Michał Kulej
Business Information Systems
The development of the potential and academic programmes of Wrocław University of
Technology
Project co-financed by European Union within European Social Fund
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 1 / 40

Introduction
Hiller F. S. Hillier, G.J. Liberman: Introduction to Operations
Research, 7th ed., McGrawHill,2001.
Taha, H. A Operations Research: An Introduction, 8th ed.,
Pearson Prentice Hall, Upper Sadle River,NJ,2007.
Winston W.L. Operations Research: Applications and Algorithms,
PWS-KENT Publishing Company, Boston, 1987.
Additional literature
Anderson D.R.,D.J.Sweeney, and T.A.Williams: An Introduction to
Management Science, 8th ed., West,St.Paul,Mn,2000.
Williams H.P.: Model Building in Mathematical Programming, 3d ed.,Wiley, New York,1990.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 2 / 40

Introduction
The application of mathematics and scientific method to military operation during World War II was called Operation Research (OR)
The term OR or often Management Science means a scientific approach to decision making which seeks to detrmine how best to design and coordinate a system usually under conditions requiring allocations of scarce resources.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 3 / 40

Introduction
Metodology of OR:
Scientific method is used to ivestigate the problem.
Observing and formulating the problem, including gathering all relevant data.
To construct a mathematical model that attemps to abstract the essence of the real problem.
Model validation.
To provide positive, understandable conclusions to the decision makers.
OR attemps to resolve the conflicts of interest among the components of organization in a way that is best for the organization as a whole.
OR attemps to find the best solution (optimal solution) for the problem.
Team approach - a group of individuals having diverse skills.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 4 / 40

Introduction
1
2
3
4
5
6
Define the problem and gather relevant data.
Formulate a mathematical model to represent the problem.
Develop a computer-based procedure for deriving solution to the problem from the model.
Test the model and refine it as needed.
Prepare for the ongoing application of the model as prescribed by management.
Implement.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 5 / 40

Introduction
This process includes determining such things as: the appropriate objectives, constraints on what can be done, interrelations between the area to be studied an the other areas of organization, possible alternative courses of action, time limits for making decision
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 6 / 40

Introduction
Mathematical model is the system of equations and related mathematical expressions that describe the essence of the problem.
Decision variables If there are n related quantifiable decisions to be made they are represented as decision variables x
1
, x
2
, . . . , x n whose values are to be determined.
Objective function The appropriate (overall) measure of performance(e.g. profit) is expressed as a mathematical function of decision variables (for example
P = 5x
1
+ 2x
2
+ · · · + 20x n
).
Constraints Any restrictions on the values that can be assigned to the decision variables expressed mathematically, typically by means of inequalities or equations (for example
( x
1
) 2
− 2x
1 x
2
+ ( x
2
) 2
≤ 25).
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 7 / 40

Introduction
Parameters of the model The constants (namely, the coefficients and right- hand sides ) in the constraints and objective function.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 8 / 40

Introduction
The problem is to choose the values of decision variables so as to optimize (maximize or minimize) the objective function, subject to the specific constraints.
An important part of model-building process:
Gathering relevant data
Sensitivity analysis
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 9 / 40

Introduction
Develop a computer-based procedure for deriving solution to the problem from the model.
Search for optimal, or best, solution . These solutions are optimal only with respect to the model being used.
Use heuristic procedures to find a good suboptimal solution in case when the time or cost required to find an optimal solution for the model is very large.
Postoptimality analysis or What-if analysis . It answers the question - what would happen to optimal solution if diffrent assumptions were made about future conditions? It involves also conducting sensitivity analysis to determine which parametres of the model are most critical in determining the solution.
Test the model and refine it as needed.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 10 / 40

Introduction
The main subjects of the lecture are the following:
Linear Programming
Introduction to Linear Programming
Solving Linear Programming Problems - The Simplex Method
Duality Theory and Sensitivity Analysis
Integer Programming
Decision Analysis
Goal Programming
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 11 / 40

Formulating of Linear Programming Problems
Operations Research (BO) means a scientific approach to decision making, which seeks to determine how to design and operate the best a system, usually under conditions requiring the allocation of scarce resources.
The major phases of OR study:
I. Define the problem and gather relevant data.
⇒
II. Formulate a mathematical model to represent the problem ⇒
III. Develop a procedure for deriving solution to the problem ⇒
IV. Test the model and refine it as needed ⇒ V. Implement
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 12 / 40

Formulating of Linear Programming Problems
I. GW plant manufactures two types of products W
1 types of row materials S
1 and S
2 and W
2 using two
. GW plant wishes to determine how many products of each type to produce so as to achieve the highest possible profit.
II. The following data are gathered:
1 unit of W
1 requires 2 kg S
1 and 1 kg S
2
.
1 unit of W
2 requires 1 kg S
1 and 1 kg S
2
.
At present, GW has in stock 100 kg of S
1 and 80 kg of S
2
.
Each unit of W
1 profit of $2.
gives a profit of $3 and each unit of W
2 gives a
Demand for W
1 equals 40 units and demand for W
2 is unlimited.
III. The decision variables amount of product W
2
: x
1
- amount of product W
1 produced, x
2 produced, Z - total profit from producing W
1
W
2
.
and
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 13 / 40

Formulating of Linear Programming Problems
max Z = 3x
1
2x
1 x
1
+ x
+ x
2
2
≤ 100
≤ 80 x
1
≤ 40
+ 2x
2 x
1 x
2
≥ 0
≥ 0
[Maximize profit]
[Constraint on row material S
1
]
[Constraint on row material S
2
]
[Demand for W
1
]
[Sign restriction]
[Sign restriction]
The problem is to choose values of x
1
Z = 3x
1
+ 2x
2 and x
2 so as to maximize subject to the restrictions. We want to derive a production plan that maximizes the GW’s profit without violating the constraints on limited resources.
IV. We use an algorithm (the next part of the lecture will focus on the simplex method for solving linear programming models). We receive: x
1
= 20, x
2
= 60, Z = 180 so GW should produce 20 units of W
60 units of W
2 with resulting total profit of $180.
1 and
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 14 / 40

Formulating of Linear Programming Problems
max ( min ) z = f ( x
1
, . . . , x n
) [Objective function] g
1
( x
1
, . . . , x n
) ≤ ( ≥
,
=) b
1
[Constraint 1]
. . .
g m
( x
1
, . . . , x n
) ≤ ( ≥
,
=) b m
[Constraint m] x
1
, . . . , x n are called decision variables (They should completely describe the decision to be made). A solution for which all constraints are satisfied is called a feasible solution . A feasible solution with the largest (smallest) objective function value is called an optimal solution .
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 15 / 40

Formulating of Linear Programming Problems
A function h ( x linear function
1
, . . . , x n
) = a
1 x
1
+ a
2 x
2
+ · · · + a n x n
. The model for which all functions f
, of x g
1
1
, x
2
, . . . , g
, . . . , m x n are is linear are called a linear model or a linear programming model(LPM) .
a max ( min ) z = c
1 x
1 a
11 x
1
+ a
12 x
2
+ · · ·
+
+ c a
2 x
1n
2
+ · · · + x n
≤ ( ≥
, c n x n
=) b
1
. . .
a
m1 x
1 x
1
≥ 0
+ a
m2
, . . . , x
2
+ · · · x r
≥ 0
, r
+ a
≤ mn x n n
≤ ( ≥
,
=) b m
[Objective function]
[Constraint 1]
[Constraint m]
[Nonnegativity constraints]
The special linear constraints (inequalities constraints) are sign constraints x i
≥ 0.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 16 / 40

Examples of Linear Programming Problems(LPP)
Diet Problem
Person X requires that all the food she/he eats come from one of the four ”basic food groups”: chocolate cake, ice cream, cola and cheesecake. She/he wants to prepare the dish only from these foods.
Each chocolate cake costs $5, each scoop of ice cream costs $2, each bottle of coca costs $3, and each piece of cheesecake costs $8. Each day X must ingest at least 500 calorie, 6 oz of chocolate, 10 oz of sugar, and 8 oz of fat. The nutritional content per unit of each food is shown in the following Table:
Chocolate cake (1 piece)
Ice cream (1 scoop)
Cola (1 bottle)
Cheesecake (1 piece)
Calories
400
200
150
500
Chocolate
3
2
0
0
Sugar
2
2
4
4
Fat
2
4
1
5
Unit price($)
5
2
3
8
Formulate a linear programming model that satisfies the daily nutritional requirements of person X at minimum cost.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 17 / 40

Examples of Linear Programming Problems(LPP)
Diet Problem
Decision variables: x
1 x
2 x
3 x
4
- amount(number of pieces) of chocolate cake (eaten daily),
- amount(number of scoops) of ice cream,
- number of bottles of cola,
- amount of cheesecake.
Linear Programming Model: min Z
400x
1
=
+
5x
1
+
200x
2
2x
+
2
+ 3x
3
150x
3
+ 8x
4
+ 500x
4
3x
2x
2x x i
1
1
1
≥
+ 2x
2
+ 2x
2
+ 4x
2
0
,
≥ 6
+ 4x
3
+ x
3
+ 4x
+ 5x
4
4
≥
≥ 8
10 i = 1
, . . . ,
4
[Minimize total cost]
≥ 500 [Calorie constraint]
[Chocolate constraint]
[Sugar constraint]
[Fat constraint]
[Sign constraint]
The optimal solution to this problem (when solved by computer) is x
1
= 0, x
2
= 3, x
3
= 1, x incurs a daily cost 90$.
4
= 0, Z = 90 $ . Thus, the minimum-cost diet
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 18 / 40

Examples of Linear Programming Problems(LPP)
Production process Model
Ch.B.Company manufactures four types of perfumes: Brute, Chanelle,
Super Brute and Super Chanelle. Production process is given in the
Figure below.
Brute
7$/oz
Super Brute
14$/oz sales
Row Material
-3$/lb sales sales
Chanelle
6$/oz
Super Chanelle
10$/oz sales
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 19 / 40

Examples of Linear Programming Problems(LPP)
Production process Model
Processing 1 lb row material requires 1 hour of working time and gives 3 oz of Brute and 4 oz of Chanelle.
1 oz of Brute processed further requires an additional 3 hours of time and yields 1 oz of Super Brute.
1 oz of Chanelle processed further requires an additional 2 hours of time and yields 1 oz of Super Chanelle.
Rylon has 6000 hours of time available and can purchase up to
4000 lb of row material.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 20 / 40

Examples of Linear Programming Problems(LPP) Production process Model
Ch.B.Company must determine how much raw material to purchase and how much of each type of perfume should be produced and wants to maximize profit(=revenues from perfume sales - processing costs).
Decision variables: x
1
- number of ounces of Brute sold, x
2
- number of ounces of Super Brute sold, x
3
- number of ounces of Chanelle sold, x
4
- number of ounces of Super Chanelle sold, x
5
- number of pounds(lb) of raw material purchased.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 21 / 40

Examples of Linear Programming Problems(LPP)
Production process Model
Linear Programming Model: max z = 7x
1
+ 14x
2
+ 6x
3
+ 10x
4
− 3x
5
[Maximize profit] subject to x
1 x
3
+ x
2
+ x
4
− 3x
5
= 0
− 4x
5
= 0
[Manufacturing Brute and Super Brute]
[Manufacturing Chanelle and Super Chanelle] x
5
≤ 4000 [Limit on raw material]
3x x i
2
≥
+ 2x
4
0
,
+ x
5
≤ i = 1
, . . . ,
6000 [Limit on working time]
5 [Sign restriction]
The optimal solution: Z=172 666.667$ (profit), x
1
Brute, x
2
= 666
.
667 oz of Super Brute, x
3
= 11333
.
333 oz of
= 16000 oz of Chanelle, x
4
= 0 oz of Super Chanelle and x
5
= 4000lb of raw material.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 22 / 40

Examples of Linear Programming Problems(LPP)
An Inventory Model - Multiperiod Decision Problem
The factory must produce a certain product during next four quarters.
The demand for each of quarter is known. Factory wants to minimize total costs and has to meet all demands on time. Appropriate data are given in the following Table:
Q I Q II Q III Q IV
Demand (units) 30 60 70 25
Maximum production level(units) 60 60
Production cost ($/unit) 55 50
Inventory cost($/unit/quarter) 2 2
60
50
3
60
55
-
Decision variables: x i
- number of units produced during quarter i, i = 1
, . . .
4.
m i
- number of units of the product on hand at (inventory)the end of quarter i, i = 1
, . . .
4.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 23 / 40

Examples of Linear Programming Problems(LPP) An Inventory Model - Multiperiod Decision Problem
30 x
1 m
1 x
2
60 70 m
2 x
3 m
3 x
4
25
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 24 / 40

Examples of Linear Programming Problems(LPP) An Inventory Model - Multiperiod Decision Problem
Linear Programming Model: min z = 55x
1 x
1
− m
1
+
= 30
50x
2
+ 50x
3
+ 55x
4
+ 2m
1
+ 2m
2
+
[Balance for I quarter]
3m x i x i m
1
+ x
2
≤ 60
,
− m
2 m
2
+ x
3
− m
3
= 70 m
3
+ x
4
= 25 i = 1
= 60
, . . .
4
≥ 0
, m i
≥ 0
, i = 1
, . . . ,
[Balance for II quarter]
[Balance for III quarter]
[Balance for IV quarter]
[Production capacity]
4 [Sign restriction]
3
The optimal solution to the problem is Z = 9615: x
1 x
3
= 60, x
4
= 25, m
1
= 10, m
2
= 10, m
3
= 0.
= 40, x
2
= 60,
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 25 / 40

Examples of Linear Programming Problems(LPP)
Multiperiod Financial Model
Finco Invest. Corp. must determine investment strategy for next three years. At present (time 0), $100000 is available for investment.
Investments A,B,C,D and E are available. The cash flow associated with investing $1 in each investment is given in the table below.
0 1 2
A -1$ +0.5$ +1$
3
-
B -1$ +0.5$ 1$
C -1$ +1.2$ -
D -1$
E -
+1.9$
-1$ +1.5$
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 26 / 40

Examples of Linear Programming Problems(LPP)
Multiperiod Financial Model
To ensure that the company’s portfolio is diversified, Finco requires that at most $75000 be placed in any single investment. In addition,
Finco can earn interest at 8% per year by keeping uninvested cash in bank. Returns from investments may be immediately reinvested.
Formulate a linear programming problem that will maximize cash on hand at time 3. Finco must decide how much money should be placed in each investment(including bank).
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 27 / 40

Examples of Linear Programming Problems(LPP)
Multiperiod Financial Model
Decision variables: x
A
, x
B
, x
C
, x
D
, x
E respectively,
- dollars invested in investment A,B,C,D, and E y
0
, y
1
, y
2
- dollars invested in bank at time t (t=0,1,2).
The Linear Programming Model: max z = x
B
+ 1
.
9x
D
+ 1
.
5x
E
+ 1
.
08y
2
[Maximize cash on hand at time 3] x
A x
A
,
+ x
C x
B
,
+ x
D x
C
, x
D
,
+ y
0 x
E
= 100000 [Balance at time 0]
0
.
x
A
5x
A
+ 1
.
2x
C
+ 0
.
5x
B
+ 1
.
08y
+ 1
.
08y
1
−
0 x
−
E x
−
B y
−
2 y
=
1
0
= 0 [Balance at time 1]
[Balance at time 2]
≤ 75000 [Limit for dollars invested ] x
A
, . . . , x
E
, y
0
, y
1
, y
2
≥ 0 [Nonnegativity constraints]
The optimal solution: Z x
D
= 40000 , x
E
=
= 218500, x
75000, x
C
= y
0
A
= y
= 60000, x
B
1
= y
2
= 30000,
= 0. Finco should not invest in C nor in the bank.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 28 / 40

The Graphical Method of Solving LPM
Only One Optimal Solution
This method can be used when the number of variables is at least three.
Solve the following linear programming problem:
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 29 / 40

The Graphical Method of Solving LPM
Only One Optimal Solution
x
2
100
80
60
40
(20,60) max Z = 3x
1
+
2x
1
+ x
2
≤ 100
2x
2 x x
1
1
+ x
2
≤ 40
≤ 80 x
1
≥ 0 x
2
≥ 0
20
Michał Kulej (Wrocł Univ. of Techn.) z=60 z=180
10 20 30 40 50 60 70 80 z=0 x
1
Operations Research
BIS 30 / 40

The Graphical Method of Solving LPM Only One Optimal Solution
Step 1 We label the variables x
1
, x
2 and coordinate axes the x
1 graph the set of points satisfying a linear inequalities (or
, x
2 and equalities)involving two variables (it is possible to do this in three dimensional space). This set of points is called a feasible region
(the shaded area on the Figure).
Step 2 We draw an objective function Z = a
1 x
1
+ a
2
= 3x
1
+ 2x
2 for a few values of Z . In our case we take Z = 0
,
Z = 60 and Z = 180.
These lines are called isoprofit lines. Once we have drawn one isoprofit line we can generate other lines by moving parallel to this line in direction that increases Z (for a max problem). The last isoprofit line intersecting(touching) the feasible region defines the largest Z -value and indicates the optimal solution to the model.
Optimal(maximal) solution is x
1
= 20, x
2 has exactly one optimal solution .
= 60, Z = 180. This problem
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 31 / 40

The Graphical Method of Solving LPM
Alternative Optimal Solutions
100
(0,80)
80 x
2
60 z=160
(20,60) x
1 x
1 x
2 max z = 2x
1
2x
1
+ x
2
+
≤ 100
2x
2 x
1
+ x
2
≤ 80
≤ 40
≥
≥
0
0
40
20
Michał Kulej (Wrocł Univ. of Techn.) z=40
10 20 30 40 50 60 70 80 z=0 x
1
Operations Research
BIS 32 / 40

The Graphical Method of Solving LPM Alternative Optimal Solutions
This problem has multiple optimal (or alternative optimal) solutions . It means that there are an infinite number of optimal solution. (Every point on the line segment connecting points (0,80) and (20,60) is optimal, with
Z = 160).
The optimal solution we could expressed as follows: x
1 x
2
=
=
0t
80t
+ t ∈ [ 0
,
1 ]
20
+
(
60
1
( 1
− t
−
) t )
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 33 / 40

The Graphical Method of Solving LPM
Unbounded Optimal Solution
x
2
Michał Kulej (Wrocł Univ. of Techn.)
80
60
40
20 z=320 z=40
10 20 30 40
z=0
Operations Research
x
1
max z
2x
1
− x
=
2
2x
1
+ 2x
2
≤ 40 x
2
≥ 20 x
1 x
2
≥ 0
≥ 0
BIS 34 / 40

The Graphical Method of Solving LPM Unbounded Optimal Solution
Problem has no optimal solution because it is possible to find points in the feasible region with arbitrary large Z -value.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 35 / 40

The Graphical Method of Solving LPM
Nonexisting (or infeasible) Solutions
x
2
60
40
20 max z = 3x
1
60x
1
+ 40x
2
+ 2x
2
≤ 240 x
1 x
2
≥ 30
≥ 20
Michał Kulej (Wrocł Univ. of Techn.)
10 20 30 40 50 60 x
1
Operations Research
BIS 36 / 40

The Graphical Method of Solving LPM Nonexisting (or infeasible) Solutions
Problem has no optimal solution - the set of feasible solution is empty
(Empty Feasible Region). The system of inequalities (and/or equalities
) defining the feasible region is inconsistent.
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 37 / 40

The Graphical Method of Solving LPM Conclusions from Graphical Method
A set of feasible solutions (feasible region) of linear programming problem is a convex set . The set D in n-dimensional space is a convex set if the line segment joining any pairs of points of D is wholly contained in
D .
There is only a finite number of extreme points(corner points) in the set of feasible solution. For any convex set D , a point P is an extreme point if each line segment that lies completely in D and contains the point P, has P as an end point of the line segment.
The optimal solution is attained (if exists) in at least one extreme point .
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 38 / 40

Properties of any Linear Programming Problem (LPP)
Every LPP must fall into one of the following four cases:
1
2
3
4
There is only one optimal solution - it is an extreme point (corner point) of the set of feasible solution.
There are alternative optimal solutions (infinity number of optimal solutions) but there is at least one optimal solution attained in the extreme point (corner point) of the set of feasible solutions LPP.
LPP is unbounded: There are points in the set of feasible solutions with an arbitrary large Z - value (max problem) or arbitrary small Z - value (min problem).
LPP is infeasible: The set of feasible solutions is an empty set
(The feasible region contains no points).
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 39 / 40

Properties of any Linear Programming Problem (LPP)
Michał Kulej (Wrocł Univ. of Techn.)
Operations Research
BIS 40 / 40