techniques and tools to make optimal
decisions.
Course Objective

The objective of this course is to introduce business

students to the basics of the linear programming approach
from both theoretical and practical perspectives.
Course Textbook

Lecture Notes on:

Introduction to Operations Research

Prof. Dr. Afaf El-Dash & Dr. Nada M. Hafez

Evaluation

Final Exam: 50%

Midterm

Classwork

Total: 100%
 Chapter (1): Introduction

1-1) Origin of operations Research (O.R.).

1-2) O.R. Models

1-3) Scope of O.R.

Origin of operations research (O.R.)
• Operations research is a scientific approach to problem solving.

• Operations research originated during World War II. The first problem tackled
systematically was determining the timing for bombs dropped from aircraft to hit
submarines

• During the Second World War, the military management in England invited a team
of experts to analyse defense-related problems, as the country faced resource
constraints and needed to maximize efficiency. These experts were tasked with
finding the most effective way to utilize available military resources. After extensive
analysis, they developed a problem-solving method, which they termed linear
programming.
After the war, O.R. techniques were applied to industrial and economic challenges,
helping optimize productivity and cost-efficiency. The U.S. military and industries later
adopted these methods, leading to widespread applications in military, civil, and
industrial sectors. Over time, O.R. evolved with various analytical techniques, with the
primary goal of optimizing profits and minimizing costs.
O. R. Model
• An O. R. model is defined as idealized (simplified) representation of a real-life system.
This system may already be in existence or may still be an idea awaiting execution. In the
first case, the models objective is to analyse the behaviour of the system in order to
improve its performance. In the second, the objective is to identify the “best” structure of
the future system.

• An O.R. model concentrates on identifying the variables and the relationships governing
them.

• There are three types of solving methods to solve the O. R. models:

• 1. Analytical methods

• 2. Iterative methods

• 3. Simulation methods as the Monto-carlo method.

Scope of O. R.

• there is no limit for the application of O.R. techniques; they may be

applied to any type of problems.
Thank You
Operations Research
Lecture(2)

BY: DR. HEMAT MOUSTAFA

Linear Programming Models (LPM)
Steps of formulation LPP

The following steps are involved in the formulation of LPP:

Step 1: Identify the decision variables of the problem that we seek to determine.

Step 2: Construct the objective (goal) that we need to optimize (maximize or minimize).

Step 3: Identify the constraints that the solution must satisfy, such as resources limitations,
interrelations between variables, etc. Formulation these constraints as linear inequalities or
equations in terms of the non-negative decision variables.

Thus, LPP is a collection of the objective function, the set of constraints. Let us study some
examples.
Example (1):
• A company produces two products, A and B, both of which require processing on two machines.
• Product A requires 10 minutes on the first machine and 15 minutes on the second machine per unit, while product
B requires 22 minutes on the first machine and 18 minutes on the second machine per unit. The total available
time on each machine is 2640 minutes per week.
• The selling prices per unit are $200 for product A and $175 for product B. The objective is to formulate a linear
programming (LP) model to maximize revenue. Additionally, the market can accommodate a maximum of 150
units of product A.
• Tabulated Form:
Tabulated Form:

Processing Time Required Selling Price Market

(minutes) ($l unit) Constraint
(units)
Machine- 1 Machine- 2
Product- A 10 15 200 150
Product - B 22 18 175 No limit
Availability of 2640 minutes 2640 minutes
Machines per week per week
• Linear Programming Formulation:

• Decision Variables:

Let 𝑋1 = Number of units of Product A produced per week.

Let 𝑋2 = Number of units of Product B produced per week.

• Objective Function:

Maximize revenue:
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 200𝑋1 + 175𝑋2

• Constraints:

1- Machine 1 Time Constraint:

10 𝑋1 + 22 𝑋2 ≤ 2640
2- Machine 2 Time Constraint:
15 𝑋1 + 18 𝑋2 ≤ 2640
3- Market Constraint for Product A:
𝑋1 ≤ 150
4- Non-Negativity Constraints:
𝑋1 ≥ 0, 𝑋2 ≥ 0

• The LPP is
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 200𝑋1 + 175𝑋2

Subject to
10 𝑋1 + 22 𝑋2 ≤ 2640
15 𝑋1 + 18 𝑋2 ≤ 2640
𝑋1 ≤ 150
𝑋1 ≥ 0, 𝑋2 ≥ 0
Example (2)

A farmer needs to mix two types of animal feed, Feed X and Feed Y, to meet the minimum daily
nutritional requirements for his livestock. Each kilogram of Feed X costs 5 and contains 3 units of
protein and 2 units of vitamins. Each kilogram of Feed Y costs 4 and contains 1 unit of protein and 4
units of vitamins. The livestock requires at least 9 units of protein and 8 units of vitamins per day. The
farmer wants to determine the optimal mix of Feed X and Feed Y to minimize costs while meeting the
nutritional requirements.
• Linear Programming Formulation:

1- Decision Variables:

Let 𝑋1 = kilograms of Feed X to use. Let 𝑋2 = kilograms of Feed Y to use.

2- Objective Function:
Minimize cost:
𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍 = 5 𝑋1 + 4 𝑋2

3- Constraints:

Protein requirement:
3 𝑋1 + 𝑋2 ≥ 9
Vitamin requirement:
2 𝑋1 + 4 𝑋2 ≥ 8
Non-negativity constraints:
𝑋1 ≥ 0, 𝑋2 ≥ 0
The LPP is
𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍 = 5 𝑋1 + 4 𝑋2

Subject to:
3 𝑋1 + 𝑋2 ≥ 9
2 𝑋1 + 4 𝑋2 ≥ 8
𝑋1 ≥ 0, 𝑋2 ≥ 0
Example (3)

A bakery produces two types of cakes: Cake X and Cake Y. Each Cake X requires 2 hours of baking
time and 1 hour of decorating time. Each Cake Y requires 1 hour of baking time and 2 hours of
decorating time. The bakery has a maximum of 10 hours of baking time and 12 hours of decorating time
available per day.

The profit from selling Cake X is 30, and the profit from selling Cake Y is 20. The bakery wants to
determine how many cakes of each type to produce to maximize profit.
Linear Programming Formulation:

1- Decision Variables:
Let 𝑋1 = number of Cake X to produce. Let 𝑋2 = number of Cake Y to produce.
2- Objective Function:

Maximize profit:
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 30 𝑋1 + 20 𝑋2
3- Constraints:

Baking time constraint:

2 𝑋1 + 𝑋2 ≤ 10
Decorating time constraint:
𝑋1 +2 𝑋2 ≤ 12
Non-negativity constraints:
𝑋1 ≥ 0, 𝑋2 ≥ 0
The LLP is
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 30 𝑋1 + 20 𝑋2

Subject to:
2 𝑋1 + 𝑋2 ≤ 10
𝑋1 +2 𝑋2 ≤ 12
𝑋1 ≥ 0, 𝑋2 ≥ 0
Remarks. The objective and the constraint function in all LPs must be linear. Additionally, all the
parameters (coefficients of the objective and constraint functions) of the model are known with
certainty.

Graphical LP Solution

The graphical solution includes two steps:

1. Determination of the feasible solution space.

2. Determination of the optimum solution from among all the points in the solution space.
Example (4)

For the following LP problems, graph the region of feasible solutions and determine the optimal
solution
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 3 𝑋1 + 5 𝑋2

​Subject to:
𝑋1 +2 𝑋2 ≤ 8,
3 𝑋1 + 2 𝑋2 ≤ 12,
𝑋1 ≥ 0, 𝑋2 ≥ 0
Graphical LP Solution

Step 1: Plot the Constraints

Constraint 1:
𝑋1 +2 𝑋2 ≤ 8

Convert to equality: 𝑋1 +2 𝑋2 = 8
Find intercepts:

When 𝑋1 =0, 𝑋2 =4. Point: (0,4).

When 𝑋2 =0, 𝑋1 =8. Point: (8,0).
Plot the line connecting (0,4) and (8,0).
Shade the region below the line (since it's ≤).
Constraint 2:
3 𝑋1 + 2 𝑋2 ≤ 12

Convert to equality: 3 𝑋1 + 2 𝑋2 = 12.

Find intercepts:

When 𝑋1 =0, 𝑋2 =6. Point: (0,6).

When 𝑋2 =0, 𝑋1 =4. Point: (4,0).
Plot the line connecting (0,6) and (4,0).
Shade the region below the line (since it's ≤).

Non-Negativity Constraints: 𝑋1 ≥ 0, 𝑋2 ≥ 0
1. Restrict the graph to the first quadrant.
Corner point points Z= 3 x1+5 x2 Optimal solution
A (0,4) 20

B (2,3) 21 X1=2, x2=3, z=21

C (4,0) 12
D (0,0) 0
Thank You
Operations Research
Lecture(3)

BY: DR. HEMAT MOUSTAFA

Graphical LP Solution

The graphical solution includes two steps:

1. Determination of the feasible solution space.

2. Determination of the optimum solution from among all the points in the solution space.
For the following LP problems, graph the region of feasible solutions and determine the optimal
solution

Example (1)

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 3 𝑋1 + 5 𝑋2

​Subject to:
𝑋1 +2 𝑋2 ≤ 8,
3 𝑋1 + 2 𝑋2 ≤ 12,
𝑋1 ≥ 0, 𝑋2 ≥ 0
Graphical LP Solution

Step 1: Plot the Constraints

Constraint 1:
𝑋1 +2 𝑋2 ≤ 8

Convert to equality: 𝑋1 +2 𝑋2 = 8
Find intercepts:

When 𝑋1 =0, 𝑋2 =4. Point: (0,4).

When 𝑋2 =0, 𝑋1 =8. Point: (8,0).
Plot the line connecting (0,4) and (8,0).
Shade the region below the line (since it's ≤).

• Test the origin (0,0):

• 0 + 0 ≤ 8 ⟹ 0 ≤ 8 (True) Since the origin satisfy the inequality, shade the region that includes the origin
(below the line).
Constraint 2:
3 𝑋1 + 2 𝑋2 ≤ 12

Convert to equality: 3 𝑋1 + 2 𝑋2 = 12.

Find intercepts:

When 𝑋1 =0, 𝑋2 =6. Point: (0,6).

When 𝑋2 =0, 𝑋1 =4. Point: (4,0).
Plot the line connecting (0,6) and (4,0).
Shade the region below the line (since it's ≤).

Non-Negativity Constraints: 𝑋1 ≥ 0, 𝑋2 ≥ 0
1. Restrict the graph to the first quadrant.
Corner point points Z= 3 x1+5 x2 Optimal solution
A (0,4) 20

B (2,3) 21 X1=2, x2=3, Z=21

C (4,0) 12
D (0,0) 0
Example (2):

𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍 = 2 𝑋1 + 3 𝑋2

​Subject to:
𝑋1 + 𝑋2 ≤ 4,
2 𝑋1 + 𝑋2 ≥ 6,
𝑋1 ≥ 0, 𝑋2 ≥ 0
Graphical LP Solution

Step 1: Plot the Constraints

Constraint 1:
𝑋1 + 𝑋2 ≤ 4

Convert to equality: 𝑋1 + 𝑋2 = 4
Find intercepts:

When 𝑋1 =0, 𝑋2 =4. Point: (0,4).

When 𝑋2 =0, 𝑋1 =4. Point: (4,0).
Plot the line connecting (0,4) and (4,0).
Shade the region below the line (since it's ≤).

• Test the origin (0,0):

• 0 + 0 ≤ 4 ⟹ 0 ≤ 4 (True) Since the origin satisfy the inequality, shade the region that includes the origin
(below the line).
Constraint 2:
2 𝑋1 + 𝑋2 ≥ 6

Convert to equality: 2𝑋1 + 𝑋2 = 6.

Find intercepts:

When 𝑋1 =0, 𝑋2 =6. Point: (0,6).

When 𝑋2 =0, 𝑋1 =3. Point: (3,0).
Plot the line connecting (0,6) and (3,0).

Shade the region above the line (since it's ≥).

Non-Negativity Constraints: 𝑋1 ≥ 0, 𝑋2 ≥ 0
1. Restrict the graph to the first quadrant.
Corner point points Z= 2 x1+3 x2 Optimal solution
A (3,0) 6 X1=3, x2=0, Z=6

B (2,2) 10
C (4,0) 8
Example (3):

𝑀𝐴𝑥 𝑍 = 350 𝑋1 + 300 𝑋2

​Subject to:
𝑋1 + 𝑋2 ≤ 200,
9 𝑋1 + 6 𝑋2 ≤ 1566
12 𝑋1 + 16 𝑋2 ≤ 2880
𝑋1 ≥ 0, 𝑋2 ≥ 0
Graphical LP Solution

Step 1: Plot the Constraints

Constraint 1:
𝑋1 + 𝑋2 ≤ 200

Convert to equality: 𝑋1 + 𝑋2 = 200

Find intercepts:

When 𝑋1 =0, 𝑋2 =200. Point: (0,200).

When 𝑋2 =0, 𝑋1 =200. Point: (200,0).
Plot the line connecting (0,200) and (200,0).
Shade the region below the line (since it's ≤).

• Test the origin (0,0):

• 0 + 0 ≤ 200 ⟹ 0 ≤ 200 (True) Since the origin satisfy the inequality, shade the region that includes the origin
(below the line).
Constraint 2:
9 𝑋1 + 6 𝑋2 ≤ 1566

Convert to equality: 9 𝑋1 + 6𝑋2 = 1566

Find intercepts:

When 𝑋1 =0, 𝑋2 =261. Point: (0,261).

When 𝑋2 =0, 𝑋1 =174. Point: (174,0).
Plot the line connecting (0,261) and (174,0).
Shade the region below the line (since it's ≤).

• Test the origin (0,0):

• 0 + 0 ≤ 1566 ⟹ 0 ≤ 1566 (True) Since the origin satisfy the inequality, shade the region that includes the
origin (below the line).
Constraint 3:
12 𝑋1 + 16 𝑋2 ≤ 2880

Convert to equality: 12 𝑋1 + 16𝑋2 = 2880

Find intercepts:

When 𝑋1 =0, 𝑋2 =180. Point: (0,180).

When 𝑋2 =0, 𝑋1 =240. Point: (240,0).
Plot the line connecting (0, 180) and (240 ,0).
Shade the region below the line (since it's ≤).

• Test the origin (0,0):

• 0 + 0 ≤ 2880 ⟹ 0 ≤ 2880 (True) Since the origin satisfy the inequality, shade the region that includes the
origin (below the line).
Point B
C1 𝑋1 + 𝑋2 = 200
𝑋1 = 200 − 𝑋2
C3 12 𝑋1 + 16𝑋2 = 2880
12 (200− 𝑋2 )+ 16𝑋2 = 2880
2400 − 12 𝑋2 + 16𝑋2 = 2880
4𝑋2 = 2880 − 2400
4𝑋2 = 480
480
𝑋2 = = 120
4
𝑋1 = 200 − 120 = 80
B (80, 120)

Point C
C1 𝑋1 + 𝑋2 = 200
𝑋1 = 200 − 𝑋2
C2 9 𝑋1 + 6𝑋2 = 1566
9 (200− 𝑋2 )+ 6𝑋2 = 1566
1800 − 9𝑋2 + 6𝑋2 = 1566
3𝑋2 = 1800 − 1566
3𝑋2 = 234
234
𝑋2 = = 78
3
𝑋1 = 200 − 78 = 122
C (122, 78)
Corner point points Z= 350 x1+300 x2 Optimal solution
A (0,180) 54000

B (80, 120) 64000

C (122, 78) 66100 X1=122, x2=78, Z=66100
D (174, 0) 60900
E (0, 0) 0
special conditions in LP models

1- alternate optimal solutions ( many optimal solutions )

There are more than one feasible point that maximize (minimize) the value
of the objective function.
Example (4): 450 rather than 350

𝑀𝐴𝑋 𝑍 = 450 𝑋1 + 300 𝑋2

​Subject to:
𝑋1 + 𝑋2 ≤ 200,
9 𝑋1 + 6 𝑋2 ≤ 1566
12 𝑋1 + 16 𝑋2 ≤ 2880
𝑋1 ≥ 0, 𝑋2 ≥ 0
Corner point points Z= 450 x1+300 x2 Optimal solution
A (0,180) 54000

B (80, 120) 72000

C (122, 78) 78300 X1=122, X2=78, Z=78300
D (174, 0) 78300 X1=174,X2=0, Z=78300
E (0, 0) 0

Note that: All the points on the line segment joining the corner point (122,78) To the corner
point at (174,0) produce the same optimal objective function value of 78,300
special conditions in LP models

2- Redundant constraints :
Constraint plays no role in determining the feasible region of the problem
Example (5):

𝑀𝐴𝑋 𝑍 = 350 𝑋1 + 300 𝑋2

​Subject to:
𝑋1 + 𝑋2 ≤ 225,
9 𝑋1 + 6 𝑋2 ≤ 1566
12 𝑋1 + 16 𝑋2 ≤ 2880
𝑋1 ≥ 0, 𝑋2 ≥ 0
Graphical LP Solution

Step 1: Plot the Constraints

Constraint 1:
𝑋1 + 𝑋2 ≤ 225

Convert to equality: 𝑋1 + 𝑋2 = 225

Find intercepts:

When 𝑋1 =0, 𝑋2 =225. Point: (0,225).

When 𝑋2 =0, 𝑋1 =225. Point: (225,0).
Plot the line connecting (0,200) and (200,0).
Shade the region below the line (since it's ≤).

• Test the origin (0,0):

• 0 + 0 ≤ 200 ⟹ 0 ≤ 200 (True) Since the origin satisfy the inequality, shade the region that includes the origin
(below the line).
Constraint 2:
9 𝑋1 + 6 𝑋2 ≤ 1566

Convert to equality: 9 𝑋1 + 6𝑋2 = 1566

Find intercepts:

When 𝑋1 =0, 𝑋2 =261. Point: (0,261).

When 𝑋2 =0, 𝑋1 =174. Point: (174,0).
Plot the line connecting (0, 261) and (174, 0).
Shade the region below the line (since it's ≤).

• Test the origin (0,0):

• 0 + 0 ≤ 1566 ⟹ 0 ≤ 1566 (True) Since the origin satisfy the inequality, shade the region that includes the
origin (below the line).
Constraint 3:
12 𝑋1 + 16 𝑋2 ≤ 2880

Convert to equality: 12 𝑋1 + 16𝑋2 = 2880

Find intercepts:

When 𝑋1 =0, 𝑋2 =180. Point: (0,180).

When 𝑋2 =0, 𝑋1 =240. Point: (240,0).
Plot the line connecting (0, 180) and (240 ,0).
Shade the region below the line (since it's ≤).

• Test the origin (0,0):

• 0 + 0 ≤ 2880 ⟹ 0 ≤ 2880 (True) Since the origin satisfy the inequality, shade the region that includes the
origin (below the line).
 Point B
C2 9 𝑋1 + 6 𝑋2 = 1566
9 𝑋1 = 1566 − 6 𝑋2
6
𝑋1 = 174 − 9 𝑋2
C3 12 𝑋1 + 16𝑋2 = 2880
6
12 (174 − 9 𝑋2 )+ 16𝑋2 = 2880
A (0,180)
2088 − 8 𝑋2 + 16𝑋2 = 2880
8𝑋2 = 2880 − 2088
8 𝑋2 = 792
792
𝑋2 = = 99
B (108,99) 8
6
𝑋1 = 174 − 99 = 108
9
B (108, 99)

D (0, 0) C (174,0)
Corner point points Z= 350 x1+300 x2 Optimal solution
A (0,180) 54000

B (108, 99) 67500 X1=108, X2=99, Z=67500

C (174, 0) 60900
D (0, 0) 0
special conditions in LP models

3- Unbounded solutions
- Objective function can be mode infinitely large (max) Or infinitely
small(min)
special conditions in LP models

4- No feasible solutions
The system of constraints in a linear programming problem may have no
points which satisfy all constraints, in such cases, there are no points in the
solution set, and linear programming problem is said to have no feasible
solution. The following Fig illustrates a problem having no feasible solution.
Constraint (1) is a “less than or equal to” type, and constraint (2) ia a “greater
than or equal to” type. A problem can certainly have both types of
constraints. In this case the set of points satisfying one constrain includes
none of the points satisfying the other.
Thank You
Operations Research
Lecture(4)

BY: DR. HEMAT MOUSTAFA

Linear Programming Using MS Excel
Example (1) : Consider the following LP model
Solution:
Example (2): Consider the following LP model

Answer the following questions

Answer:
Example (3)
Solution:

LP formulation
𝑴𝒂𝒙 𝑷 = 𝟏𝟎 𝑺 + 𝟗 𝑫

S.t.

𝟕
𝑺 + 𝑫 ≤ 𝟔𝟑𝟎
𝟏𝟎
𝟏 𝟓
𝑺 + 𝑫 ≤ 𝟔𝟎𝟎
𝟐 𝟔
𝟐
𝑺 + 𝑫 ≤ 𝟕𝟎𝟖
𝟑
𝟏 𝟏
𝑺 + 𝑫 ≤ 𝟏𝟑𝟓
𝟏𝟎 𝟒
𝑺, 𝑫 ≥ 𝟎
Excel solution
Example (4)
𝑴𝒊𝒏 𝑪=𝟖 𝑿+ 𝟔𝒀

Subject to:
𝟑 𝑿 + 𝟐 𝒀 ≥ 𝟏𝟔𝟎
𝟓 𝑿 + 𝟐 𝒀 ≥ 𝟐𝟎𝟎
𝑿 + 𝟐 𝒀 ≥ 𝟖𝟎
𝑿 ≥ 𝟎, 𝒀 ≥ 𝟎
Solution:
Thank You
Operations Research
Lectures(5 &6)

BY: DR. HEMAT MOUSTAFA

Sensitivity Analysis (post-optimality
analysis)
It is often desirable to study the effect of discrete changes in the different coefficients of the
problem on the current optimal solution. One way to accomplish this is to solve the problem
anew, but this may be computationally inefficient. If one makes use of properties of the
simplex solution, it is possible to reduce additional computations considerably. This will be
the objective of post-optimality analysis.

The changes in the linear programming problem usually studied by post-optimality analysis
include.

1. Tightness of the constraints, that is, changes in the right-hand side of the constraints.

2. Coefficients of the objective function (profit or cost).

LP and Sensitivity Analysis Using Microsoft
Excel
Example (1) : Consider the following LP model

Subject to:

Suppose that 𝒁 represents the profit of producing 3 different products, where 𝒙𝒋 is the
number of units produced of product 𝒋, 𝒋 = 𝟏, 𝟐, 𝟑 .
Solution:
The Answer Report
The Sensitivity Report

- the Sensitivity Report determines how the optimal solution is sensitive to changes in the
original parameters. It displays the following:

The sensitivity analysis for the changes in the objective function coefficients, represented by the
Variable cells table.

The sensitivity analysis for the changes in the RHS parameters, represented by the Constraints
table.
- The aim of the sensitivity analysis is to determine within what range the model parameters
could change without changing the optimal solution. Now, let's interpret the sensitivity report.
1. Changes in the objective function coefficients

In the Variable cells table;

- The column Objective Coefficient displays the original values of the coefficients of the objective
function that are already used to solve the problem.
- The column Final Value illustrates the optimal solution of the decision variables.
- The column Reduced Cost illustrates the change in the objective function per unit increase in
the decision variable. In other words, the reduced cost is the amount by which the coefficient of
the objective function of certain variable needs to change before that variable will become non-
zero (before the variable enters the solution) .
For instance, 𝒙𝟏 = 𝟎 and its reduced cost = -7.5 , this means that forcing 𝒙𝟏 to become 1
decreases the optimal objective value by 7.5. That is, (i.e., profit per unit on 𝒙𝟏) would need to
increase by 7.5 before it would be profitable to produce any of product 1.

-The two columns Allowable Increase and Allowable Decrease indicate by how much the
coefficients of the objective function 𝒄𝒋 ( could increase or decrease, respectively, without
changing the optimal solution of the decision variables.

For instance, 𝒄𝟏 = 𝟓𝟎 , it is allowed to increase by 7.5 or decrease infinitely without changing the
optimal solution. Note that Excel usually represents very large values with 1E+30, thus, the
value 1E+30 could be considered as infinity.
It should be noted that any change in the coefficients of the objective function does not change
the feasible area, therefore, the sensitivity analysis indicates by how much the coefficients of
the objective function could change without changing the optimal solution of decision
variables. However, the optimal value of the objective function will change. For instance, 𝒄𝟑 =
𝟓𝟓, it is allowed to increase by 5 or decrease by 15 without changing the optimal solution of
decision variables. That is 𝒄𝟑 could range between lower and upper limits of (40, 60) without
changing the optimal solution, while any additional change beyond these limits will change the
optimal solution. For example, If 𝒄𝟑 is increased to 57, then the optimal solution of the
decision variables will not change but the objective value would become 5910 = 50 (0) + 60 (70)
+ (57) (30) .
2- Changes in the RHS parameters

In the Constraints table;

- The column Final value indicates the obtained LHS values of the constraints.

- The column Constraint R.H.Side displays the original RHS parameters that are used to solve
the model.

- The two columns Allowable Increase and Allowable Decrease indicate by how much the RHS
parameters (𝒃′𝒊 𝒔) could increase or decrease, respectively, without changing the current
optimal basis. For instance, 𝒃𝟏 = 𝟏𝟎𝟎 , it is allowed to increase by 10 or decrease by 6.667
without changing the optimal basis (i.e., x2 and x3).
- The Shadow price column represents the change in the value of the objective function per unit
increase in the RHS parameter within the allowable range. If a RHS coefficient is changed
beyond its allowable range, the model should be resolved again with the new change.

For instance , 𝒃𝟏 = 𝟏𝟎𝟎 and the corresponding shadow price is 60, this means that if 𝒃𝟏 is
increased by 1 unit (i.e., increased to 101) then the objective value would increase by 60.
Similarly, if 𝒃𝟒 is increased to 65 then the objective value would decrease by 12.5 = −𝟐. 𝟓 ×
𝟔𝟓 − 𝟔𝟎 .However, if 𝒃𝟒 is increased by 41 units, then the user needs to resolve the problem in
order to calculate the change in the optimal objective value.
It should be noted that changes in the RHS parameters leads to changes in the feasible area.
Therefore, there are two cases:

1) If the RHS parameter of a non-binding constraint is changed within the allowable range, then
the optimal solution will remain as it is without any change, that is why the shadow prices of non-
binding constraints are zeros. For instance, the shadow prices of the non-binding constraints 2
and 3 are zeros. Which means that changing the RHS parameters of constraints 2 or 3 would not
affect the optimal solution as long as 𝒃𝟐 is decreased by no more than 30 (i.e., its slack value),
or is increased by no more than 20 (i.e., its surplus value).

2) If the RHS parameter of a binding constraint is changed within the allowable range, then the
optimal basis remains as it is without any change, but both the optimal values of the basic
decision variables and the optimal objective value will change, that is why the shadow prices of
binding constraints are non-zero values.
Example (2): Consider the following LP model

Answer the following questions

Answer:
The Answer report
The Sensitivity report
Thank You