Operational research ppt chapter 2

1. Linear Programming
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2. Linear Programming
Linear Programming Method is a technique for
choosing the best alternative from a set of feasible
alternatives, in situations in which the objective
function as well as the constraints can be expressed as
linear mathematical functions.
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3. Areas of Application
i. Product Mix
ii. Blending
iii. Distribution Problems
iv. Purchasing
v. Portfolio Selection
vi. Advertising Media Mix
vii. Production and Inventory Scheduling
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4. Total Man-hours
Available
Description Man-hours per unit
Desk Bookcase
640 Assembly 8 4
540 Finishing 4 6
100 Inspection 1 1
Profit per unit $30 $20
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A small company in office furnishings field produces both desks and
bookcases. There are three stages of production –assembly, finishing and
inspection. A desk requires 8 man-hours, and bookcase 4 man-hours of
assembly time. The finishing time needed is 4 man-hours for a desk and 6 man-
hours for a bookcase. Finally, each product requires one man-hour per unit of
inspection time. For a working week, this company has available 640 man-hours
in assembly, 540 man-hours in finishing, and 100 man-hours in inspection. The
accountants have determined that, within the range of its production
possibilities, each desk can be sold with a profit of $30, and each bookcase for a
profit of $20.
Objective: Maximum possible Profit!!!

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6. Structure of LP model
It consists of:
1. Decision variables: The courses of action, various
activities (alternatives) to achieve optimal values of
the objective function. They are denoted as X1, X2,
X3, ...., Xn.
2. Objective function: It is expressed in terms of the
decision variables to optimize the criteria of
optimality; it is a measure of performance and
denoted by Z. Z=C1X1 + C2X2 + .... + CnXn.
Where C1, C2, ...,Cn are parameters which represent
contributions of X1, X2, .., Xn to measure
performance.
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7. Cont...
3. constraints/limitations: Resources are of limited
supply, thus the mathematical relationship to
express this limitation is inequality of the decision
variables. i.e. X1 + X2 <= b1.
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8. General model
Max./Min. Z= C1X1 + C2X2 + ... + CnXn Objective function
Subject to constraints:
a11X1 + a12X2 + ...+ a1nXn (<=,=,>=) b1
a21X1 + a22X2 + ...+ a2nXn (<=,=,>=) b2
X1, X2, Xn>=O non-negativity constraint
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9. Definitions
1. Decision Variables: Quantities whose values can be chosen by a
decision maker
2. Objective Function: A function of the decision variables whose
value is to be made as large (or, in some problems, as small) as
possible
3. Constraints: Restrictions on the values which can be taken by
the decision variables
4. Non-negativity Constraints: Requirements that the decision
variables not take negative values
5. Feasible solution: Any set of values of the decision variables
which satisfies all the constraints
6. Optimal Solution: Any feasible solution such that the value of
the objective function is as high(or, in minimization problems, as
low) as possible
7. Linear Program: A setup in which the objective function and
the constraints involve only linear functions of the decision
variable
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10. Some assumptions of LPP
1. Proportionality: proportionality must exist in the
objective function and the constraint inequalities. i.e.
If unit contributes $10 to profit, total contribution must
be equal to 10X1.
2. Additivity: both in the objective function and
constraint inequalities, the total of all activities is given
by the sum total of each activity conducted separately.
(total profit is sum of individual profits).
3. Continuity: all decision variables are continuous. i.e.
Fractional values are possible.
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11. Cont...
4. Certainty: coefficients of objective function,
coefficients of constraint inequalities, and constraints
(resources) are known with certainty.
5. Finite choices: LP model assumes that a limited
number of choices are available to a decision maker and
decision variables do not assume negative values. Only
non-negative levels of activity are feasible.
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12. Formulation of LPP: Max. case
Example 2.1:
A firm is engaged in producing two products A & B. Each
unit of product A requires 2 kg of raw materials and 4
hours of labour for processing, while each unit of B
requires 3 kg of raw materials and 3 hours of labour of the
same type. Every week, the firm has an availability of 60 kg
of raw materials and 96 labour hours. One unit of product
A sold yields $40 and B gives $35 as profit.
Formulate the LPP to determine how many units of each
product should be produced per week so that the firm
earns max. Profit. Assume that there is no marketing
constraint so that all that is produced can be sold.
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13. Formulation cont...
Solution:
Max. Z= 40X1 +35X2 profit
S.T.
2X1 + 3X2<=60 raw materials constraint
4X1 + 3X2<=96 labour hours constraint
X1,X2>=0 non-negativity constraint
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14. Cont... Min. case
Example 2.2:
Agricultural research institute suggested to a farmer to
spread out at least 4800 kg of a special phosphate
fertilizer and not less than 7200 kg of a special nitrogen
fertilizer to raise productivity of crops in field. There are
two sources for obtaining these mixture A & B. Both of
these are available in bags weighting 100kg each and they
cost $40 and $24 respectively. Mixture A contains
phosphate and nitrogen equivalent of 20kg and 80kg
respectively, while mixture B contains these ingredients
equivalent of 50kg each.
Formulate this as LPP and determine how many bags of
each type should the farmer buy in order to obtain the
required fertilizer at min. Cost.
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15. formulation
Solution:
Min. Z = 40X1 + 24X2 cost
S.T.
20X1 + 50X2>=4800 phosphate requirement
80X1 + 50X2>=7200 nitrogen requirement
X1,X2>=0 non-negativity constr.
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16. Solution of LPP
LPP can be solved either by graphical method or
algebraic (simplex) method.
graphical method can be used only when two variables
are involved.
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17. Graphical method
To use graphical method for solving LPP, the following
steps are required:
1. Identify the problem. The decision variables, the
objective function and the constrain restrictions.
2. Draw a graph that includes all the
constraints/restrictions and identify the feasible
region.
3. Obtain the point on the feasible region that
optimizes the objective function---optimal solution.
4. Interpret the result.
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18. Max. case
Consider example 2.1 again.
For this problem, decision variables are X1 and X2 for
products A and B.
Thus: max. Z= 40X1 + 35X2
S.T.
2X1 + 3X2<=60
4X1 + 3X2<=96
X1,X2>=0
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20. Cont...
N.B: the region of acceptable values of the decision
variables in relation to the given constraints ( and non-
negativity restriction) is called feasible region.
Interpretation:
the optimal solution is to produce 18 units of product
A and 8 units of product B every week to get the profit
of $1000. No other product mix under the given
conditions could yield more profit than this.
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21. Cont...
alternately
Extreme point method is possible
Point X1 X2 Z
O 0 0 0
P 0 20 700
Q 18 8 1000 maximum (optimum)
R 24 0 960
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22. Minimization Case
Example 2.2 reconsidered
Min. Z = 40X1 + 24X2 total cost
S.t
20X1 + 50X2 >= 4800 phosphate requirement
80X1 + 50X2 >= 7200 nitrogen requirement
X1 , X2 >= 0
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24. Cont...
Decision variables X1 & X2 represent , respectively, the
number of bags of mixture A and of mixture B to be
bought.
Interpretation:
the total cost is minimum at the optimal point and the
optimal solution is to buy 144 bags of mixture B only
and none of mixture A. This would entail a total cost of
$3456.
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25. Binding and non-binding constraints
Constraints is termed as binding if the left hand side of the
right hand side of it are equal when optimal values of
decision variables are substituted into the constraint it is
termed non-binding otherwise.
e.g
2X1 + 3X2 2 x 18 + 3 x 8 = 60 = RHS binding
4X1 + 3X2 4 x 18 + 3 x 8 = 96 = RHS binding
20X1 + 50X2 20x0 + 50x144 =7200 ≠4800 RHS non binding
80X1 + 50X2 80x0 + 50x144=7200=RHS binding
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26. Redundant Constraint
If and when a constraint, when plotted, doesn`t form
part of the boundary making the feasible region of the
problem, it is said to be redundant.
The inclusion or exclusion of a redundant constraint
obviously doesn`t affect the optimal solution to the
problem.
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27. Cont..
Example:
Continuing with example 2.1, suppose packaging is
required. Every unit of product A requires 4 hours
while every unit of B needs 3.5 hours for packaging.
Suppose that in packaging department, 105 hours are
available every week. What product mix would
maximize the profit?
Soln.:
max. Z=40X1 + 35X2
S.T. 2X1 + 3X2<=60 raw materials constraint
4X1 + 3X2<=96 labour hours constraint
4X1 + 3.5X2<=105 packaging hours constr.
X1,X2>=0 non-negativity constraint
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29. Cont...
Interpretation:
The inclusion of the packaging hours constraint does
not provide a side of the polygon representing the
feasible region. Thus this constraint is of no
consequence and hence redundant.
optimal solution of the problem remains the same as
that of example 2.1.
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