Problem solving techniques and applications of operations research

19 March 2020 Dr. Abdulfatah A. Salem 2
Problem solving approaches
Conventional
approach
Observational
approach
Scientific
approach
Using previous experience in solving problems * Differ
environment (economical, social, legal) complicate solution
Develop new tools and methods
Study differences between the ancient and recent cases
Take into considerations the new environments
Problems solving techniques
Break even point method, financial analysis, decision theory
and computers

Problem solving Technique
 The First 5 steps are the process of decision making
 The Quantitative Approach to decision making based on the scientific method of problem
solving called The Management Science.
 Some times they call it Decision Science
 Operations Research is the synonymous term of the Management Science.
 Operations Research is the application of analytical methods designed to help the decision
makers choose between various courses of action available to accomplish specified objectives

What is Operations Research ?
Operations
The activities carried out in an organization.
Research
The process of observation and testing characterized by the scientific method.
Operations Research is
The use of quantitative methods to assist decision-makers (in designing,
analyzing, and improving the operation of systems) to make the better decisions.
•The scientific study of operations for the purpose of making better decisions
•For this purpose objectives of the organization are defined and analyzed.
•These objectives are then used as the basis to compare the alternative courses of
action.
“the science of better”

HISTORY OF OPERATION RESEARCH
The ideas and methods of Operations Research began to take shape during World War II, and thereafter
have been put to good use in a wide variety of industrial, financial, and scientific endeavors.
 It is generally accepted that the field originated in England during the World War II.
 Some say that Charles Babbage (1791-1871) is the Father of OR because his research into the cost
of transportation and sorting of mail led to England’s University Penny Post in 1840.
 Modern Operations Research originated at the Bowdsey Research Station in U.K. in 1937 to analyze
and improve the working of the UK’s Early Warning Radar System.
 During the Second World War about 1000 Men and Women were engaged to work for British Army.
 After World War II, Military Operational Research in U.K. became Operational Analysis (OA) within
the U.K. Ministry of Defense with expanded techniques and graving awareness.
The ambiguous term operations research (O.R.) was coined during World War II, when the British
military management called upon a group of scientists together to apply a scientific approach in the
study of military operations to win the battle. The main objective was to allocate scarce resources in an
effective manner to various military operations and to the activities within each operation.

MODELS of OPERATIONAL RESEARCH
building models is the central to the practice of Operations Research. A model is a simplified,
idealized representation of a real object, a real process, or a real system. The models used here
are called mathematical models because the building blocks of the models are mathematical
structures such as equations, inequalities, matrices, functions, and operators. In developing a
model, these mathematical structures are used to capture and describe the most salient features
of the. Entity that is being modeled.
Deterministic Models Stochastic Models
Linear Programming Discrete-Time Markov Chains
Network Optimization Continuous-Time Markov Chains
Integer Programming Queuing Theory (waiting lines)
Nonlinear Programming Decision Analysis
Dynamic Programming

NATURE of OPERATIONAL RESEARCH
System Orientation
of Operation
Research
Reduces complexity
by
use of computers
Helpful in improving
the quality of
solution
Goal oriented
optimum
solution
Quantitative
Solutions
Application
of
Scientific Method
Require
willing
executives
Use
of
models
Human
Factor
The Use of
Interdisciplinary
Team.

LIMITATIONS of OPERATIONAL RESEARCH
Magnitude of
Computation
Non-
Quantifiable
Factors
Distance
between User
and Analyst
Time and
Money Costs
Implementation

Operations Research phases
Problem
Orientation
Problem
Definition
Data
collection
Model
Formulation
Model
Solution
Model
Validation
Output
Analysis
Implementing
and
Monitoring

Applications of Operations Research
In Industry
• Production, blending, product mix, inventory control, demand forecast,
sale and purchase, transportation, repair and maintenance, scheduling
and sequencing, planning and control of projects etc.
In Defence
• Air force, army and navy.
In Planning
In Agriculture
• Allocation of land to various crops in accordance with the climatic
conditions
• Optimum distributions of water from various resources like canal for
irrigation purposes.
In Public Utilities
• Petroleum, paper, chemical, metal processing, aircraft, transport and
distribution, mining and textile industries.

19 March 2020 Dr. Abdulfatah Salem 11
Model Formulation
is the process of translating a
verbal statement of a problem
into a mathematical statement.
lt is an art that can only be mastered with practice and experience.
Understand
the problem
thoroughly
Describe the
objective
Describe
each
constraint
Define the
decision
variables
Write the
objective in
terms of the
decision
variables
Write the
constraints
in terms of
the decision
variables
Guidelines for Model Formulation

LINEAR PROGRAMMING
A method to achieve the best outcome or extreme
values (such as maximum profit or lowest cost)
in a mathematical model whose requirements
are represented by linear relationships
• Provides a powerful tool in modeling many applications.
• Has attracted most of its attention in optimization during the last
six decades for two main reasons:
• Applicability: There are many real-world applications that can be
modeled as linear programming;
• Solvability: There are theoretically and practically efficient
techniques for solving large-scale problems.

DECISION VARIABLES
• mathematical symbols
representing levels of
activity of a firm.
OBJECTIVE FUNCTION
• a linear mathematical
relationship describing
an objective of the firm,
in terms of decision
variables, that is to be
maximized or minimized
CONSTRAINTS
• restrictions placed on the
firm by the operating
environment stated in
linear relationships of the
decision variables.
PARAMETERS
- numerical coefficients
and constants used in the
objective function and
constraint equations.
Basic Components of an LP

Model Components Definitions
Decision variables:
describe our choices that are under our control;
Objective function:
describes a criterion that we wish to minimize (e.g., cost) or maximize
(e.g., profit);
Constraints:
describe the limitations that restrict our choices for decision variables.
Formally
We use the term “linear programming (LP)” to refer to an
optimization problem in which the objective function is linear
and each constraint is a linear inequality or equality.

Assumptions of Linear Programming Model
1. Conditions of certainty exist.
2. Linearity
3. Proportionality in objective function and constraints (1 unit – 3 hours, 3
units- 9 hours).
4. Additivity (total of all activities equals sum of individual activities).
5. Divisibility assumption that solutions need not necessarily be in whole
numbers (integers); ie.decision variables can take on any fractional value.

Problem Statement:
Toys Manufacturing company produces two types of wooden toys: soldiers and
trains. A soldier sells for $27 and uses $10 worth of raw materials. Each soldier that is
manufactured increases labor and overhead cost by $14. A train sells for $21 and
uses $9 worth of raw materials. Each train built increases labor and overhead cost by
$10. The manufacture of wooden soldiers and trains requires two types of skilled
labor: carpentry and finishing. A soldier requires 2 hours of finishing labor and 1 hour
of carpentry labor. A train requires 1 hour of finishing and 1 hour of carpentry labor.
Each week, the company can obtain all the needed raw material but only 100 finishing
hours and 80 carpentry hours. Demand for trains is unlimited, but at most 40 soldiers
are bought each week. The company wants to maximize weekly profit.
MODEL FORMULATION
Product Mix
Formulate a linear programming model of the situation that can
be used to maximize the weekly profit.

1. Decision variables:
the decision variables should completely describe the decisions to be made.
– how many soldiers and trains should be manufactured each week.
X1 = number of soldiers produced each week
X2 = number of trains produced each week
2. Objective function:
Maximize the net profit (weekly revenues – raw materials cost – labor and overhead costs).
 Weekly revenues = weekly revenues from soldiers + weekly revenues from train
= 27 X1 + 21 X2
 Weekly raw materials costs = 10 X1 + 9 X2
 Weekly variable costs = 14 X1 + 10 X2
Therefore, the Company wants to
Maximize 27 X1 + 21 X2) – (10 X1 + 9 X2) – (14 X1 + 10 X2) = 3 X1 + 2 X2
Hence, the objective function is:
Maximize Z = 3 X1 + 2 X2

3.Constraints:
as X1 and X2 increase, the company objective function grows larger.
This means that the company were free to choose any values of X1
and X2, the company could make an arbitrarily large profit by choosing
X1 and X2 to be very large. Unfortunately, the values of X1 and X2 are
limited by the following three restrictions (often called constraints):
Constraint 1: each week, no more than 100 hours of finishing time may be used.
2 X1 + X2  100
Constraint 2: each week, no more than 80 hours of carpentry time may be used.
X1 + X2  80
Constraint 3: because of limited demand, at most 40 soldiers should be produced.
X1  40
4. Sign restrictions (non negativity)
X1  0 and X2  0

 Combining the non negativity constraints with the
objective function and the structural constraints yield the
following optimization model:
– Max Z = 3 X1 + 2 X2 (objective function)
 2 X1 + X2  100 (finishing constraint)
 X1 + X2  80 (carpentry constraint)
 X1  40 (soldier demand constraint)
 X1  0 and X2  0 (no negativity constraint)

Problem Statement:
A dairy production factory produce two types of foods, cheese and
yoghurt. The products undergo two major processes: fermentation and
packaging operations. The profits per unit are EGP 7 and EGP 4
respectively. Each cheese item requires 3 minutes for fermentation
operation and 2 minutes for packaging operation, whereas each yoghurt
item requires 1 minute for fermentation operation and 4 minutes for
packaging operation. The available operating time is 120 minutes and
100 minutes for fermentation operation and packaging operation
machines. The manager has to determine the optimum quantities to be
manufacture the two products to maximize the profits.
Formulate the aboveproblemas a linear programming

Solution
Decision variables
Manager must decide how many items from cheese and yoghurt product should
be produced each week. With this in mind, he has to define:
X1 be the number of cheese product to be produced .
X2 be the number of yoghurt product to be produced .
Objective function
The decision maker wants to maximize (revenue or profit) or minimize (costs).
Manager can concentrate on maximizing the total weekly profit (Z).
Here profit equals to:
(weekly revenues) – (raw material purchase cost) – (other variable costs).
Hence Manager’s objective function is:
Maximize Z = 7X1 + 4X2

Constraints
It show the restrictions on the values of the decision variables. Without constraints
manager could make a large profit by choosing decision variables to be very large.
Here there are three constraints:
•Available machine-hours for each machine and time consumed by each product
• 3X1 + 2X2 ≤ 120
• X1 + 4X2 ≤ 100
•Sign restrictions are added if the decision variables can only assume nonnegative
values, (Manager can not use negative number machine and time never negative
number )
• X1, X2 ≥ 0
So, the LP model will be:
Max Z = 7X1 + 4X2
S.T.
3X1 + 2X2 ≤ 120
X1 + 4X2 ≤ 100
X1, X2 ≥ 0

Problem Statement:
Task
Formulate a linear programming model for this problem, to determine how many containers of each
product to produce tomorrow in order to maximize the profits.
juice Price/quart Cost/quart Fruit needed
Guava 3 1 1 Kg.
pomegranate 2 0.5 2 Kg.
Grape 2.5 1.5 1.25 Kg.
tangerine 2.5 2 2
Mix 4 2 0.25 Kg each
Company makes five types of juice using Guava, pomegranate, Grape and tangerine. The
following table shows the price and cost per quart of juice (one container of juice) as well as the
number of kilograms of fruits required to produce one quart of juice. On hand there are 400 Kg
of Guava, 300 Kg. of pomegranate, 200 Kg. of Grape and 150 Kg. of tangerine. The manager
wants Grape juice to be used for no more than 20 percent of the number of containers
produced. He wants the ratio of the number of containers of pomegranate juice to the number of
containers of Guava juice to be at least 5 to 6. tangerine juice should not exceed half of the total
product.. Mix juice should not exceed one quarter of the total product.

Step 1: Defining the Decision Variables
Step 2: Choosing an Objective Function
Solution
X1 = # of containers of Guava juice
X2 = # of containers of pomegranate juice
X3 = # of containers of Grape juice
X4 = # of containers of tangerine juice
X5 = # of containers of mix juice
Max Z = 2X1 + 1.5X2 + 1X3 + 0.5X4 + 2X5
Guava constraints X1 + 0.25X5 ≤ 400
Pomegranate constraints 2X2 + 0.25X5 ≤ 300
Grape constraints 1.25X3 + 0.25X5 ≤ 200
Tangerine constraints 2X4 + 0.25X5 ≤ 150
Max of Grape juice constraints X3 ≤ 0.2(X1 + X2 + X3 + X4 + X5)
Ratio of pomegranate to guava constraints X2/X1 ≥ 5/6
Max of tangerine juice X4 ≤ 0.5(X1 + X2 + X3 + X4 + X5)
Max of Mix juice X5 ≤ 0.25(X1 + X2 + X3 + X4 + X5)
Non negativity constraints X1,2,3,4,5 ≥ 0
Step 3: Identifying the Constraints

Breakfast Food Calories
Fat
(g)
Cholesterol
(mg)
Iron
(mg)
Calcium
(mg)
Protein
(g)
Fiber
(g)
Cost
($)
1. Bran cereal (cup)
2. Dry cereal (cup)
3. Oatmeal (cup)
4. Oat bran (cup)
5. Egg
6. Bacon (slice)
7. Orange
8. Milk-2% (cup)
9. Orange juice (cup)
10. Wheat toast (slice)
90
110
100
90
75
35
65
100
120
65
0
2
2
2
5
3
0
4
0
1
0
0
0
0
270
8
0
12
0
0
6
4
2
3
1
0
1
0
0
1
20
48
12
8
30
0
52
250
3
26
3
4
5
6
7
2
1
9
1
3
5
2
3
4
0
0
1
0
0
3
0.18
0.22
0.10
0.12
0.10
0.09
0.40
0.16
0.50
0.07
A diet breakfast must include at least 420 calories, 5 milligrams of iron, 400 milligrams of calcium,
20 grams of protein, 12 grams of fiber, and must have no more than 20 grams of fat and 30
milligrams of cholesterol. Using the table below
Formulate the LP model that Minimize the cost of the diet breakfast meal.
MODEL FORMULATION
Diet Problem

Decision variables
X1 = Cups of bran cereal
X2 = Cups of dry cereal
X3 = Cups of oatmeal
X4 = Cups of oat bran
X5 = Eggs
X6 = Slices of bacon
X7 = Oranges
X8 = Cups of milk
X9 = Cups of orange juice
X10 = Slices of wheat toast
Objective function
MINIMIZE Z = 0.18X1 + 0.22X2 + 0.10X3 + 0.12X4 + 0.10X5 + 0.09X6+ 0.40X7 + 0.16X8 + 0.50X9
+ 0.07X10

I. 90X1 + 110X2 + 100X3 + 90X4 + 75X5 + 35X6 + 65X7 + 100X8 + 120X9 + 65X10  420
II. 2X2 + 2X3 + 2X4 + 5X5 + 3X6 + 4X8 + X10  20
III. 270X5 + 8X6 + 12X8  30
IV. 20X1 + 48X2 + 12X3 + 8X4+ 30X5 + 52X7 + 250X8 + 3X9 + 26X10  400
V. 6X1 + 4X2 + 2X3 + 3X4+ X5 + X7 + X10  5
VI. 3X1 + 4X2 + 5X3 + 6X4 + 7X5 + 2X6 + X7+ 9X8+ X9 + 3X10  20
VII. 5X1 + 2X2 + 3X3 + 4X4+ X7 + 3X10  12
VIII. XI  0
constraints

Problem Statement:
Compartment
Capacity
Weight (tons) Space (cubic meters)
Front 10 6800
Centre 16 8700
Rear 8 5300
Furthermore, the weight of the cargo in
the respective compartments must be
the same proportion of that
compartment's weight capacity to
maintain the balance of the plane.
Cargo
Weight
(tons)
Volume
(Cubic meters/ton)
Profit
($/ton)
C1 18 480 310
C2 15 650 380
C3 23 580 350
C4 12 390 285
Any proportion of these cargoes can be accepted. The objective is to determine how
much (if any) of each cargo C1, C2, C3 and C4 should be accepted and how to
distribute each among the compartments so that the total profit for the flight is
maximized.
• Formulate the above problemas a linear programming
A cargo plane has three compartments
for storing cargo: front, centre and rear.
These compartments have the following
limits on both weight and space:

Solution
We need to decide how much of each of the four cargoes to put in each of the three compartments.
Hence let:
xij be the number of tons of cargo i
(i=1,2,3,4 for C1, C2, C3 and C4 respectively)
that is put into compartment j
(j=1 for Front, j=2 for Centre and j=3 for Rear)
where i=1,2,3,4; j=1,2,3
Decision Variables
Cargo Front Center Rear
Max
Weight
Volume
/ ton
$ Profit
/ ton
C1 x11 x12 x13 18 480 310
C2 x21 x22 x23 15 650 380
C3 x31 x32 x33 23 580 350
C4 x41 x42 x43 12 390 285
Max Capacity (tons) 10 16 8
Max Capacity (C.merters) 6800 8700 5300

Objective Function
The objective is to maximize total profit Z,
i.e.
Max Z = 310[X11+ X12+X13]
+ 380[X21+ X22+X23]
+ 350[X31+ X32+X33]
+ 285[X41+ X42+X43]

Constraints
X11 + X12 + X13 <= 18
X21 + X22 + X23 <= 15
X31 + X32 + X33 <= 23
X41 + X42 + X43 <= 12
X11 + X21 + X31 + X41 = 10
X12 + X22 + X32 + X42 = 16
X13 + X23 + X33 + X43 = 8
480X11 + 650X21 + 580X31 + 390X41 <= 6800
480X12 + 650X22 + 580X32 + 390X42 <= 8700
480X13 + 650X23 + 580X33 + 390X43 <= 5300
XIJ >=0
 Cannot pack more of each of the four cargoes than we have available
The weight capacity of each compartment must be respected
The volume (space) capacity of each compartment must be respected
Non negativity constraints

Industry firm has plants in two cities one at Porg-elarab and the other at assiut,
there are three warehouses located near the larger market areas of
Alexandria, Cairo, and Aswan.
The sales requirements for next year at the Alexandria warehouse are 10000
items, at the Cairo warehouse 8000 items, and at the Aswan warehouse
15000 items. The plant capacity at each location is limited. Porg-elarab plant
can produce and ship 20000 items per year; the Assiut plant can produce
15000 items per year. The cost of shipping one item from each plant to each
warehouse differs, and these unit shipping costs are:
19 March 2020 Dr. Abdulfatah A. Salem
Alexandria Cairo Aswan
Porg-elarab 1 3 9
Assiut 7 4 2
32
The company wishes to develop a shipping schedule that will
minimize its total annual transportation cost
MODEL FORMULATION
Transportation Problem

To formulate this problem using LP, we again employ the concept of double subscribed variables. We let the
first subscript represent the origin (plant) and the second subscript the destination (warehouse). Thus, in
general, Xij refers to the number of items shipped from origin i to destination j. Therefore, we have six
decision variables as follows:
Decision variables:
X11 = # of items shipped from Porg-elarab to Alexandria
X12 = # of items shipped from Porg-elarab to Cairo
X13 = # of items shipped from Porg-elarab to Aswan
X21 = # of items shipped from Assiut to Alexandria
X22 = # of items shipped from Assiut to Cairo
X23 = # of items shipped from Assiut to Aswan
Objective function:
Min Z = X11 + 3 X12 + 9 X13 + 7 X21 + 4 X22 + 2 X23
Constraints:
X11 + X21 = 10000 (Alexandria demand)
X12 + X22 = 8000 (Cairo demand)
X13 + X23 = 15000 (Aswan demand)
X11 + X12 + X13  20000 (Porg-elarab Supply
X21 + X22 + X23  15000 (Assiut Supply)
Xij  0 for i = 1, 2 and j = 1, 2, 3.

The Edfina company operates two plants. There are three farmers Ahmed, Mustafa and kamel, they are
growing the Mango fruits. The growers are willing to supply fresh mango in the following amounts:
•Ahmed : 200 tons at EGP 11/ton
•Mustafa : 310 tons at EGP 10/ton
•Kamel: 420 tons at EGP 9/ton
Shipping costs in $ per ton are:
Plant capacities and labor costs are:
The mango juice are sold at EGP 50/ton to the distributors. The company can sell at this price all they can
produce. The objective is to find the best mixture of the quantities supplied by the three growers to the two
plants so that the company maximizes its profits.
Plant A Plant B
Ahmed 3 3.5
Mustafa 2 2.5
Kamel 6 4
Plant A Plant B
Capacity (ton) 460 560
Labor Cost (EGP/ton) 26 21
MODEL FORMULATION
Product Mix

Decision Variables
We need to decide how much to supply from each of the three growers to each of the two plants.
Hence let xij be the number of tons supplied from grower i (i=1,2,3 for Ahmed, Mustafa and Kamel
respectively) to plant j (j=1 for Plant A and j=2 for Plant B)
Solution
Plant A Plant B
Ahmed X11 X12
Mustafa X21 X22
Kamel X31 X32
Price/ton
11
10
9
Max QTY
200
310
420
Capacity 460 560
Labor cost/ton 26 21

Objective Function
The objective is to maximize total profit,
i.e. Maximize (revenue - grower shipping cost - plant labor cost - grower supply cost)
Max Z = (50-11-26-3)X11 + (50-11-21-3.5)X12 + (50-10-26-2)X21
+ (50-10-21-2.5)X22 +(50-9-26-6)X31 + (50-9-21-4)X32
Constraints
1) Cannot supply more than a grower has available - a supply constraint
X11 + X12  200
X21 + X22  310
X31 + X32  420
2) The capacity of each plant must be respected - a capacity constraint
X11 + X21 + X31  460
X12 + X22 + X32  560
3) The non negativity constraints
Xij  0

Careem Taxi company has four taxis locating at Smouha area and there are
four passengers, Ahmed, Ali, Husein and Khaled requiring taxis. The distance
between the taxis and the passengers are given in the table below, in
Kilometers'. The Taxi company wishes to assign the taxis to passengers so that
the distance traveled is a minimum.
Ahmed Ali Husein Khaled
Renault 3 4 6 5
Toyota 4 1 5 3
Lanos 8 3 2 5
Optra 2 5 3 4
MODEL FORMULATION
Assignment Problem

Renault 3 4 6 5
Toyota 4 1 5 3
Lanos 8 3 2 5
Optra 2 5 3 4
Renault X11 X12 X13 X14
Toyota X21 X22 X23 X24
Lanos X31 X32 X33 X34
Optra X41 X42 X43 X44
Decision variables:
1 if taxi i is assigned to passenger j
Let Xij =
0 otherwise
Where : i = 1, 2, 3, 4 stands for Renault, Toyota, Lanos, and Optra respectively.
j = 1, 2, 3, 4 stands for Ahmed, Ali, Husein, and Khaled respectively
The Linear Programming formulation will be as follows:

1 1 1 1
1
1
1
1
Objective function:
Min Z = 3X11 + 4X12 + 6X13 + 5X14 + 4X21 + X22 + 5X23 + 3X24 + 8X31 + 3X32 + 2X33 + 5X34 +
2X41 + 5X42 + 3X43 + 4X44
Constraints
X11 + X12 + X13 + X14 = 1 (Renault)
X21 + X22 + X23 + X24 = 1 (Toyota)
X31 + X32 + X33 + X34 = 1 (Lanos)
X41+ X42 + X43 + X44 = 1 (Optra)
X11 + X21 + X31 + X41 = 1 (Ahmed)
X12 + X22 + X32 + X42 = 1 (Ali)
X13 + X23 + X33 + X43 = 1 (Husein)
X14 + X24 + X34 + X44 = 1 (Khaled)

El Nagah group for investment invests in Project A, Project B, Project C and Project D. The
group strategy has placed limits on the amount that can be committed to any one type of
investment. The group has EGP 5 million available for immediate investment and wishes to do
two things:
1. Maximize the interest earned on the investments made over the next six months
2. Satisfy the diversification requirements as set by the group strategy.
The specifics of the investment possibilities are:
Project Interest earned % Maximum investment
(EGP Million)
A 8 1
B 12 2.25
C 17 1.75
D 14 1.6
In addition, the group specifies that at least 40% of the funds invested must be in Project C
and Project D , and that no less than 10% be invested in Project A.
MODEL FORMULATION
Portfolio selection problem

To formulate El Nagah group’s for investment problem as a linear programming model, we assume the
following decision variables:
Decision variables:
X1 = Quantity invested in Project A
X2 = Quantity invested in Project B
X3 = Quantity invested in Project C
X4 = Quantity invested in Project D
Objective function:
Max Z = 0.08 X1 + 0.12 X2 + 0.17 X3 + 0.14 X4
Constraints:
X1  1
X2  2.25
X3  1.75
X4  1.6
X3 + X4  0.4(X1 + X2 + X3 + X4)
X1  0.1(X1 + X2 + X3 + X4)
X1 + X2 + X3 + X4  5
Xi  0 , i = 1, 2, 3, 4

A company has budgeted up to $8000 per week for local advertisement. The money is to be
allocated among four promotional media: TV spots, newspaper ads, and two types of radio
advertisements. The company goal is to reach the largest possible high-potential audience
through the various media. The following table presents the number of potential customers
reached by making use of advertisement in each of the four media. It also provides the cost per
advertisement placed and the maximum number of ads that can be purchased per week.
Medium
Audience
Reached per ad
Cost per ad
Max. ads per
week
TV spot (1 minute) 5000 800 12
Daily newspaper (full-page ad) 8500 925 5
Radio spot (30 second, prime time) 2400 290 25
Radio spot (1 minute, afternoon) 2800 380 20
The company arrangements require that at least five radio spots be placed each week. To ensure a board-scoped
promotional campaign, management also insists that no more than $1800 be spent on radio advertising every week.
42
MODEL FORMULATION
Media Selection Problem

1. Decision variables:
X1 = number of 1-miute TV spots taken Each week
X2 = number of full-page daily newspaper ads taken each week.
X3 = number of 30-second prime-time radio spots taken each week.
X4 = number of 1-minute afternoon radio spots taken each week.
2. Objective function
Max Z = 5000 X1 + 8500 X2 + 2400 X3 + 2800 X4
3. Constraints
Constraint 1 (maximum TV spots/week) X1  12
Constraint 2 (maximum newspaper ads/week) X2  5
Constraint 3 (maximum 30-second radio spots/week) X3  25
Constraint 4 (maximum 1-minute radio spots/week) X4  20
Constraint 5 (weekly budget) 800 X1 + 925 X2 + 290 X3 + 380 X4  8000
Constraint 6 (minimum radio spots contracted) X3 + X4  5
Constraint 7 (maximum dollars spent on radio) 290 X3 + 380 X4  1800
4. Non-negative constraints:
X1, X2, X3, X4  0
43

A Mango juice company operates two plants. The growers are willing to supply fresh mango in the following
amounts:
•S1: 200 tons at $11/ton
Shipping costs in $ per ton are:
Plant capacities and labor costs are:
The mango juice are sold at $50/ton to the distributors. The company can sell at this price all they can
produce. The objective is to find the best mixture of the quantities supplied by the three growers to the two
plants so that the company maximizes its profits.
Plant A Plant B
S1 3 3.5
S2 2 2.5
S3 6 4
Plant A Plant B
Capacity (ton) 460 560
Labor Cost ($/ton) 26 21
MODEL FORMULATION
Product Mix Problem

Decision Variables
We need to decide how much to supply from each of the three growers to each of the two plants.
Hence let xij be the number of tons supplied from grower i (i=1,2,3 for S1, S2 and S3 respectively)
to plant j (j=1 for Plant A and j=2 for Plant B)
Solution
Plant A Plant B
S1 X11 X12
S2 X21 X22
S3 X31 X32
Price/ton
11
10
9
Max QTY
200
310
420
Capacity 460 560
Labor cost/ton 26 21

Objective Function
The objective is to maximize total profit,
i.e. Maximize (revenue - grower shipping cost - plant labor cost - grower supply cost)
Max Z = (50-11-26-3)X11 + (50-11-21-3.5)X12 + (50-10-26-2)X21
+ (50-10-21-2.5)X22 +(50-9-26-6)X31 + (50-9-21-4)X32
Constraints
1) Cannot supply more than a grower has available - a supply constraint
X11 + X12  200
X21 + X22  310
X31 + X32  420
2) The capacity of each plant must be respected - a capacity constraint
X11 + X21 + X31  460
X12 + X22 + X32  560
3) The non negativity constraints
Xij  0

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