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