Transcript LINEAR PROGRAMMING

LINEAR PROGRAMMING
Linear Programming
 Technique for economic allocation of
scarce resources like man, machine,
material etc to several competing activities
such as products, services etc on the basis
of given criterion of optimality.
 Linear relationships used in mathematical
modeling
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General Structure of LP
 LP model, as in any OR model, includes
three basic elements:
• Decision variables that we seek to determine
• Objective (goal) that we aim to achieve
• Constraints that we need to satisfy
 Maximize or Minimize a Linear objective
subject to linear equalities or nonequalities (constraints).
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Terminology
 Objective Function - Value measure used to rank
alternatives, Seek to maximize or minimize this
objective, examples: maximize NPV, minimize
cost. Example 3x + 5y , x & y are decision
variables.
 Decision Variables - Quantities you can control
to improve your objective which should
completely describe the set of decisions to be
made.
 Constraints - Limitations on the values of the
decision variables. Example 2x + 3y ≤24, x ≥0, y
≥0
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Mathematical Model
Optimize (max or min) Z
z  c 1 x 1  c 2 x 2  ...  c nx n
Z = measure of performance variable
c1, c2, …cn represent contribution of a unit of the
respective variable to performance of z
x1, x2, …. xn are decision variables
subject to linear , non-negative constraints.
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Algebraic Formulation
Optimize (max or min) the objective function
n
Z 
 cx
j
j
j 1
subject to linear constraints
n

a ij x j   b j ;
i = 1,2,…m constraints
j 1
x j  0;
non-negativity constraints
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Guidelines on LP model Formulation
 Step 1 – Identify decision variables.
• Express each constraint in words. Identify the equality
condition
• Express objective function verbally
• Identify the decision variables verbally
 Step 2 – Identify the problem data
 Step 3 – Formulate the constraints
• Convert the verbal constraint equation in
mathematical form.
 Step 4 – Formulate the objective function in
mathematical form
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Example of LP Model
 Reddy Mikks produces both interior and exterior paints
from two raw materials, M1 and M2. The following table
provides the basic data of the problem
Tonnes of raw Material per ton of
Exterior paint
Interior paint
Raw Material M1
6
4
Raw Material M2
1
2
Profit per ton ($1000)
5
4
Max daily
availability (tons)
24
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A market survey restricts the maximum demand of
interior paint to 2 tons. Additionally, the daily demand
of interior paint cannot exceed that of exterior paint by
more than 1 ton. Reddy Mikks wants to determine the
optimum (best) product mix of interior and exterior
paints that maximize daily profit.
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Reddy Mikks – Objective Function
 We need to determine the amount of interior and
exterior paints to be produced.
 The decision variables in this problem are
defined as
• x1 – Exterior paint produced per day
• x2 – Interior paint produced per day
 The objective is to maximize profit. Contribution
of exterior is $5000/Ton and interior paint is
$4000/Ton.
 Objective function will be
• Maximize Z (profit) = 5x1 + 4x2
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Reddy Mikks - Constraints
 The first set of constraints is the raw material
 Usage of raw material by both paints must be
less than available raw materials.
 Usage of raw material M1 is 6x1+4x2 for exterior
and interior paints.
 Similarly for M2 it is 1x1+2x2 for exterior and
interior paints.
 Available raw material is 24 Tons of M1 and 6
Tons of M2
 Hence constraint equations are
• 6x1 + 4x2 24 (Raw material M1)
• x1 + 2x2 6 (Raw material M2)
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Reddy Mikks - Constraints
 The second set of constraints is the demand.
 Maximum daily demand of interior paint is 2 T
 And the demand for interior paint over exterior
paint cannot exceed 1 T.
 Expressed mathematically these are
• x2  2
• x2 – x1  1
 Finally the non-negativity constraint gives us
• x1, x2 0
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The Complete Model
 Objective Function
• Maximize Z (profit) = 5x1 + 4x2
 Constraints
•
•
•
•
6x1 + 4x2 24 (Raw material M1)
x1 + 2x2 6 (Raw material M2)
x2  2
-x1 + x2  1
• x1, x2 0
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LP Assumptions
 Proportionality requires the contribution of each decision
variable in both the objective function and the constraints
to be directly proportional to the value of the variable. In
the RM example, if quantity discounts are offered then
the revenues will not be proportional to sales.
 Additivity stipulates that the total contribution of all the
variables in the objective function and in the constraints
to be the direct sum of individual contribution of each
variable.
 Certainty stipulates that each linear coefficient of the
objective function and constraints is known.
 Divisibility stipulates that the variable is assumed to take
fractional values.
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One More example
A tape recorder company manufactures models A, B & C,
which have profit contribution per unit of Rs 15, Rs 40
and Rs 60 respectively. The wekly minimum production
requirements are 25 units of model A, 130 units for
model B, and 55 units of model C. A dozen units of
model A requires 4 hours for manufacturing, 3 hours for
assembling and 1 hour for packaging. The corresponding
figures for dozen units of B & C are (2.5, 4,2) and (6,9,4)
respectively. During the forthcoming week the company
has available 130 hours of manufacturing, 170 hours of
assembling and 52 hours of packaging. Formulate the LP
model.
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Solution
 Ten minutes to solve.
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