MIDTERM LECTURE NOTE#1
Linear Programming
Linear Programming (LP) refers to family of mathematical techniques that can be
used to solutions to constrained optimization problems. The programming aspect of
linear programming refers to the use of algorithms. Algorithm is a well defined
sequence of steps that will lead to a solution.
LP models are based on linear relationships. LP involves the use of mathematical
models to provide optimum solution to certain types of problems with the following
characteristics:
LINEAR PROGRAMMING PARAMETERS
A) The objective function:
that expresses the objective of the problem. It is a set of mathematical
statements of the restrictions. Every firm’s major objective is to maximise
profit and minimise cost.
B) The constraints :
Are the limitations or restrictions of resources such s the availability of raw
materials, processing time, market limitations and other resource restrictions.
The elements of constraints are;
 Right hand side (RHS) quantity that specifies the limit of for that
constraint. It must be a constant, not a variable.
 An algebraic sign that indicates whether the limit is an upper bound (≤),
and a lower bound (≥), or (=) that must be met exactly.
 He decision variables to which the constraints apply.
 The impact that one unit of each decision variable will have on the right
hand side quantity of the constraint.
C) Optimal Solution :
is the feasible solution that yields the best value in terms of the objective.
D) Parameters:
Are fixed values that specify the impact that one unit of each decision variable
will have on the objective and on any constraints it pertains to and to the
numerical value of each constraint.
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ASSUMPTIONS OF LP MODELS
A) Proportionality :
is that each decision variable has a linear impact in the objective function and
in each constraint in which it appears. This means that if, the profit of X1 is $4
per unit, and it must be the same, regardless of the quantity of X1.
B) Additivity :
pertains to potential values of decision variables. It is assumed that non
integer values are acceptable.
C) Certainty Relates to the model parameter (numerical values). It is assumed that these
values are known and constant. All relevant constraints have been identified
and represented in the model. Eg: there is a limit on the demand for some
item, that limit should be recorded as a constraint (eg: X≤100)
D) Non-negativity:
negative values of variables are unrealistic and will not be considered in any
potential solutions; only positive values and zero will be allowed. X1,X2 ≥ 0
OTHER LP PROPERTIES

Binding Constraints – whose intersection determines the optimal
solution to a problem.

Redundant Constraint – a constraint that does not form a unique part
of the boundary of the feasible solution space.

The Extreme point Approach – involves finding the coordinates of
each corner point that borders the feasible solution space and then
determining which corner point provides the best value of the objective
function.

The highest value for the maximisation problem or the lowest value for
the minimization problem.
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
The extreme point theorem states that if a problem has an optional
solution, at least one optional solution will occur at a corner point of the
feasible solution space.

Corner points represent intersections of constraints. Finding the
corner points involves finding the coordinates of the constraints.
Steps in Formulating LP Models:

Define decision variables

Determine the objective function

Determine the appropriate values for parameters and determine
whether an upper limit or lower limit or equality is called for

Use this information to build a model

Validate the model
Eg: Ramee Furniture Mart makes two kinds of products, cabinets and dressers that
pass through Assembly and Finishing and inspection departments. The Assembly
department has 100 hours of work available in each week, while the Finishing
department has 22 hours and Inspection department has 39 hours available in each
week. Making one cabinet needs 4hours in assembly and 2 hours in Finishing and
department, 3 hours in Inspection department while it takes 10 hours to assemble a
dresser and 1 hour to finish and 3 hours to inspect it. If the profit generated per
cabinet is $60 and a dresser is $50. How many units of cabinets and dressers should
be produced in order to obtain the maximum profit?
Solution:
Cabinets (X1)
Dressers(X2)
Time available
Time in Assembly
4hrs
10hrs
100hrs
Time in Finishing
2hrs
1hr
22hrs
Time in Inspection
3hrs
3hrs
39hrs
Profit per unit
$60
$50
Step1: Decision variables:
Let,
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X1 = Number of units of cabinets to be produced and sold to maximise profit.
X2 = Number of units of Dressers to be produced and sold to maximise profit
Step: 2 Objective function:
Maximise Profit, P= 60X1 + 50X2
Subject to:
Constraints: 4X1 + 10X2 ≤ 100;
2X1 + 1X2 ≤ 22;
3X1 + 3X2 ≤ 39
Further using graph or analytical (simultaneous equations) method, the equation
should be solved to find feasible solution.
Slack

Is the amount of scarce resources that is unused by a given solution

This amount can range from zero, for a case in which all of a particular
resource is used to the original amount of the resource that was available.

can exist in a ≤ constraint

it can be used to represent slack in a constraint
Computing the slack for Ramee Furniture Mart Problem:
Amount used with
Originally
Amount of Slack
X1and X2 =4
Available
= (Available- used)
Time in Assembly
4(9)+ 10(4) =76
100
100 - 76 =24hrs
Time in Finishing
2(9) + 1(4) =22
22
22-22 =0 hrs
Time in Inspection
3(9) + 3(4) = 39
39
39 - 39 = 0hrs
Surplus –is the amount by which optional solution causes a ≥ constraint to exceed
the required minimum amount.
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To determine the surplus:

Substitute the optimal values of the decision variables into the left side of the
constraint and solve.

The difference between the resulting value and the original right-hand side
amount of surplus.
Minimization Example:

The constraints are the greater than or equal to variety which causes the
feasible solution.

Space to be restricted to an area away from the origin instead of close to the
origin.

The optimum is the point which the smallest possible value of the objective
function, instead of the largest.
Eg: Determine the values of decision variables, X1, and X2 that will yield the
minimum cost using the objective function approach.
Minimize Cost, C= 10X1+2X2 ≥ 18
s.t
6X1+2X2 ≥ 18
8X1+10X2 ≥ 40
X2 ≥ 1
X1 and X2 ≥ 0
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EXAMPLES
Al Jazeera Industries produces two types of ice creams. Its popular chocolate ice
creams that costs $ 2.2 to make a five hundred grams box
and sells at $3.5
whereas fruit salad ice creams that costs $4 for a five hundred grams box and sells
at $5.5. The company has 15000 boxes of raw materials in stock, and the manager
has specified that a minimum of 1000 boxes of the chocolate ice creams and 1200
boxes of fruit salad ice cream boxes should be made.
(a) What are the decision variables in this problem?
(b) What are the constraints?
(c) Formulate a linear programming that will enable the manager to determine
how many units of each product to produce i order to satisfy the conditions
specified with the maximum profit.
Exercises:
1. Consider the following LP problem and find out the optimum solution. What is
the value of objective function? Compute the slack.
(a) Max 3X1 + 2X2
s.t
2X1 + 2X2 ≤ 8
3X1 + 2X2 ≤ 12
1X1 + 0.5X2 ≤ 3; X1 and X2 ≥ 0
(b) Minimize 6X1 + 24X2
s.t
10X1 + 2X2 ≥32
2X1 + 6X2 ≥ 12
X1 and X2 ≥ 0
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