LINEAR PROGRAMMING

FORMULATION AND GRAPHIC METHOD

DEFINITIONMathematical technique for

optimizing the use of constrained resources, where the relationship between variables is linear.

Historical Perspective1928 – John von Neumann published related central theorem

of game theory.1944 – Von Neumann and Morgenstern published Theory of

Games and Economic Behavior.1936 – W.W. Leontief published "Quantitative Input and

Output Relations in the Economic Systems of the US”

which was a linear model without objective function.1939 – Kantoravich (Russia) actually formulated and solved a

LP problem.1941 – Hitchcock poses transportation problem (special LP)WWII – Allied forces formulate and solve several LP problems

related to military.A breakthrough occurred in 1947... 3

SCOOP US Air Force wanted to investigate the feasibility of applying

mathematical techniques to military budgeting and planning. George Dantzig had proposed that interrelations between

activities of a large organization can be viewed as a LP model and that the optimal program (solution) can be obtained by minimizing a (single) linear objective function.

Air Force initiated project SCOOP (Scientific Computing of Optimum Programs)

NOTE:SCOOP began in June 1947 and at the end of the same

summer, Dantzig and associates had developed:1) An initial mathematical model of the general linear

programming problem.2) A general method of solution called the simplex method.

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Simplex Today

A large variety of Simplex-based algorithms exist to solve LP problems.

Other (polynomial time) algorithms have been developed for solving LP problems:

Khachian algorithm (1979) Kamarkar algorithm (AT&T Bell Labs, mid 80s)

BUT none of these algorithms have been able to beat Simplex in actual practical applications.

HENCE Simplex (in its various forms) is and will most likely remain the most dominant LP algorithm for at least the near future

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Linear – relationship between two or more variables which is directly and precisely proportional.

Programming – refers to use of certain mathematical techniques to get the best possible solution to a given problem involving limited resources.

LP

Linear Programming

Basic Requirements Decision variables and their relationships should

be linear. Objective function should be well defined Presence of Constraints Alternative course of action Non negative restrictions Linearity: Both Objective and Constraints must

be expressed in Linear equations/inequalities.

Terminology Decision variables – variables directly under the

control of decision maker. Objective function – An expression showing

relationship between decision variables and organization's objectives.

Constraints – Limit or restriction on the availability of resources.

Non – negativity constraint – The decision variables must not assume negative values which is impossible situation.

Feasible region – The region which is common to all the constraints of given LPP.

Basic Solution – A solution of basic

variables, where the values of non-basic variables are zero.

Feasible Solution – A solution satisfying all constraints and non-negativity restriction of a LPP.

Basic Feasible Solution – A basic solution, where the values of basic variables are non-negative.

Optimal Solution – A basic feasible solution which also optimizes the objective function.

Assumptions Proportionality:

Proportionality/Linearity exists in Objective and Constraints

Additivity:         Total resources used must equal sum of resources

used in individual activity Divisibility:       

Solution need not be whole numbers and may take up fractional value.

Certainty:        Co-efficients are known (Deterministic)

Finiteness:      Finite number of alternative solutions

Applications of LPP Industry

Product Mix Production

Smoothing Blending Production

Scheduling Trim Loss

Miscellaneous Diet Problems Farm Planning Airline Routing Facility Location

Management Portfolio Selection Financial Mix Media Selection Manpower Planning Transportation Job Allocation Travelling Salesman

Problem Formulation

Problem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement.

LP Model FormulationMax/min z = c1x1 + c2x2 + ... + cnxn

subject to:a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1

a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2

: am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm

xj = decision variablesbi = constraint levelscj = objective function coefficientsaij = constraint coefficients

The Linear Programming Model (1) Let: X1, X2, X3, ………, Xn = decision variables

Z = Objective function or linear function

Requirement: Maximization of the linear function Z.

Z = c1X1 + c2X2 + c3X3 + ………+ cnXn

subject to the following constraints: …..Eq (2)

where aij, bi, and cj are given constants.

The Linear Programming Model (2)

The linear programming model can be written in more efficient notation as:

…..Eq (3)

The decision variables, xI, x2, ..., xn, represent levels of n competing activities.

Examples of LP Problems (1) 1. A Product Mix Problem

A manufacturer has fixed amounts of different resources such as raw material, labor, and equipment.

These resources can be combined to produce any one of several different products.

The quantity of the ith resource required to produce one unit of the jth product is known.

The decision maker wishes to produce the combination of products that will maximize total income.

Examples of LP Problems (2)2. A Blending Problem

Blending problems refer to situations in which a number of components (or commodities) are mixed together to yield one or more products.

Typically, different commodities are to be purchased. Each commodity has known characteristics and costs.

The problem is to determine how much of each commodity should be purchased and blended with the rest so that the characteristics of the mixture lie within specified bounds and the total cost is minimized.

Examples of LP Problems (3)3. A Production Scheduling Problem

A manufacturer knows that he must supply a given number of items of a certain product each month for the next n months.

They can be produced either in regular time, subject to a maximum each month, or in overtime. The cost of producing an item during overtime is greater than during regular time. A storage cost is associated with each item not sold at the end of the month.

The problem is to determine the production schedule that minimizes the sum of production and storage costs.

Examples of LP Problems (4)4. A Transportation Problem

A product is to be shipped in the amounts al, a2, ..., am from m shipping origins and received in amounts bl, b2, ..., bn at each of n shipping destinations.

The cost of shipping a unit from the ith origin to the jth destination is known for all combinations of origins and destinations.

The problem is to determine the amount to be shipped from each origin to each destination such that the total cost of transportation is a minimum.

Examples of LP Problems (5)5. A Flow Capacity Problem

One or more commodities (e.g., traffic, water, information, cash, etc.) are flowing from one point to another through a network whose branches have various constraints and flow capacities.

The direction of flow in each branch and the capacity of each branch are known.

The problem is to determine the maximum flow, or capacity of the network.

Developing LP Model (1)

The variety of situations to which linear programming has been applied ranges from agriculture to zinc smelting.

Steps Involved: Determine the objective of the problem and describe it by

a criterion function in terms of the decision variables. Find out the constraints. Do the analysis which should lead to the selection of

values for the decision variables that optimize the criterion function while satisfying all the constraints imposed on the problem.

Guidelines for Model Formulation

Understand the problem thoroughly. Describe the objective. Describe each constraint. Define the decision variables. Write the objective in terms of the decision

variables. Write the constraints in terms of the

decision variables.

Formulating LP Problems

One of the most common LP applications is the product mix problem

Two or more products are produced using limited resources such as personnel, machines, and raw materials

The profit that the firm seeks to maximize is based on the profit contribution per unit of each product

The company would like to determine how many units of each product it should produce so as to maximize overall profit given its limited resources

Flair Furniture Company The Flair Furniture Company produces inexpensive

tables and chairs Processes are similar in that both require a certain

amount of hours of carpentry work and in the painting and varnishing department

Each table takes 4 hours of carpentry and 2 hours of painting and varnishing

Each chair requires 3 of carpentry and 1 hour of painting and varnishing

There are 240 hours of carpentry time available and 100 hours of painting and varnishing

Each table yields a profit of $70 and each chair a profit of $50

Flair Furniture Company The company wants to determine the best

combination of tables and chairs to produce to reach the maximum profit

HOURS REQUIRED TO PRODUCE 1 UNIT

DEPARTMENT(T)

TABLES(C)

CHAIRSAVAILABLE HOURS THIS WEEK

Carpentry 4 3 240

Painting and varnishing 2 1 100

Profit per unit $70 $50

Flair Furniture Company The objective is toMaximize profit The constraints are

1. The hours of carpentry time used cannot exceed 240 hours per week

2. The hours of painting and varnishing time used cannot exceed 100 hours per week

The decision variables representing the actual decisions we will make are

T = number of tables to be produced per weekC = number of chairs to be produced per week

Flair Furniture Company

We create the LP objective function in terms of T and C

Maximize profit = $70T + $50C Develop mathematical relationships for the two

constraints For carpentry, total time used is

(4 hours per table)(Number of tables produced)+ (3 hours per chair)(Number of chairs produced)

We know thatCarpentry time used ≤ Carpentry time available4T + 3C ≤ 240 (hours of carpentry time)

Flair Furniture Company Similarly

Painting and varnishing time used ≤ Painting and varnishing time available

2 T + 1C ≤ 100 (hours of painting and varnishing time)

This means that each table produced requires two hours of painting and varnishing time

Both of these constraints restrict production capacity and affect total profit

Flair Furniture Company The values for T and C must be nonnegative

T ≥ 0 (number of tables produced is greater than or equal to 0)C ≥ 0 (number of chairs produced is greater than or equal to 0)

The complete problem stated mathematically

Maximize profit = $70T + $50Csubject to

4T + 3C ≤ 240 (carpentry constraint)2T + 1C ≤ 100 (painting and varnishing constraint)T, C ≥ 0 (nonnegativity constraint)

Linear Programming Problem A firm producing 2 products P1 and P2.

Each unit of P1 requires 1 kg material A, 2 kg of material B and 1 unit of labor. Each unit of P2 requires 3 kg of material B and 1 unit of labor. Each week firm has availability of 6 kg of material A, 19 kg of material B and 8 units of labor. One unit of P1 sold earn profit of Rs. 5 and P2 earns profit of Rs 7. How many units of each product is to be produced so as to give the maximum profit.

A Maximization Problem LP Formulation Max Z = 5x1 + 7x2 (Profit function)

s.t. x1 < 6 (Material A Constraint)

2x1 + 3x2 < 19 (Material B Constraint)

x1 + x2 < 8 (Labor Constraint)

x1, x2 > 0 (Non-negativity constraint)

Graphic Method of solving LPP Formulate linear programming problem. Each inequality in the constraints may be

written as equality. Plot the constraint lines considering them as

equations. Identify the feasible solution region. Locate the corner points of the feasible region. Calculate the value of objective function on the

corner points. Choose the point where the objective function

has optimal value.

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Graphical Solution Constraint #1 Graphed

x2

x1

x1 < 6

(6, 0)

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Graphical Solution Constraint #2 Graphed

2x1 + 3x2 < 19

x2

x1

(0, 6 1/3)

(9 1/2, 0)

Graphical Solution Constraint #3 Graphed

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x2

x1

x1 + x2 < 8

(0, 8)

(8, 0)

Graphical Solution Combined-Constraint Graph

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2x1 + 3x2 < 19

x2

x1

x1 + x2 < 8

x1 < 6

Graphical Solution Feasible Solution Region

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FeasibleRegion

x2

Example 1: Graphical Solution

The Five Extreme Points8

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FeasibleRegion

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C

B

A

x2

Evaluation of objective function at feasible region corner points

Corner points

Coordinates Objective function Z = 5x1 + 7x2

Value

A (0,6.33) 5(0)+7(6.33) 44.31

B (5,3) 5(5)+7(3) 46

C (6,2) 5(6)+7(2) 44

D (6,0) 5(6)+7(0) 30

E (0,0) 5(0)+7(0) 0

Since Z has maximum value at point B (5,3), so

x1 = 5, x2 = 3Where profit is Rs 46.

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Alternate Solution by extreme points

Objective Function Line

x1

x2

(7, 0)

(0, 5)Objective Function5x1 + 7x2 = 35

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Alternate Graphical Solution Optimal Solution

x1

x2

Objective Function5x1 + 7x2 = 46

Optimal Solution(x1 = 5, x2 = 3)

Extreme Points and the Optimal Solution

The corners or vertices of the feasible region are referred to as the extreme points.

An optimal solution to an LP problem can be found at an extreme point of the feasible region.

When looking for the optimal solution, you do not have to evaluate all feasible solution points.

You have to consider only the extreme points of the feasible region.

Summary of the Graphical Solution Procedure for Maximization Problems

Prepare a graph of the feasible solutions for each of the constraints.

Determine the feasible region that satisfies all the constraints simultaneously.

Draw an objective function line. Move parallel objective function lines toward

larger objective function values without entirely leaving the feasible region.

Any feasible solution on the objective function line with the largest value is an optimal solution.

Minimization Problem

Min Z = 5x1 + 2x2

Subject to constraints 2x1 + 5x2 > 10

4x1 - x2 > 12 x1 + x2 > 4

x1, x2 > 0

Graphical Solution Graph the Constraints Constraint 1: When x1 = 0, then x2 = 2; when x2

= 0, then x1 = 5. Connect (5,0) and (0,2). The ">" side is above this line.

Constraint 2: When x2 = 0, then x1 = 3. But setting x1 to 0 will yield x2 = -12, which is not on the graph. Thus, to get a second point on this line, set x1 to any number larger than 3 and solve for x2: when x1 = 5, then x2 = 8. Connect (3,0) and (5,8). The ">" side is to the right.

Constraint 3: When x1 = 0, then x2 = 4; when x2 = 0, then x1 = 4. Connect (4,0) and (0,4). The ">" side is above this line.

Graphical Solution Constraints Graphed

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x2

4x1 - x2 > 12

x1 + x2 > 4

2x1 + 5x2 > 10

x1

Feasible Region

Graphical Solution Graph the Objective Function

Set the objective function equal to an arbitrary constant (say 20) and graph it. For 5x1 + 2x2 = 20, when x1 = 0, then x2 = 10; when x2= 0, then x1 = 4. Connect (4,0) and (0,10).

Move the Objective Function Line Toward Optimality

Move it in the direction which lowers its value (down), since we are minimizing, until it touches the last point of the feasible region, determined by the last two constraints.

Graphical Solution Objective Function Graphed

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x2 Min z = 5x1 + 2x2

4x1 - x2 > 12

x1 + x2 > 4

2x1 + 5x2 > 10

x1

Solve for the Extreme Point at the Intersection of the Two Binding Constraints

4x1 - x2 = 12 x1+ x2 = 4 Adding these two equations gives:

5x1 = 16 or x1 = 16/5. Substituting this into x1 + x2 = 4 gives: x2

= 4/5

Graphical Solution

Solve for the Optimal Value of the Objective FunctionSolve for z = 5x1 + 2x2 = 5(16/5) + 2(4/5) = 88/5.

Thus the optimal solution is x1 = 16/5; x2 = 4/5; z = 88/5

Graphical Solution Optimal Solution

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x2 Min z = 5x1 + 2x2

4x1 - x2 > 12

x1 + x2 > 4

2x1 + 5x2 > 10

Optimal: x1 = 16/5 x2 = 4/5x1

Special Cases Multiple Optimal Solutions Infeasible problem Unbounded problem Redundant Constraint

Multiple optimal solutions

If objective function falls on more than one optimal point, the solution is said to be multiple optimal.

It exists when slope of the objective function and one of the constraint equation are same.

Infeasible problem

No feasible solution. It means there are no points that

simultaneously satisfy all constraints in the problem.

Common region do not develop in the first quadrant.

Objective function do not pass through any point of all constraints.

Example of Infeasible Problem

Max Z = 2x1 + 6x2

subject to constraints 4x1 + 3x2 < 12 2x1 + x2 > 8

x1, x2 > 0

Infeasible Problem There are no points that satisfy both

constraints, hence this problem has no feasible region, and no optimal solution.

x2

x1

4x1 + 3x2 < 12

2x1 + x2 > 8

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Unbounded Problem

When feasible solution space formed by the constraints is not confined to the first quadrants only.

If constraints passes through 2nd, 3rd or 4th quadrant.

No limits on the constraints. Common feasible region is not bounded in

any respect.

Example of Unbounded Problem

Max Z = 3x1 + 4x2

subject to x1 + x2 > 5 3x1 + x2 > 8

x1, x2 > 0

Unbounded Problem The feasible region is unbounded and the

objective function line can be moved parallel to itself without bound so that z can be increased infinitely. x2

x1

3x1 + x2 > 8

x1 + x2 > 5

Max 3x1 + 4x2

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Redundant Constraint

When one constraint is less restrictive than other constraint, it is said to be redundant constraint.

Advantages & DrawbacksS. No. Advantages  Drawbacks 1 Problem Insight Fraction Values2 Considers all possible Solution Co-efficients must be

Deterministic3 Better Decisions Function must be Linear

4 Better Tools Time & Un-certainty effect not considered

5 Highlights Bottlenecks Approximation of Complex problems

6 Clear specification of Management problems