2. › Linear programming (LP
, also called linear optimization) is a
method to achieve the best outcome (such as maximum
profit or lowest cost) in a mathematical model whose
requirements are represented by linear relationships. Linear
programming is a special case of mathematical
programming (mathematical optimisation).
› Many practical problems in operations research can be
expressed as linear programming problems. Certain special
cases of linear programming, such as network flow problems
and multicommodity flow problems are considered important
enough to have generated much research on specialized
algorithms for their solution. A number of algorithms for
other types of optimization problems work by solving LP
problems as sub-problems.
3. 1.Limited resources :limited number of
labour, material equipment and finance
2. Objective :refers to the aim to optimize
(maximize the profits or
minimize the costs).
3. Linearity :increase in labour input will have a proportionate
in
output.
:the products, workers' efficiency, and machines are
assumed to be identical.
increase
4.
Homogeneity
5.
Divisibility
:it is assumed that resources and products can be
divided into fractions. (in case the fractions are not
possible, like production of one-third of a
computer, a modification of linear programming
called integer programming can be used).
4. • There must be well defined objective function.
• There must be a constraint on the amount.
• There must be alternative course of action.
• The decision variables should be interrelated
and non-negative.
• The resource must be limited in supply.
8. ADVANTAGE
S:
1. It helps in attaining optimum use of productive
factors.
2. Itimproves the quality of the decisions.
3. It provides better tools for meeting the changing
conditions.
4. Ithighlights the bottleneck in the production process.
• problems the computational difficulties
are enormous.
2. It may yield fractional value answers to decision
variables.
3. Itis applicable to only static situation.
4. LP deals with the problems with single objective.
Limitations:
1. For large
9. Decision variables and parameters: Decision variables describe
the quantities that the decision makers would like to determine. They
are the unknowns of a mathematical programming model. Typically
we will determine their optimum values with an optimization method.
Ina general model, decision variables are given algebraic
designations such as x1,x2,x3,…..xn.
The collection of coefficients for all values of the indices i and jare
called the parameters of the model. For the model to be completely
determined all parameter values must be known.
Objective function: The objective function evaluates some
quantitative criterion of immediate importance such as cost, profit,
utility,or yield.
10. CONSTRAINTS: A constraint is an inequality or equality defining limitations on decisions.
Constraints arise from a variety of sources such as limited resources, contractual
obligations, or physical laws. In general, an LP is said to have m linear constraints that can
be stated as:
One of the three relations shown in the large brackets must be chosen for each constraint.
The number is called a "technological coefficient," and the number is called the "right-
side" value of the i-th constraint. Strict inequalities (<, >, and ) are not permitted. When
formulating a model, it is good practice to give a name to each constraint that reflects its
purpose.
NONNEGATIVITY RESTRICTIONS:
Inmost practical problems the variables are required to be nonnegative.
This special kind of constraint is called a nonnegativity restriction. Sometimes variables
are
required to be nonpositive or, in fact, may be unrestricted (allowing any real value).
11. The solution of an LP problem, (also known as FEASIBLE
SOLUTION) is actually the optimum level of the decision
variables where all the constraints are satisfied.
There are two methods for reaching an optimal feasible
solution:
1. Graphical Method
2. Simplex Method
13. STEPSINVOLVED IN DEVELOPING A
LINEAR PROGRAMMINGPROBLEM:
• Formulation followed by the construction of the
mathematical model.
• Solution by graphical or simplex method.
• Interpretation and sensitivity analysis.
14. GRAPHICAL
METHOD:
The graphical method is applicable to solve the LPP involving two decision
variables x1, and x2, we usually take these decision variables as x, y instead
of x1, x2. To solve an LP, the graphical method includes two major steps.
a)The determination of the solution space that defines the feasible solution.
Note that the set of values of the variable x1, x2, x3,....xn which satisfy all the
constraints and also the non-negative conditions is called the feasible
solution of the LP.
b) The determination of the optimal solution from the feasible region.
15. THREE TERMS TO
BE KEPT IN MIND:
• Feasible Region: The set of points that satisfy all the
constraints.
• Corner Point Property: The optimum answer is always lying
at the corner points of the feasible region.
• Optimal solution: The corner point with the best objective
function value is optimal.
16. EXAMPLE:
A manufacturing plant makes two types of inflatable boats, a two-person boat
and a four-person boat. Each two-person boat requires 0.9 labor hours from
the cutting department and 0.8 labor-hours from the assembly department.
Each four person boat requires 1.8 labor-hours from the cutting department
and 1.2 labor-hours from the assembly department. The maximum labor-
hours available per month in the cutting department and the assembly
department are 864 and 672, respectively. The company makes a profit of 25
on each 2-person boat and 40 on each 4-person boat.
How many of each type should be manufactured each month
to maximize profit? What is the maximum profit?
17. SOLUTIO
N:
T
osolve such a problem involves:
1.Constructing a mathematical model of the problem, and
2.using the mathematical model to find the solution.
Step 1
:Identify the decision variables.
Inthis problem, we have
x=number of 2-person boats to manufacture
y=number of 4-person boats to manufacture
Step 2:Summarize relevant information in table form,
relating
the decision variables with the rows in the table, if possible.
18. RECALL:EACH TWO-PERSON BOAT REQUIRES 0.9LABOR HOURS FROM THE
CUTTING DEPARTMENT AND 0.8LABOR-HOURSFROM THEASSEMBLY
DEPARTMENT. EACH FOUR PERSON BOAT REQUIRES 1.8LABOR-HOURSFROM THE
CUTTING DEPARTMENT AND 1.2LABOR-HOURS FROM THEASSEMBLY
DEPARTMENT.
THE MAXIMUM LABOR-HOURS AVAILABLE PER MONTH IN THE CUTTING DEPARTMENT
AND THE ASSEMBLY DEPARTMENT ARE 864 AND 672, RESPECTIVELY. THE COMPANY
MAKES A PROFIT OF 25 ON EACH 2-PERSON BOAT AND 40 ON EACH FOUR-PERSON
BOAT.
19. Step 3:Determine the objective and write a linear objectivefunction
P =25x + 40y
Step 4:Write problem constraints using linear equations and/or inequalities.
0:9x + 1:8y ≤ 864
0:8x+1:2y≤672
Step 5:Write nonnegativeconstraints.
x>0
y> 0
We now have a mathematical model of the given problem. We
need to find the production schedule which results in maximum
profit for the company and to find that maximum profit.
Now we follow the procedure for geometrically solving a linear
programming problem with two decision variables.
20. APPLYING THE FUNDAMENTAL THEOREM OF
LINEAR PROGRAMMING
• Graph the feasible region. Be sure to find all corner points.
• Construct a corner point table listing the value of the
objective function at each corner point.
• Determine the optimal solution(s) from the table.
• For an applied problem, interpret the optimal solution(s) in
terms of the original problem.
21.
22.
23.
24.
25. NOW WE
SOLVE THE
PROBLEM:
Corner Point
(0; 0)
(0; 480)
(480; 240)
P =25x+40y
0
19,200
21,600
(860; 0) 21,500
We conclude that the company can make a
maximum profit of 21,600by producing 480
two-person boats and 240 four-person
boats.
26. SIMPLEX
METHOD:
• Applicable when there are more than two variables.
• The simplex method is not used to examine all the feasible
solutions.
• Itdeals 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 non-negative variable which is added to the left hand side of
the constraint to convert it into an equation is called slack variable.
• The non-negative variable which is subtracted from the left hand
side of a constraint to convert it into an equation is called surplus
variable.
27. SOMEIMPORTANT THEOREMS IN SIMPLEXMETHOD
• Theorem1:The set of feasible solutions to an LPP is a convex set.
• Theorem2: [Fundamental theorem of LP] Ifthe feasible region of an LPP is a convex polyhedron, then
there exists an optimal solution to the LPP and at least one basic feasible solution must be optimal.
• Theorem3: [Reduction of feasible solution to a basic feasible solution] Ifan LPP has a feasible solution,
then it also has a basic feasible solution.
• Theorem4: [Replacement of a Basic vector] Let an LPP have a basic feasible solution. Ifwe drop one of the
basic vectors and introduce a non-basic vector in the basic set, then the new solution obtained is also a
basic feasible solution, provided these vectors are suitably selected.
Theorem 5:[Conditions of optimality] A sufficient condition for a basic feasible solution to an LPP to be an
optimum (maximum) is that zj-cj ≥0for all jfor which the column vector ajЄA is not, in the basis B.
• Theorem 6:[Unbounded solution] Let there exist a basic feasible solution to a given LPP.Iffor at least one
j,for which yij≤0 (I=1,2,…,m),zj-cj is negative, then there does not existany optimum solution to this LPP.
28.
29. STANDARD FORM OF LPP: (IN SIMPLEX
METHOD)
• All the constraints should be converted to equations
except for the non-negativity restrictions which remain
inequalities.
• The right side element of each constraint should be
made non-negative (if not).
• All variables must be positive.
• The objective function should always be in
maximisation form. If it is given as minimisation form, it
should be converted into maximisation form, i.e. Min
f(x)= -Max[-f(x)]
• Coefficient of the slack or surplus variable in the
objective function to be considered as zero.
35. Cj 3 2 0 0 Mini
mum
Ratio
Basic
V
ari
abl
es
Cb Xj X Y S
1 S
2
S
1 0 4 1 1 1 0 4
S
2 0 2 (
1
) -
1 0 1 2
Zj
=Cb
Xj 0 0 0 0
j
=
Zj
-
Cj
-3 -2 0 0
TABLE1
36. Cj 3 2 0 0 Mini
mum
Ratio
Basic
V
ari
abl
es
Cb Xj X Y S
1 S
2
S
1 0 2 0 (
2
) 1 -
1 2
/
2
=
1
X 3 2 1 -
1 0 1
Zj
=Cb
Xj 3 -3 0 0
j
=
Zj
-
Cj
0 -5 0 0
TABLE2
37. Cj 3 2 0 0 Mini
mum
Ratio
Basic
V
ari
abl
es
Cb Xj X Y S
1 S
2
Y 2 1 1 0 1
/
2 -
1
/
2
X 3 6 2 0 1 1
Zj
=Cb
Xj 6 2 4 2 All
jis
+
v
e
.
j
=
Zj
-
Cj
3 0 4 2
TABLE3