LINEAR PROGRAMMING
Vassilis Kostoglou
E-mail: vkostogl@it.teithe.gr
URL: www.it.teithe.gr/~vkostogl/en
LINEAR PROGRAMMING - V. Kostoglou
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The problem of limited resources
•
Every business activity requires resources
•
There are several types of resources, such as:
–
Capital
–
Human resources
–
Equipment
–
Available time
–
Raw materials
•
In real life there are constraints in all types of resources making them limited
•
The problem: the optimal allocation of limited resources
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What is Mathematical Programming?
Mathematical programming is used to find the optimal or best available solution to a
problem that requires a decision or set of decisions about how best to use a set of
limited resources to achieve an objective.
Steps involved in Mathematical Programming
-
Conversion of a stated problem into a mathematical model that abstracts all the
essential elements of the problem
-
Exploration of different solutions of the problem
-
Finding out the optimal of most suitable solution
Linear programming requires that all the mathematical functions of the model are
linear functions.
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What is Linear Programming (LP)?
A mathematical technique helping to plan and make decisions relative to the necessary
balances to allocate resources.
Calculates find the minimum or maximum value of the objective function.
Guarantees the optimal solution to the model formulated.
Linear programming is a widely used mathematical modeling technique to determine
the optimum allocation of resources among competitive requirements.
Resources typically include raw materials, manpower, machinery, time, capital, space,
volume etc.
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LP is a powerful tool which has been proved very useful due to its application in many
different types of real business problems in areas like finance, production, sales and
distribution, personnel management, marketing and several others.
As its name implies, the LP model consists of linear objective function and linear
constraints, implying that the variables of the model have a proportionate relationship.
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Historical overview
 1928 – John von Neumann published the Relative Central Theorem of Game
Theory
 1944 – Von Neumann and Morgenstern published the Theory of Games and
Economic Behavior
 1936 – W.W. Leontief published "Quantitative Input and Output Relations in the
Economic Systems of the US" which was a linear model without objective function
 1939 – Kantoravich (Russia) actually formulated and solved a LP problem
 1941 – Hitchcock poses the Transportation Problem (special case of LP)
 WW II – Allied forces formulate and solve several LP problems related to military
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A breakthrough occurred in 1947...
 US Air Force wanted to investigate the feasibility of applying mathematical
techniques to military budgeting and planning.
 George Dantzig proposed that the interrelations among activities in a large
organization can be viewed as a LP model and that the optimal program (solution)
can be obtained by minimizing a single linear objective function.
 Air Force initiated the project SCOOP (Scientific Computing of Optimum Programs).
SCOOP began in June 1947 and at the end of that summer Dantzig and
associates had developed:
1) An initial mathematical model of the general linear programming problem.
2) A general method of solution called the Simplex method.
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The essentials of the Linear Programming model
For a given problem situation, there are certain essential conditions that need to be
solved by using Linear Programming:
1. Limited resources: limited number of labour, material equipment and finance
2. Objective: refers to the optimization aim (maximize profits or minimize costs)
3. Linearity: the increase of labour input will cause a proportionate increase of output
4. Homogeneity: products, workers, efficiency, and machines are assumed to be
identical
5. Divisibility: it is assumed that resources and products can be divided into fractions
(in practice these fractions are not possible, like production of one-third of a
computer - in these cases a modification of Linear Programming called Integer
Programming can be used)
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Some characteristic LP applications
1. Scheduling school buses to minimize the total distance traveled.
2. Allocating police patrol units to high crime areas in order to minimize response time
to calls.
3. Scheduling bank tellers so that needs are met during each hour of the day while
minimizing the total labour cost.
4. Selecting the product mix in a factory to make best use of machine and labour
hours available while maximizing the company's net profit.
5. Picking blends of raw materials in animal feed production plants to produce finished
feed combinations at minimum cost.
6. Determining the distribution system that minimizes total shipping cost.
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7. Developing a production schedule that will satisfy future demands for a firm’s
product and at the same time will minimize the total production and inventory costs.
8. Allocating the available renting space in a new shopping mall so that to maximize
the revenues of a leasing company.
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Requirements of an LP problem
1. LP problems seek to maximize or minimize a quantity (usually profit or cost)
expressed as an objective function.
2. The presence of restrictions or constraints limits the degree to which an objective
can be pursued.
3. There must be alternative courses of action to choose from.
4. The objective function and the constraints in linear programming problems must be
expressed in terms of linear equations or inequalities.
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A Linear Programming model seeks to maximize or minimize a linear function, subject
to a set of linear constraints.
The linear model consists of the following components:
-
A set of decision variables
-
An objective function
-
A set of constraints
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The importance of Linear Programming
Many real world problems lend themselves to linear programming modeling.
Also many of these real problems can be approximated by linear models.
There are well-known successful LP applications in:
– Manufacturing
– Marketing
– Finance and investment
– Advertising
– Agriculture
There are efficient techniques that solve LP models.
The outputs generated from LP computer packages provide a very useful
“what if” analysis called sensitivity analysis.
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Assumptions of the linear programming model
 The parameters' values are known with certainty.
 The objective function and the constraints exhibit constant scale returns.
 Τhe additivity assumption: There are no interactions between the decision variables.
 The continuity assumption: Variables can take any value within a given feasible
range.
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Integer and mixed-integer problems
A LP problem in which all the decision variables must have integer values is called an
integer programming problem.
A problem in which only some of the decision variables must have integer values is
called a mixed-integer programming problem.
Sometimes, some (or all) of the decision variables must have the value of either 0 or 1.
Such problems are then called zero-one mixed-integer programming problems.
Simplex method cannot be used to such problems. Advanced methods are available for
this purpose.
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Generally about LP

Problems of resources’ optimal allocation.

Expressed with a linear first degree function.

Optimization of a function which is subjected to constraints.

For non linear functions “non linear programming” is used.

Other cases: Mathematical programming, Integer programming, Programming of
separable variables.
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A model example

A chemical factory produces three independent products A, B and C.

Each product is processed sequentially by the machines Χ, Y and Ζ as following:
- Product Α: processing by the machines Χ and Y
- Product Β: processing by the machines Χ, Y and Ζ
- Product C: processing by the machines Y and Ζ
Graphical representation of the production process
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
Each product liter requires one unit of machine capacity

Daily machine capacities:
- Χ: 100 units
- Υ: 200 units
- Ζ: 400 units

Net profits for each liter of Α, Β and C are in proportion 3:4:2

Unlimited demand for products Α and C

Maximum daily demand of Β: 80 liters
According to current management’s decision, the following amounts are produced
daily: 80 liters of product Β, 20 liters of product Α and 120 liters of product C.
Is this the most appropriate policy regarding profit maximization?
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Configuration of the mathematical model
max F = 3*x1+4*x2+2*x3
with structure constraints:
x1+x2  100 (for machine Χ)
x2+x3  200 (for machine Y)
x1+x2+x3  400 (for machine Ζ)
x2  80 (from daily maximum demand)
and non-negativity constraints
x1, x2, x3  0
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The standard LP model
max f (X)  c1 x1  c2 x...  cn xn
with structure constraints:
a11 x1  a12 x2  ...  a1n xn  b1
a21 x1  a22 x2  ...  a2 n xn  b2
............................................
am1 x1  am 2 x2  ...  amn xn  bm
and non-negativity constraints
x1  0, x2  0, ... , xn  0
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Model variations

Minimization of the objective function

Some of the constraints may have > or >= or < or =

Some of the decision variables are not necessarily non-negative
In every case, a mathematical representation equivalent to the standard LP model
has to be determined.
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Alternative LP model formulation
max f (X)  C * X
with constraints
A * X  P0
X0
where
 a11 a12 ... a1n 
a

a
...
a
22
2n 
 21
.
.
.
. 

A
.
.
.
. 


.
.
. 
 .
an1 an 2 ... ann 
 x1 
x 
 2
. 

X 
.
 
 . 
 xn 
 b1 
b 
 2
. 

P0   
.
 
 . 

bm 

LINEAR PROGRAMMING - V. Kostoglou
C  [c1 c2 ... cn]
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LP terminology
 Restrictive line
A straight line corresponding to a constraint of the model.
 Vertex or extreme point
A point at which two restrictive lines are intersecting.
 Solution
Every combination of the decision variables’ values.
 Feasible solution
A solution satisfying all the constraints.
 Non feasible solution
A solution not satisfying at least one of the constraints.
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 Feasible region
The (curved) region of feasible solutions formed by the restrictive lines.
 Extreme point feasible solution
Is one of the vertexes of the feasible region.
 Neighboring feasible solutions
They are connected with an edge (boundary) of the feasible region.
 Basic solution (extreme point solution)
A solution corresponding to a vertex (has non-zero variables equal to the number
of the constraints).
 Non basic solution
A solution that is not on a vertex of the feasible region and can be feasible or non
feasible.
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
Basic feasible solution
A basic solution that corresponds to a vertex of the feasible region and all its
variables are non-negative.

Slack value
Any excess of an available resource (constraint symbol < or <).

Surplus value
Any "surpassing" (surplus) of a requirement (constraint symbol > or >).

Auxiliary variables
Variables corresponding to slack values and surplus values.

Basic (non basic) variable
A non-zero (zero) variable in a solution containing decision and auxiliary variables.

Active (or mandatory) constraint
When the resource is completely consumed or there is no surplus (zero slack or
surplus respectively).
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
Non mandatory constraint
When a resource is not exhausted (<) or a requirement is being surpassed (>)
(non-zero slack or surplus respectively).

Optimal solution
The feasible solution of an extreme point that gives to the objective function the
optimal value (maximum or minimum).
The optimal solution can be only one; however there are a case with unlimited
optimal solutions or no optimal solution or the value of the objective function tends
to infinity. In every case the amount of feasible solutions of an extreme point is
finite. When a feasible solution of an extreme point is better than all its neighboring
then this solution is the optimal.

Optimal value
The value of the objective function that corresponds to the optimal solution.
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Example 1
A manufacturing industrial unit produces three conventional television models
(standard, deluxe and super) and a high-tech model (high-tech). Management is certain
that all production can be purchased in the market. Each device passes in the
production process from all three relevant departments: the workshop, the assembly,
and the final control.
The number of working hours required for each device type in each department is listed
in the following table.
Departments /TV types
Standard Deluxe
Super
High-tech
Workshop
12
15
15
25
Assembly
10
12
13
20
Control
1/2
3/5
2
2
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The total capacity of the unit and the capacities of the three departments do not allow
more than 2500, 3000 and 240 man-hours per day in the workshop, assembly and
control respectively. Moreover, due to existing signed contracts at least 50 standard
and 50 deluxe devices must be produced daily. The net profit (selling price - total unit
cost) from the sale of one unit of each device is:
Standard Deluxe Super
Profit (€)
25
30
40
Hightech
100
Questions:
1. Formulate a linear programming model to determine the optimal production mix.
2. Investigate the possibility of reducing the dimensions of the Simplex tableau by
removing any redundant constraints.
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Problem analysis

Decision variables:
The production quantities of the four TV types: x1, x2, x3, x4

Objective function:
Maximization of the total profit - F = 25*x1 + 30*x2 + 40*x3 + 100*x4

Constraints for each department

Constraints for the production quantities
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Model formulation
max F  25x1  30x2  40x3  100x4
with constraints
12 x1  15 x2  15 x3  25 x4  2500
10 x1  12 x2  13x3  20 x4  3000
(1 / 2) x1  (3 / 5) x2  2 x3  2 x4  240
x1  50
x2  50
and
x1 , x2 , x3 , x4  0
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Observing the first two structure constraints it can be seen that:
10 x1  12 x2  13x3  20 x4  12 x1  15x2  15x3  25x4  2500  3000
i.e. the left side of the second constraint is obviously smaller than the right side.
So, the second constraint is surplus and therefore it can be removed.
The problem can be simplified more if the following transformation of variables is
carried out:
x1  50  x1  50  0  x1  0
x2  50  x2  50  0  x2  0
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Substituting in the constraints
x1  x1  50
and
x2  x2  50
indicates that
12 x1  15 x2  15 x3  25 x4  1150
(1 / 2) x1  (3 / 5) x2  2 x3  2 x4  185
Dividing the first constraint by ~ 6.22 (= 1150/185) indicates that
1.93x1  2.4 x2  2.4 x3  4 x4  185
This relation 'covers' the second constraint of the problem
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Final LP model of the problem
max F   25x1  30x2  40x3  100x4
with just one structure constraint
12x1  15x2  15x3  25x4  1150
and
x1, x2 , x3 , x4  0
where
x1  x1  50
x2  x2  50
F   F  2750
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Example 2
A housewife faces the problem of purchasing quantities of meat, potatoes and
vegetables, which constitute her main personal diet. She has already been informed by
the dietitian of a women's magazine that her full weekly diet nutrition should contain as
a minimum requirement 8 units of carbohydrates, 15 units of protein and 6 units of
vitamins.
The unit numbers of these three constituents contained in each unit of weight (kg) of
the three foods and the food unit costs are shown in the following table.
Foods
Meat
Potatoes
Vegetables
Carbohydrates
3
1
1
Proteins
4
3
4
Vitamins
1
3
1
Cost (€/kg)
10
1
2
Nutrients
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The housewife would like to know the quantities of each food she should purchase, so
that to meet the minimum requirements of these basic ingredients and minimize the
total purchasing cost.
After some time a new nutrient requirement consisting of 7 units of calcium has been
added to housewife’s dietary. Each kg of meat, potatoes and vegetables contain 2, 1
and 4 units of calcium, respectively.
How this additional requirement of the new ingredient affects the quantities that must
be purchased with a minimum cost?
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Problem analysis
 Decision variables:
The quantities (kg) of meat, potatoes and vegetables - x1, x2, x3
 Objective function:
Minimization of total cost:
F = 10*x1 + 1*x2 + 2*x3
 Constraints of nutrients (minimum required quantities)
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Model formulation
min F  10x1  x2  2 x3
with constraints
3 x1  1x2  1x3  8
4 x1  3 x2  4 x3  15
1x1  3 x2  1x3  6
and
x1 , x2 , x3  0
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For the new additional nutrient (calcium) the addition of a new constraint is required:
2 x1  1x2  4 x3  7
When the problem with the new constraint is solved the influence of the new nutrient
to the cost and composition of the diet can be determined.
 What is sensitivity analysis
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Example 3
A hydraulic components production small company manufactures two types of safety
valves. For this purpose the company buys matrices (molds) from an external supplier,
and then processes, holes and smoothes them. The respective production rates expressed in number of valves per hour - of the above three phases are presented in
the following table.
Valves production rates per phase
Type Α
Type Β
Phase
Processing
30
40
Holing
28
35
Smoothing
30
25
Valve type
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Each matrix for type A valves costs €2, while that for type B valves costs €3. The ready
valves of type A and B are sold €5 and €6 respectively. The use of the three basic
production machines (lathe, drill, and grinder) implies a current operating cost, which
equals €20, €14 and €20 per working hour respectively. It is considered that there is no
restriction in the amounts or combination of the valves of the two types that can be sold
in the market.
Develop the appropriate mathematical model to determine the mix of production that
maximizes the total net profit.
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Problem analysis
 Decision variables:
The number of valves of type Α and Β produced per hour - x1, x2
 Constraints:
There are not obvious constraints originating neither from the production process
nor from the capacities of the 3 phases.
The only constraint is time; essentially the hourly rates of valves production.
Lathe (phase 1):
In one hour it produces 30 valves Α or 40 valves Β 
one unit of type Α is produced in 1/30 of the hour, while one unit of type Β in 1/40 of
the hour  x1 units of type Α in x1/30 hours and x2 units of type Β in x2/40 hours 
total required time: x1/30 + x2/40 
The time constraint of lathe is: x1/30 + x2/40  1 (hour)
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Drill (phase 2):
In one hour it produces 28 valves of type Α or 35 valves of type Β 
one unit of type Α is produced in 1/28 of the hour and the 1Β in 1/35 
x1 units of A in x1/28 and x2 units of B in x2/35 
total time: x1/28 + x2/35 
The time constraint of lathe is:
x1/28 + x2/35  1 (hour)
Grinder (phase 3):
In one hour produces 30 valves Α or 25 valves Β 
one unit of type Α in 1/30 of the hour, while one unit of type Β in 1/25 
x1 units of A in x1/30 and x2 units of B in x2/25 
total time: x1/30 + x2/25 
The time constraint of lathe is: x1/30 + x2/25  1 (hour)
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Cost of each valve of type Α:
If in 1 hour the lathe produces 30 units and the hourly operating cost is €20,
this means that the cost of each unit is €20/30.
In the drill (with the same reasoning) the cost is €14/28.
In the grinder the production cost of each unit is €20/30.
Total cost of producing a unit of type A: 20/30 +14/28 +20/30 = €11/6
Cost of each valve of type B:
With the same analysis and by summing the individual costs on the 3 machines,
the total production cost of each unit is: 20/30 +14 / 28 +20 / 30 = €11/6
Net profit from the sale of each unit:
Profit from the sale of each unit of type A:
€5 - (€2 + €11/6) = €7/6
Profit from the sale of each unit of type B:
€6 - (€3 + €17/10) = €13/10
Total hourly production profit:
(7/6)*x1 + (13/10)*x2
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The LP model
max F 
7
13
x1 
x2
6
10
with constraints
x1
x2

1
30
40
x1
x2

1
28
35
x1
x2

1
30
25
and
x1 , x2  0
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Example 4
A company manufactures two new models of modern refrigerators with corresponding
names "no frost'' and" freezer ". The production area is divided in three departments:
the forming part (F) where the final surfaces are formed, the assembly (A), and the
painting (P) where the final finishing and painting of the refrigerators are made.
Each refrigerator type needs a different number of working hours in each department.
So, each batch of 12 refrigerators "no frost" needs 60, 80 and 20 working hours in
departments F, A and P respectively, while the respective batch of 12 refrigerators
"freezer" needs 70, 85 and 10 working hours in each of the three departments
respectively. The staffing of the departments is such, that the working hours available
to each of them are 2400, 3000 and 600 respectively during a working month.
Design a monthly production schedule minimizing the total idle time of the staff.
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Problem analysis

Decision variables:
Number of batches of the 2 refrigerators types - xNF, xF

Constraints in the available working hours in each department:
Forming (F): 60xNF + 70xF  2400
Assembly (A): 80xNF + 85xF  3000
Painting (P): 20xNF + 10xF  600
The difficulty of the problem lies in the creation of the objective function as there are no
economic data, the aim being to minimize the unused time.
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Observing the composition of the three constraints referred to man-hours, it comes up
that the unused time in each department is the difference between the right (available
working hours) and the left (actual working hours) side of the corresponding constraint.
Thus we can define new variables x1, x2, and x3, symbolizing the unused (idle) times of
departments F, A and P respectively.
Idle times of each department:
Department F: 60xNF + 70xF  2400  60xNF + 70xF + x1 = 2400
Department A: 80xNF + 85xF  3000  80xNF + 85xF + x2 = 3000
Department P: 20xNF + 10xF  600  20xNF + 10xF + x3 = 600
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LP model formulation
min F  6000 160xNF 165xF
with constraints
60 x NF  70 xF  2400
80 x NF  85 xF  3000
20 x NF  10 xF  600
and
xNF , xF  0
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Example 5
Nori & Leets Co., one of the largest steel producers in the world, is located next to
Steeltown and is nearly the sole employer of its inhabitants. Steeltown grew and
progressed rapidly along with the company, which now employs nearly 50.000 of its
inhabitants. For this reason the mentality of its residents for many years has been:
"whatever is good for Nori & Leets is good for the city too." However this has now
changed, because uncontrolled air pollution from factory furnaces has altered the city
environment and endangered the health of the population.
A recent uprising of the shareholders led to the election of a new enlarged managing
board. New management is determined to pursue a responsible social policy and has
already discussed with officials of the municipality and representatives of the citizens
about what should be done concerning the problem of air pollution. They have mutually
concluded to strict specifications of air quality for city’s atmosphere.
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The three main types of pollution factors in the atmosphere of the region are
particulates, sulfuric oxides and hydrocarbons. The new measures require that the
company must reduce its annual emissions of these pollution factors by the following
amounts (expressed in thousands of tons).
Required annual reductions of contaminants’ emissions
Pollution factor
Particulates
Required reduction of annual
emission rate (thousand tons)
60
Sulfuric oxides
150
Hydrocarbons
125
The managing board instructed company’s engineers to determine how the required
decreases can be achieved in the most economical manner.
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The operation of the factory implies two main pollution sources: a) the blast furnaces
used to produce cast iron, b) the open hearth furnaces used to convert iron into steel.
In both cases, engineers suggested that the most effective ways of mitigating pollution
are: 1) increasing the height of the furnaces, 2) using filters (including gas traps) in the
furnaces, and 3) improving the quality of fuels (using cleaner high-quality ingredients)
used in the furnaces.
All these methods have some specific technological limits on the amount of the
emission of the pollution factors that can wipe out as shown in the table below.
Nevertheless, these methods can be used in any fraction (percentage) of their potential
of pollution mitigation. Moreover, as the function of the three methods is independent,
the emission reduction achieved by each of them is not significantly affected by
simultaneous use of anyone of the others.
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51
Reductions of emission rates by the full use of each method
of pollution mitigation (in thousand tons per year)
Higher furnaces
Pollution factors
Filter use
Improved fuels
Blast
Open hearth
Blast
Open hearth
Blast
furnaces
furnaces
furnaces
furnaces
furnaces
Open hearth
furnaces
Particulates
12
9
25
20
17
13
Sulfate oxides
35
42
18
31
56
49
Hydrocarbons
37
53
28
24
29
20
After the collection of the above data it became obvious that none of the three methods
was sufficient itself to achieve all necessary reductions in environmental pollution.
However, the combination of all methods in full capacity on the one hand would be
more than enough, and on the other hand the total cost would be prohibitive for the
company if, as natural, the prices of its products should continue to be competitive.
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52
For these reasons the company's engineers concluded that a combination of these
methods should be used, probably with fractions (percentages) of their full capacity of
pollution mitigation based on the corresponding costs. Moreover, because of the
differences between the blast furnaces and the open hearth furnaces, probably should
not be used the same combination of methods in these two types of furnaces.
It was also specified that the cost of a method’s reduced use is directly proportional to
the percentage of its capacity. Thus, for any given fraction used, the corresponding
annual cost is equal to the appropriate percentage of the full cost. The company then
conducted detailed costing to estimate the total annual cost for the implementation of
each of the three pollution mitigating methods. Apart of the increased operating and
maintenance costs, it was also given emphasis in both the fixed cost of each method
(which was converted to its equivalent on annual basis) and at any intermediate loss in
the efficiency of the production process. This analysis resulted in the following
estimates of the annual cost for using each method to full capacity.
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53
Annual cost of using each pollution mitigation method (in millions of €)
Mitigation method
Blast furnaces
Open hearth furnaces
Higher furnaces
8
10
Filter use
7
6
Improved fuels
11
9
The final stage of the preliminary study was the development of a general frame for the
overall plan of the company on pollution mitigation. The project consists of determining
the methods which will be used for the reduction of the pollution and the possible
fractions of their use.
Due to the combined nature of the problem of determining the plan that meets all the
requirements at minimum cost, an Operational Research group was formed to solve it.
The group adopted an approach using Linear Programming. Which was it?
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Problem analysis

Decision variables:
x1: heighten the blast furnaces
x2: heighten the open hearth furnaces
x3: filter use in the blast furnaces
x4: filter use in the open hearth furnaces
x5: fuels improvement in the blast furnaces
x6: fuels improvement in the open hearth furnaces
where
xi (i = 1, 2, …6) the fraction used of the corresponding method

Total cost:
8x1+10x2+7x3+6x4+11x5+9x6
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
Constraints from the reductions of the pollution factors:
For the particulates:
12x1+9x2+25x3+20x4+17x5+13x6  60
For the sulfate oxides:
35x1+42x2+18x3+31x4+56x5+49x6  150
For the hydrocarbons:
37x1+53x2+28x3+24x4+29x5+20x6  125
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The LP model
min F  8x1  10x2  7 x3  6 x4  11x5  9 x6
with nine structure constraints
12 x1  9 x2  25 x3  20 x4  17 x5  13x6  60
35 x1  42 x2  18 x3  31x4  56 x5  49 x6  150
37 x1  53x2  28 x3  24 x4  29 x5  20 x6  125
x1 , x2 , x3 , x4 , x5 , x6  1
and
x1 , x2 , x3 , x4 , x5 , x6  0
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Example 6
A paper industry produces among other products white high resolution paper, which is
wrapped in two meters long pieces in standardized cylinders of one meter height. The
white paper is cut into a number of smaller and probably different lengths, with fixed
height of one meter, in order to meet all customers’ orders.
The paper industry received last week the following orders. The lengths are expressed
in palms (tenths of meter, i.e. 10 cm), which is the usual measure of this type of paper.
Weekly orders
Length (palms) Number of items
9
30
7
150
5.5
65
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The industry would like to cut the fixed length of 20 palms white papers so that to
satisfy all the above orders, and to minimize the total wastage (loss).
Design an appropriate mathematical model for solving the problem of the paper
industry with the use of Linear Programming.
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Problem analysis
It is not possible to cut more than three pieces of white paper from one cylinder
(e.g. 3*5.5 = 16.5 and 3.5 palms as wastage - this way can be symbolized as: 5.5, 5.5,
5.5 – wastage 3.5).
Each cylinder can be cut to one, two or three pieces by the following 13 possible ways
(lengths of pieces - wastage):
(9, 0, 0 - 11) - (9, 9, 0 - 2) - (9, 7, 0 - 4) - (9, 5.5, 0 - 5.5) - (9, 5.5, 5.5 - 0) (7, 0, 0 - 13) - (7, 7, 0 - 6) - (7, 7, 5.5 - 0.5) - (7, 5.5, 0 - 7.5) - (7, 5.5, 5.5 - 2) (5.5, 0, 0 – 14.5) - (5.5, 5.5, 0 - 9) and (5.5, 5.5, 5.5 - 3.5).
Every possible way of cutting the blotting paper can be represented by a variable xi.
Consequently 13 variables are defined:
x1, x2, …x13 (i = 1, 2,.. 13): number of cylinders that are cut by the ith way
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Total loss:
Minimization of the objective function
F = 11*x1 + 2*x2 + 4*x3 + 5.5*x4 + 0*x5 + 13*x6 + 6*x7 + 0.5*x8 + 7.5*x9 +
2*x10 + 14.5*x11 + 9*x12 + 3.5*x13
Constraints:
For the 30 pieces of 9 palms: 1x1 + 2x2 + 1x3 + 1x4 + 1x5 = 30
For the 150 pieces of 7 palms: 1x3 + 1x6 + 2x7 + 2x8 + 1x9 + 1x10 = 150
For the 65 pieces of 5,5 palms: 1x4 + 2x5 + 1x8 + 2x9 + 2x10 + 1x11 + 2x12 + 3x13 = 65
Also, for all variables non-negativity constraints apply.
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Organization of the mathematical model
1) Conversion of the objective function and constraints in the LP model format.
2) Change the inequality direction of the required constraints, so that the right hand
sides of all constraints are non-negative numbers. Consequently, if the right hand
side of a constraint is negative, a change of the sign and the inequality direction is
needed (multiplying both sides by -1).
3) Conversion of the structure constraints which are inequalities to equivalent
equalities. This is achieved by the addendum (addition if the direction of the
inequality is < or  or deduction if the direction is of the format > or ) of new nonnegative variables, which are called dummy variables (slack or surplus).
4) Construction in the left hand side of the structure constraints of a complete unary
table by adding the appropriate number of new variables, called artificial variables.
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Methodology of graphical solution
1. Each constraint is initially considered as equality and is plotted by a line (or a level
in three-dimensional problems).
2. The part of the level (or of the space in three-dimensional problems), which’s all
points verify a specific constraint is identified for every constraint.
[The fastest way is to check whether the intersection of the axes (coordinate 0,0)
verifies the constraint (what is valid for this coordinate will also be valid for all points
belonging to the same semi-level)]
3. The intersection of all semi-levels is defined, which verifies all the constraints.
Consequently all vertices of the intersection are being defined.
4. The objective function f(x) is plotted, consisting of the family of parallel lines with a
standard direction coefficient. The different values of the objective function are
represented by the corresponding parallel lines.
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63
5. The objective function line moves in parallel to herself to the direction where her
value increases (or decreases if the aim is its minimization), until it meets with the
part of the level which is the intersection of all constraints. From the family of lines
meeting the intersection of the constraints, only the lines crossing the intersection
vertices are considered.
6. The values of the lines crossing the intersection vertices are calculated. The vertex,
from which passes the line with the highest (smallest in minimization problems)
value is the optimum solution of the problem. The values of the variables consisting
the optimum solution are the coordinates of this vertex.
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Graphical solution of maximization problems
max f(x) = x1 + 2x2
with structure constraints
2x1 + x2  20
x1 + x2  12
x1 + 3x2  15
and non-negativity constraints
x1, x2  0
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Graphical solution of maximization problem
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66
Vertex
Coordinates
f(x) value
Α
(0,0)
0
Β
(0,5)
10
C
(9,2)
13
D
(10,0)
10
Thus, the maximum value of f(x) is 13, the corresponding line meets vertex C, and the
values of the variables are x1 = 9 and x2 = 2.
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Graphical solution of minimization problems
min z(x) = 24x1 + 18x2
with structure constraints
120x1 + 60x2  480
56x1 + 112x2  440
103x1 + 120x2  720
and non-negativity constraints
x1 , x2  0
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Graphical solution of minimization problem
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69
Solving the system of the two equations results in:
x1 = 1.752 and x2 = 4.496
Thus, the minimum value of the objective function is:
Min Ζ = 24*1.752 + 18*4.496 = 122.98
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70
The Simplex method
A large variety of Simplex-based algorithms exist to solve LP problems.
Other (polynomial type) algorithms have been developed for solving LP problems:
- Khachian algorithm (1979)
- Kamarkar algorithm (AT&T Bell Labs, mid 80s)
None of these algorithms has been able to beat Simplex in actual practical applications;
hence Simplex (in its various forms) is and will most likely remain the most dominant
LP algorithm for at least the near future.
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71
Fundamental theorem
Extreme point (or Simplex filter) theorem:
If the maximum or minimum value of a linear function defined over a polygonal convex
region exists, then it is certain that this value is at the boundary of the region.
A finite number of extreme points imply a finite number of solutions.
Hence, the search is reduced to a finite set of points.
However, a finite set can still be too large for practical purposes.
Simplex method provides an efficient systematic search guaranteed to converge in a
finite number of steps.
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72
General form of the initial Simplex tableau
Base x1
x2 …
xn
xn+1 xn+2 ...
xn+m
R.H.
xn+1
a11
a12 ...
a1n
1
0
...
0
b1
xn+2
a21
a22 ...
a2n
0
1
...
0
b2
.............................................................
...
...
xn+m am1
-f
c1
am2 ... amn
0
0
...
1
bm
c2 ...
0
0
...
0
0
cn
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Previous Simplex tableau
Base
x1
...
xm
xm+1
...
Xs
...
xn+m
R.H.
x1
1
...
0
a1,m+1
...
a1,s
...
a1,n
b1
...
...
...
...
...
...
...
...
0
0
...
...
...
...
...
...
...
...
...
...
...
xr
0
0
ar,m+1
ar,s
ar,n
br
...
...
...
...
...
...
...
...
0
0
...
...
...
...
...
...
...
...
...
...
...
xm
0
...
1
ar,m+1
...
am,s
...
am,n
bm
...
0
...
0
cm+1
...
cs
...
cn
-f0
...
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74
New (current) Simplex tableau
Base
x1
…
…
...
xr
…
...
...
xm
-f
x1 ..
xr .. ..
1
0 - a1sarr
0
...
...
...
1
0 arr=
ars
…
…
...
...
...
...
0
-amsarr
0
.... csarr.....
xm ....
xm+1 ......... xs ............ xn+m
0.. ...a1,m+1 -a1,sar,m+1 0
a1,m+1 -a1,sar,m+1
...
...
…
...
...
...
...
...
ar,m+1
ar,n
0 ar,m+1 =
1
arn =
ars
ars
...
...
…
…
...
...
...
...
...
...
...
...
1 am,n+1 -am,sar,m+1
0
am,n -am,sar,n
0
cm+1 -csar,m+1 .... 0 ....
cn-csar,n
LINEAR PROGRAMMING - V. Kostoglou
R.H.
b1-a1,sb’r
…
…
br
= b’r
ars
…
…
…
bm-amsbr
-f0-csar,n
75
Basic steps of Simplex
1. Begin the search at an extreme point (i.e. a basic feasible solution).
2. Determine if the movement to an adjacent extreme point can improve
the optimization of the objective function.
If not, the current solution is optimal.
If improvement is possible, then proceed to the next step.
3. Move to the adjacent extreme point which offers (or at least appears to offer)
the best improvement in the objective function.
4. Continue with steps 2 and 3 until the optimal solution is found
or it can be shown that the problem is either unbounded or infeasible.
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76
Synopsis of Simplex method steps
(after the determination of the initial basic feasible solution)
Step 1:
Identification of the incoming non-basic variable.
Step 2:
Identification of The outgoing non-basic variable.
Step 3:
By elementary operations between the lines of the current tableau the next
Simplex tableau is indicated, in which are reflected the changes that
practically lead to the next vertex of the feasible region.
Step 4:
The new (current) basic feasible solution is being tested whether it is optimal.
If this solution is optimal then the procedure stops,
otherwise: Move to step 1.
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77
Steps of LP problem solution - Maximization problems
1.
Introduction to the base of the variable xs with the largest positive coefficient cs
(lowest negative in minimization problems).
2. Removal from the base of the variable xr with the smallest positive quotient br/ars,
where ars > 0.
3. Formulation of a new Simplex tableau replacing into the base variable xr with xs.
4. Calculation of the elements of variable xs new line by dividing the corresponding
elements of variable xr previous row of with the pivot element ars.
5. Replacement of all remaining elements according to the equation:
New element = respective previous element –
[(respective element of previous column of xs)*(respective element of new row of xr)]
6.
Back to step 1
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78
The above procedure is repeated continuously until one of the followings occurs:
a) All cost coefficients become negative or zero.
In this case the optimal solution has been found.
or
b) All the ais coefficients become negative or zero.
In this case the problem is unbounded and there is no finite solution.
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79
Solution of the model example (see p. 17)
max F = 3*x1+4*x2+2*x3
with structure constraints
x1+x2  100
x2+x3  200
x1+x2+x3  400
x2  80
and non-negativity constraints
x1, x2, x3  0
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80
Adding the appropriate number of slack variables (artificial variables are not needed as
all structure constraints are of the type ) the model example is formulated as:
maxf = 3x1 + 4x2 + 2x3
x1 + x2
+ x4
x2 + x3
+ x5
= 100
= 200
x1 + x2 + x3
= 400
x2
+ x6
+ x7
= 80
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81
Sequential Simplex tableaux (maximization problem)
Base
x4
x5
x6
x7
-f
x1
1
0
1
0
3
x2
1
1
1
1*
4
x3
0
1
1
0
2
x4
1
0
0
0
0
x5
0
1
0
0
0
x6
0
0
1
0
0
x7
0
0
0
1
0
R.H.
100
200
400
80
0
x4
x5
x6
x2
-f
1*
0
1
0
3
0
0
0
1
0
0
1
1
0
2
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
-1
-1
-1
1
-4
20
120
320
80
-320
x1
x5
x6
x2
-f
1
0
0
0
0
0
0
0
1
0
0
1*
1
0
2
1
0
-1
0
-3
0
1
0
0
0
0
0
1
0
0
-1
-1
0
1
-1
20
120
300
80
-380
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82
x1
x3
x6
x2
-f
1
0
0
0
0
0
0
0
1
0
x1
x3
x6
x7
-f
1
0
0
0
0
1
1
-1
1
-1
0
1
0
0
0
1
0
-1
0
-3
0
1
-1
0
-2
0
0
1
0
0
-1
-1
1
1*
1
20
120
180
20
-620
0 1
1 0
0 -1
0 0
0 -3
0
1
-1
0
-2
0
0
1
0
0
0
0
0
1
0
100
200
100
80
-700
Optimal solution:
x1 = 100, x2 = 0, x3 = 200, x4 = 0, x5 = 0, x6 = 100, x7 = 80 and maxf = 700
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83
Steps of LP problem solution – Minimization problems
1. Formulation of the artificial function F = Σxi, where xi the artificial variables
expressed as functions of the other variables of the corresponding constraint.
2. Construction of the first Simplex tableau, putting an extra line at the bottom with the
coefficients of the artificial function F under those of the initial function f.
3. Minimization of F with the known procedure of Simplex method (steps 1 to 6 of the
algorithm), inserting each time into the base, the variable with the smallest negative
coefficient ci.
4. Repetition of the minimization procedure for function F, until it takes the value zero
(where all the cost coefficients become negative or zero).
5. Removal of the coefficient row of F and of the columns of all artificial variables.
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84
6. Minimization of the initial function f, inserting each time in the base the variable with
the smallest negative coefficient.
7. Repetition of the minimization procedure for function f, until all cost coefficients
become positive or zero.
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85
Solution of a minimization problem
The following minimization problem is given:
minf = x1 + 2x2
with structure constraints
2x1 + x2  6
x1 + x2 = 4
x1 + x2  8
and non-negativity constraints
x1  0, x2  0
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86
The existing inequalities are transformed to equalities (by adding slack variables x3 and
x4) and the unary table is created (by adding the artificial variables x5 and x6). Thus, the
problem is formulated as following:
minf = x1 + 2x2
with constraints
2x1 +
x2 + x3
x1 +
x2
x1 + 3x2
=6
+ x5
- x4
=4
+ x6 = 8
and
x1, x2, x3, x4, x5, x6  0
According to the algorithm, at first the artificial function F has to be created.
F = x5 + x6 = (4 - x1 - x2) + (8 - x1 - 3x2 + x4)  F = 12 - 2x1 - 4x2 + x4
Simplex tableaux are following, up to the determination of the optimal solution,
minimizing successively functions F and f.
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87
Sequential tableaux of Simplex method (minimization problem)
Base
x3
x5
x6
-f
-F
x1
2
1
1*
1
-2
x2
1
1
3
2
-4
x3
1
0
0
0
0
x4
0
0
-1
0
1
x5
0
1
0
0
0
x6
0
0
1
0
0
R.H.
6
4
8
0
-12
x3
x5
x2
-f
5/3
2/3*
1/3
1/3
0
0
1
0
1
0
0
0
1/3
1/3
-1/3
2/3
0
1
0
0
-1/3
-1/3
1/3
-2/3
10/3
4/3
8/3
-16/3
-F
x3
x1
x2
-f
-F
-2/3
0
1
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
-1/3
-1/2
1/2
-1/2
1/2
0
0
-5/2
3/2
-1/2
-1/2
1
4/3
1/2
-1/2
1/2
-1/2
1
-4/3
0
2
2
-6
0
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88
Final Simplex tableau
Base
x3
x1
x2
-f
x1
0
1
0
0
x2
0
0
1
0
x3
1
0
0
0
x4 R.H.
-½
0
½
2
-½
2
½
-6
Optimal solution: x1 = 2, x2 = 2, x3 = x4 = x5 = 0 and min f = 6
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89
Models without unique optimal solutions
•
Infeasibility:
Occurs when a model has no feasible point.
•
Unboundness:
Occurs when the objective function can become infinitely large (max) or infinitely
small (min).
•
Alternate solution:
Occurs when more than one point optimizes the objective function
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90
LP application case study
A product mix problem
A manufacturer, named Giga Ltd., has fixed amounts of different resources such as
raw material, labour, and equipment.
These resources can be combined to produce anyone of several different products.
The quantity of the ith resource required to produce one unit of the jth product is known.
The decision maker wishes to produce the combination of products that will maximize
total income.
The Giga company produces two large products: I and II.
The raw material requirements, space needed for storage, production rates, and selling
prices for these products are given in the following table.
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91
Production data for Giga company
Product
Storage space (m2 /unit)
Raw material (kg/unit)
Production rate (units/hr)
Selling price ($/unit)
I
4
5
60
13
II
5
3
30
11
The total amount of raw material available per day for both products is 1575 kg.
The total storage space for all products is 1500 m2, and a maximum of 7 hours per day
can be used for production.
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92
Graphical solution
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Analysis of the problem
When the decision variables are more than two, it is always advisable to use Simplex
method to avoid a lengthy graphical procedure.
The Simplex method is not used to examine all the feasible solutions.
It deals only with a small and unique set of feasible solutions, the set of vertex points
(i.e. extreme points) of the convex feasible space that contains the optimal solution.
The application of the Simplex method yielded the following optimum production
programme for Giga:
The company can maximize its sale income to € 4335 by producing 270 units of
product I and 75 units of product II. There will be no surplus of raw materials or
production time, but there will be 45 units of unused storage space.
The managers are interested to know whether it is worthwhile to increase the
production by purchasing additional units of raw materials and by either expanding its
production facilities or working overtime.
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The critical questions are:
•
Which is the income value (or marginal value) of each additional unit of each type of
resources?
•
Which is the maximum cost (or marginal cost) that they should be willing to pay for
each additional unit of resources?
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Answers to these questions can be obtained from the objective function in the last
tableau of the Simplex solution:
Z
15
16
S2 
S 3  $4335
7
7
that is,
Z  $4335 
15
16
S2 
S3
7
7
As S1, S2 and S3 represent the surplus resources, negative values of these variables
(i.e., -S1, -S2, -S3) represent additional units of these resources that can be made
available. The income values (or marginal values) of additional units of these resources
can be obtained by taking the partial derivatives of Z with respect to -S1, -S2 and -S3.
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Therefore, the marginal value of one additional unit of each resource is:
Storage space 
Z
 $0
( S1 )
Z
15
Raw materials 
$
 ( S )
7
2
Z
16
Production time 
$
 (  S3 )
7
Thus, the marginal values of additional units of resources can be obtained directly from
the coefficients of the objective function in the last tableau of a Simplex solution.
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The Giga Company should be willing to pay up to $15/7 for an additional unit of raw
materials and up to $16/7 for an additional unit of production time.
If the actual cost of an additional unit of these resources is smaller than the marginal
value, the company should be able to increase its income by increasing production.
However, the marginal values above are valid only as long as there is surplus storage
space available.
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Sensitivity analysis
This analysis helps to test the sensitivity of the optimal solution with respect to changes
of the objective function coefficients, the coefficients in the constraints’ inequalities, or
the constant terms (right hand sides) of the constraints. For example in the above case
study:

The actual selling prices (or market values) of the two products may vary from time
to time. Over which ranges can these prices change without affecting the optimality
of the present solution?

Will the present solution remain the optimal solution if the amount of raw materials,
production time or storage space is suddenly changed because of shortages,
machine failures or other unexpected events?

The amount of each resource type needed to produce one unit of product type can
be either increased or decreased. Will such charges affect the optimal solution?
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