-
Linear Programming Problems - FormulationLinear Programming is a
mathematical technique for optimum allocation of limited or scarce
resources, such as labour, material, machine, money, energy and so
on , to several competing activities such as products, services,
jobs and so on, on the basis of a given criteria of optimality.
-
Linear Programming Problems - FormulationThe term Linear is used
to describe the proportionate relationship of two or more variables
in a model. The given change in one variable will always cause a
resulting proportional change in another variable.
-
Linear Programming Problems - FormulationThe word , Programming
is used to specify a sort of planning that involves the economic
allocation of limited resources by adopting a particular course of
action or strategy among various alternatives strategies to achieve
the desired objective.
-
Structure of Linear Programming modelThe general structure of
the Linear Programming model essentially consists of three
components. i) The activities (variables) and their relationships
ii) The objective function and iii) The constraints
-
i) The activities are represented by X1, X2, X3 ..Xn. These are
known as Decision variables.ii) The objective function of an LPP
(Linear Programming Problem) is a mathematical representation of
the objective in terms a measurable quantity such as profit, cost,
revenue, etc.
-
(Maximize or Minimize)Z=C1X1 +C2X2+ ..Cn XnWhere Z is the
measure of performance variable X1, X2, X3, X4..Xn are the decision
variablesAnd C1, C2, Cn are the parameters that give contribution
to decision variables.iii) Constraints are the set of linear
inequalities and/or equalities which impose restriction of the
limited resources
-
General Mathematical Model of an LPP(Maximize or Minimize) Z=C1
X1 + C2 X2 ++CnXnSubject to constraints,a11X1+ a 12X2++ a 1nXn ()
b1a21X1+ a 22X2++ a 2nXn () b2a31X1+ a 32X2++ a 3nXn () b3am1X1+ a
m2X2++ a mnXn () bmand X1, X2 .Xn > 0
-
Guidelines for formulating Linear Programming modeli) Identify
and define the decision variable of the problemii) Define the
objective function
-
iii) State the constraints to which the objective function
should be optimized (i.e. either Maximization or Minimization) iv)
Add the non-negative constraints from the consideration that the
negative values of the decision variables do not have any valid
physical interpretation
-
Assumptions of Linear ProgrammingCertaintyIn all LP models it is
assumed that, all the model parameters such as availability of
resources, profit (or cost) contribution of a unit of decision
variable and consumption of resources by a unit of decision
variable must be known with certainty and constant.Divisibility
(Continuity)The solution values of decision variables and resources
are assumed to have either whole numbers (integers) or mixed
numbers (integer or fractional). However, if only integer variables
are desired, then Integer programming method may be employed.
-
AdditivityThe value of the objective function for the given
value of decision variables and the total sum of resources used,
must be equal to the sum of the contributions (Profit or Cost)
earned from each decision variable and sum of the resources used by
each decision variable respectively. /The objective function is the
direct sum of the individual contributions of the different
variables LinearityAll relationships in the LP model (i.e. in both
objective function and constraints) must be linear.
-
Example 1Vitamins A and B are found in foods F1 and F2 . 1 unit
of F1 contains 3 units of vitamin A and 4 units of vitamin B. 1
unit of F2 contains 6 units of vitamin A and 3 units of vitamin B.
1 unit of F1 and F2 cost Rs. 4 and Rs. 5 respectively. The minimum
daily requirement (for a person) of vitamin A and vitamin B is 80
and 100 units. Assuming that anything in excess of the daily
minimum requirement of vitamin A and vitamin B is harmful, find out
the optimum mixture of foods F1 and F2 at minimum cost which meets
the daily requirement of vitamin A and B. Formulate this as L. P.
P.
-
Solution: The given data is expressed in the form of table
as:Food Vitamins (No. unit of food)cost of food ABF1344F2635Min.
requirement80 100The problem is to find the quantities of foods F1
and F2 so as to meet the minimum requirement of vitamins A and B.
Decision variables: These are unknowns to be found out. Hence, Let
X1 the quantity (number of units) of F1 required X2 the quantity
(number of units) of F2 requiredObjective function: The objective
is to minimize the cost of the food. Now, the cost of x1 unit of
food F1 is ( x1* per unit cost = 4x1). Similarly cost of x2 unit of
food F2 is ( x2* per unit cost = 5x2). Hence the total cost is (
4x1 + 5x2 =C (say)). Hence the objective function is to minimize, C
= 4x1 + 5x2 .
-
Constraints: a) First constraint: The minimum requirement of
vitamin A is 80 units. The number of units of vitamin A contained
in X1 unit of food F1 are ( X1 * number of units of A contained in
I unit of F1) i. e. (3X1). Similarly number of units of vitamin A
contained in X2 unit of food F2 are ( X2 * number of units of A
contained in I unit of F2) i. e. (6X2). Thus the total intake of
vitamin A will be ( 3X1 + 6X2) which must be at least 80 units. 3X1
+ 6X2 80 ..(1)(vitamin A constraint) b) Similarly, minimum
requirement of vitamin B is 100 units. 4X1 + 3X2 100 ..(1)(vitamin
B constraint) c) Since number of units of X1 and X2cannot be
negative, hence X1, X2 0 (non negativity condition)
-
5. Thus the LPP format will be Minimize C = 4X1 + 5X2 Subject to
3X1 + 6X2 80 (vitamin A constraint) 4X1 + 3X2 100 (vitamin B
constraintX1, X2 0 (non negativity condition)
-
Example 1A city hospital has the following minimal daily
requirements for nurses.
PeriodClock time (24 hours day)Minimum number of nurses
required16 a.m. 10 a.m.2210 a.m. 2 p.m.732 p.m. 6 p.m.1546 p.m. 10
p.m.8510 p.m. 2 a.m.2062 a.m. 6 a.m.6
-
Nurses report at the hospital at the beginning of each period
and work for 8 consecutive hours. The hospital wants to determine
the minimal number of nurses to be employed so that there will be a
sufficient number of nurses available for each period. Formulate
this as a linear programming problem by setting up appropriate
constraints and objective function.
-
i) Identify and define the decision variable of the problemLet
X1, X2, X3, X4, X5 and X6 be the number of nurses joining duty at
the beginning of periods 1, 2, 3, 4, 5 and 6 respectively. ii)
Define the objective functionMinimize Z = X1 + X2 + X3 + X4 + X5 +
X6
-
iii) State the constraints to which the objective function
should be optimized. The above objective function is subjected to
following constraints.X1 + X2 7X2 + X3 15X3 + X4 8X4 + X5 20X5 + X6
6X6 + X1 2X1, X2, X3, X4, X5, X6 0
-
Linear Programming: Graphical SolutionExample 2.Solve the
following LPP by graphical methodMaximize Z = 5X1 + 3X2 Subject to
constraints2X1 + X2 1000 X1 400 X1 700 X1, X2 0
-
Solution:The first constraint 2X1 + X2 1000 can be represented
as follows.We set 2X1 + X2 = 1000When X1 = 0 in the above
constraint, we get,2 x 0 + X2 = 1000 X2 = 1000Similarly when X2 = 0
in the above constraint, we get,2X1 + 0 = 1000X1 = 1000/2 = 500
-
The second constraint X1 400 can be represented as follows,We
set X1 = 400 The third constraint X2 700 can be represented as
follows,We set X2 = 700
-
The constraints are shown plotted in the above figure.
PointX1X2Z = 5X1 +3X20000A0700Z = 5 x 0 + 3 x 700 =
2,100B150700Z = 5 x 150 + 3 x 700 = 2,850* MaximumC400200Z = 5 x
400 + 3 x 200 = 2,600D4000Z = 5 x 400 + 3 x 0 = 2,000
-
The Maximum profit is at point B When X1 = 150 and X2 = 700 Z =
2850
-
Example 3Solve the following LPP by graphical methodMaximize Z =
400X1 + 200X2 Subject to constraints18X1 + 3X2 800 9X1 + 4X2 600 X2
150 X1, X2 0
-
Solution:The first constraint 18X1 + 3X2 800 can be represented
as follows.We set 18X1 + 3X2 = 800When X1 = 0 in the above
constraint, we get,18 x 0 + 3X2 = 800X2 = 800/3 = 266.67
-
Similarly when X2 = 0 in the above constraint, we get,18X1 + 3 x
0 = 800X1 = 800/18 = 44.44
-
The second constraint 9X1 + 4X2 600 can be represented as
follows,We set 9X1 + 4X2 = 600When X1 = 0 in the above constraint,
we get,9 x 0 + 4X2 = 600 X2 = 600/4 = 150X1 = 600/9 = 66.67
-
Similarly when X2 = 0 in the above constraint, we get,9X1 + 4 x
0 = 600 The third constraint X2 150 can be represented as
follows,We set X2 = 150
-
PointX1X2Z = 400X1 + 200X20000A0150Z = 400 x 0+ 200 x 150 =
30,000* MaximumB31.1180Z = 400 x 31.1 + 200 x 80 = 28,444.4C44.440Z
= 400 x 44.44 + 200 x 0 = 17,777.8
-
The Maximum profit is at point AWhen X1 = 150 and X2 = 0 Z =
30,000
-
Example 4Solve the following LPP by graphical methodMinimize Z =
20X1 + 40X2 Subject to constraints36X1 + 6X2 108 3X1 + 12X2 36 20X1
+ 10X2 100X1 , X2 0
-
Solution:The first constraint 36X1 + 6X2 108 can be represented
as follows.We set 36X1 + 6X2 = 108When X1 = 0 in the above
constraint, we get,36 x 0 + 6X2 = 108X2 = 108/6 = 18
-
Similarly when X2 = 0 in the above constraint, we get,36X1 + 6 x
0 = 108X1 = 108/36 = 3The second constraint 3X1 + 12X2 36 can be
represented as follows,We set 3X1 + 12X2 = 36
-
When X1 = 0 in the above constraint, we get,3 x 0 + 12X2 = 36 X2
= 36/12 = 3Similarly when X2 = 0 in the above constraint, we
get,3X1 + 12 x 0 = 36X1 = 36/3 = 12
-
The third constraint 20X1 + 10X2 100 can be represented as
follows, We set 20X1 + 10X2 = 100When X1 = 0 in the above
constraint, we get,20 x 0 + 10X2 = 100X2 = 100/10 = 10
-
Similarly when X2 = 0 in the above constraint, we get,20X1 + 10
x 0 = 100X1 = 100/20 = 5
-
The Minimum cost is at point CWhen X1 = 4 and X2 = 2 Z = 160
PointX1X2Z = 20X1 + 40X20000A018Z = 20 x 0 + 40 x 18 = 720B26Z =
20 x2 + 40 x 6 = 280C42Z = 20 x 4 + 40 x 2 = 160* MinimumD120Z = 20
x 12 + 40 x 0 = 240
-
Example 5Solve the following LPP by graphical methodMaximize Z =
2.80X1 + 2.20X2 Subject to constraintsX1 20,000 X2 40,000 0.003X1 +
0.001X2 66X1 + X2 45,000X1, X2 0
-
Solution:The first constraint X1 20,000 can be represented as
follows.We set X1 = 20,000The second constraint X2 40,000 can be
represented as follows,We set X2 = 40,000
-
The third constraint 0.003X1 + 0.001X2 66 can be represented as
follows, We set 0.003X1 + 0.001X2 = 66When X1 = 0 in the above
constraint, we get,0.003 x 0 + 0.001X2 = 66X2 = 66/0.001 =
66,000Similarly when X2 = 0 in the above constraint, we get,0.003X1
+ 0.001 x 0 = 66X1 = 66/0.003 = 22,000
-
The fourth constraint X1 + X2 45,000 can be represented as
follows, We set X1 + X2 = 45,000When X1 = 0 in the above
constraint, we get,0 + X2 = 45,000X2 = 45,000Similarly when X2 = 0
in the above constraint, we get,X1 + 0 = 45,000X1 =45,000
-
PointX1X2Z = 2.80X1 + 2.20X20000A040,000Z = 2.80 x 0 + 2.20 x
40,000 = 88,000B5,00040,000Z = 2.80 x 5,000 + 2.20 x 40,000 =
1,02,000C10,50034,500Z = 2.80 x 10,500 + 2.20 x 34,500 = 1,05,300*
MaximumD20,0006,000Z = 2.80 x 20,000 + 2.20 x 6,000 =
69,200E20,0000Z = 2.80 x 20,000 + 2.20 x 0 = 56,000
-
The Maximum profit is at point CWhen X1 = 10,500 and X2 = 34,500
Z = 1,05,300
-
Example 6Solve the following LPP by graphical methodMaximize Z =
10X1 + 8X2 Subject to constraints2X1 + X2 20 X1 + 3X2 30 X1 - 2X2
-15X1 X2 0
-
Solution:The first constraint 2X1 + X2 20 can be represented as
follows.We set 2X1 + X2 = 20 When X1 = 0 in the above constraint,
we get,2 x 0 + X2 = 20X2 = 20
-
Similarly when X2 = 0 in the above constraint, we get,2X1 + 0 =
20X1 = 20/2 = 10The second constraint X1 + 3X2 30 can berepresented
as follows,We set X1 + 3X2 = 30
-
When X1 = 0 in the above constraint, we get,0 + 3X2 = 30X2 =
30/3 = 10Similarly when X2 = 0 in the above constraint, we get,X1 +
3 x 0 = 30X1 = 30
-
The third constraint X1 - 2X2 -15 can be represented as follows,
We set X1 - 2X2 = -15When X1 = 0 in the above constraint, we get,0
- 2X2 = -15X2 = -15/2 = 7.5Similarly when X2 = 0 in the above
constraint, we get,X1 2 x 0 = -15X1 = -15
-
The Maximum profit is at point CWhen X1 = 6 and X2 = 8 Z =
124
PointX1X2Z = 10X1 + 8X20000A07.5Z = 10 x 0 + 8 x 7.5 = 60B39Z =
10 x 3 + 8 x 9 = 102C68Z = 10 x 6 + 8 x 8 = 124* MinimumD100Z = 10
x 10 + 8 x 0 = 100