Chapter 1_LPP_TPS

What is Operations Research?

Meaning Of OR:

According to T.L.Saaty “OR is the art of giving bad results to problems to wkich otherwise worse results are given”.

According to Churchman “OR is the application of scientific methods, techniques and tools to problems involving the operations of systems so as to provide these in control of the application with optimum solutions to the problem”.

According to H.M.Wagner “ OR is the scientific approach to problem solving for executive management”.

Scope of OR:

• In Agriculture• In Finance• In Industry• In Marketing• In Production Management• In Personnel Management• In Production Management• In L.I.C

Methodology Of OR:

Define the Problem

Develop the Model

Obtain Input Data

Solve the Model

Model Validation

Analyze the results

Implement the Solution

What is Linear Programming Problem?

(LPP)

Formulation of LPP:

Step 1:Determine the objective function as a linear function of the variables.

Step 2: Formulate the other conditions of the problem such as resources limitations as linear equations or inequations interms of the variables.

Step 3: Add the non-negativity constraints.

Production Allocation Problem:A manufacturer produces two types of models A &

B. Each model A requires 4 hours of grinding and 2 hours of polishing, where as model B requires 2 hours of grinding and 5 hours of polishing. The manufacturer has 2 grinders and 3 polishers. Each grinder works for 40 hours a week and each polisher works for 60 hours a week. Profit on model A is Rs.3 and on model B is Rs.4. Whatever is produced in a week is sold in the market. How should the manufacturer allocate his production capacity to the types of models so that he may make the maximum profit in week?

2. A manufacturing firm needs 5 component parts. Due to inadequate resources, the firm is unable to manufacture all its requirements. So the management is interested in determining as to how many , if any , units of each component should be purchased from the outside and how many should be produced internally. The relevant data are given below.

Component M A T TR PP PC

C1 4 1 1.5 20 48 30

C2 3 3 2 50 80 52

C3 1 1 0 45 24 18

C4 3 1 0.5 70 42 31

C5 2 0 0.5 40 28 16

WhereM: Per unit milling time in hoursA: Per unit assembly time in hoursT: Per unit testing time in hoursTR: Total requirements in unitsPP: Price per unit quoted in the marketPC: Per unit direct costsResources available are as follows:Milling hours: 300 Assembly hours: 160Testing hours: 150

Formulate this as an LPP, taking the objective function as maximization of saving by producing the components internally.

3. The marketing department of Everest company has collected information on the problem of advertising for its product. This relates to the advertising media available, the number of families expected to be reached with each alternative, cost per advertisement, the maximum availability of each medium and the expected exposure of each one. The information is as given below:

Advertising Media

No.of families expected to

cover Cost per Ad (Rs)

Maximum availability (No.of

times)

Expected exposure

(units)

TV(30 sec) 3000 8000 8 80

Radio(15 sec) 7000 3000 30 20

Sunday edition of a daily (1/4 page) 5000 4000 4 50

Magazine(1 page) 2000 3000 2 60

Other information and requirements:

a) The advertising budget is Rs 70,000

b) At least 40,000 families should be covered.

c) At least 2 insertions be given in Sunday edition of a daily.

Formulate this as LPP to maximize the expected exposure.

Graphical Method of solving a GLPP:

Step 1: Consider each inequality as equation.

Step 2: Plot each equation on the graph.Step 3: Identify the feasible solution or

common region satisfying all the constraints simultaneously.

Step 4: Determine the corner points of the feasible region.

Step 5: Find the optimal solution by substituting the corner points in the objective function.

Two products: Chairs and Tables

Decision: How many of each to make this month?

Objective: Maximize profit

Flair Furniture Co. DataTables

(per table)

Chairs(per chair)

Hours Available

Profit Contribution

7 5

Carpentry 3 hrs 4 hrs 2400

Painting 2 hrs 1 hr 1000

Other Limitations:• Make no more than 450 chairs• Make at least 100 tables

Decision Variables:

T = No. of tables to make

C = No. of chairs to make

Objective Function: Maximize Profit

Z = 7 T + 5 C

Constraints:

• Have 2400 hours of carpentry time available

3 T + 4 C < 2400 (hours)

• Have 1000 hours of painting time available

2 T + 1 C < 1000 (hours)

More Constraints:• Make no more than 450 chairs

C < 450 (no. chairs)

• Make at least 100 tables T > 100 (no. tables)

Non-negativity:Cannot make a negative number of chairs ortables

T > 0C > 0

Max Z = 7T + 5C (Profit)

Subject to the constraints:

3T + 4C < 2400 (carpentry hrs)

2T + 1C < 1000 (painting hrs)

C < 450 (max no.of chairs)

T > 100 (min no.of tables)

T, C > 0 (non-negativity)

Carpentry

Constraint Line

3T + 4C = 2400

Intercepts

(T = 0, C = 600)

(T = 800, C = 0)

0 800 T

C

600

0

Feasible

< 2400 hrs

Infeasible

> 2400 hrs

3T + 4C = 2400

Painting

Constraint Line

2T + 1C = 1000

Intercepts

(T = 0, C = 1000)

(T = 500, C = 0)

0 500 800 T

C1000

600

0

2T + 1C = 1000

0 100 500 800 T

C1000

600

450

0

Max Chair Line

C = 450

(T=0, C=450)

Min Table Line

T = 100

(T=100, C=0)Feasible

Region

The corner points of the feasible region are:

A(100,0)

B(500,0)

C(320,360)

D(200,450)

E(100,450)

Z value at A(100,0) = 700

Z value at B(500,0) =3500

Z value at C(320,360) =4040

Z value at D(200,450) =3650

Z value at E(100,450) =2950

Z value maximum at C(320,360) hence an optimum solution is given by

T=320 & C=360 and max profit is Rs.4040

2. Solve graphically:

0x,0 x

18x2 x

84x73x

60x45x Subject to

x146x ZMinimize

21

21

21

21

21

and

3. Solve graphically

0x,0 x

3x x

1x xSubject to

x23x ZMinimize

21

21

21

21

and

4. Solve graphically

0x,0 x

3x3x-

1x xSubject to

x x ZMaximize

21

21

21

21

and

Simplex method of solving GLPP

Standard from of LPP:

1.Check whether the objective function is to be maximized or minimized. If minimized convert it into maximization form

Min Z= -{Max(-Z)}

EX:Min Z= 2x-3y+7z

Min Z= -{Max(-[2x-3y+7z ])}

Min Z= -{Max(-2x+3y-7z)}

2. The right hand side constant of each constraint should be non-negative. If not made it positive by multiplying it by

-1(one).

Ex:

0y)(x, and

7y-2x-7))(1(y]-(-1)[2x

265y-8xs.t

5y2x Max Z

0y)(x, and

-7y-2x

265y-8xs.t

5y2x Max Z

3. All the decision variables should be non-negatively restricted i.e if the variable x is unrestricted in sign then such a variable is expressed as the difference between two variables which are non-negative

0,0 where xxxxx

0 x

7 y -3x

12 9y2x s.t

7y-2xZ Max:EX

0y0,y 0, x

7 yy-3x

7 )y-y(-3x

12 y9-y92x

12 )y-y9(2x s.t

y7y7-2x)y-y7(-2xMax Z

4.Convert all the inequalities into equations by introducing non-negative variables on the left hand side of each constraint called slack or surplus variables.

.constrainteach of

side handleft on the variablesurplus a substract then of is

inequality of type theIf .constrainteach of side handleft the

on ableslack vari a add then of is inequality of type theIf

0y),(x and

254y7x

84 2yx-

60 y2xs.t

3y2x Max Z

0y),(x and

254y7x

84 s-2yx-

60 sy2xs.t

3y2x Max Z

2

1

Simplex method of solving LPP

• Solve by simplex method

0)x,(x and

2 x

1 xx-

6 2x x

24 4x6x .

x45x Max Z

21

2

21

21

21

21

ts

Step 1: Convert the LPP into standard form.

)0s0,,0,0,0x,0(x and

2 s x

1 xx-

6 2x x

24 4x6x .

x45x Max Z

432121

42

321

221

121

21

sss

s

s

sts

)0s0,,0,0,0x,0(x and

2 s x

1 xx-

6 2x x

24 4x6x .

.0.0.0.0x45x Max Z

432121

42

321

221

121

432121

sss

s

s

sts

ssss

Step 2: Prepare simplex table

10001020s

01001-110s

00102160s

000146240s

ssssxxbSolB CB.V

000045c

4

3

2

1

432121i

j

jj

j

s ).(z where

compute Now

orxBC

cz

j

jj

000045Δ

10001020s

01001-110s

00102160s

000146240s

ssssxxbSolB CB.V

000045c

j

4

3

2

1

432121i

j

step.next the ootimal.Gotnot is

solution the then negative is j Δoneatleast If : II Case

optimal. be willsolution the then 0j Δall If: I Case:3 Step

)infinitive & ratios negative ignore ( them of minimum

chose and Ratios compute variable outgoing determine To

Δnegativemost having variable The :(I.V) variable Incoming

variable outgoing & incoming Determine :4 Step

j

00004Δ

10001020s

01001-110s

00102160s

000146240s

ssssxxSolB CB.V

000045c

j

4

3

2

1

432121

j

5

(I.V)

000045Δ

10001020s

01001-110s

00102160s

000146240s

ssssxxSolB CB.V

000045c

j

4

3

2

1

432121

j

/xRatios(Sol 1)4624

(I.v)

ignore

ignore

(o.v)

16 6

000045Δ

10001020s

01001-110s

00102160s

000146240s

ssssxxSolB CB.V

000045c

j

4

3

2

1

432121

j

Ratios

Step 5: Mark the element in the incoming variable corresponding to the outgoing variable called key element or pivotal element.

Step 6: Divide each element of the Key row (including bi) by the key element to get the corresponding values in the new table.

1st row (new)

0 0 0 1/6 4/6 1 4

s s s s x xbSol 432121i

For each row other than the key row,

value row treplacemen ingCorrespond

column key the in element Row

-element row Old

element row New

Corresponding replacement row value:

2 nd old row:

Old row element- Row element in key column* corresponding replacement value

6 1 4 =21 1 1 =02 1 4/6 =8/60 1 1/6 =-1/61 1 0 =10 1 0 =00 1 0 =0

0 0 0 1/6 4/6 1 4

s s s s x xbSol 432121i

0 0 1 0 2 1 6

s s s s x xbSol 432121i

2 nd row (new):

0 0 1 1/6- 8/6 0 2

s s s s x x bSol 432121i

Corresponding replacement row value:

3 rd old row:

Old row element- Row element in key column* corresponding replacement value

1 -1 4 =5-1 -1 1 =01 -1 4/6 =10/60 -1 1/6 =1/60 -1 0 =01 -1 0 =10 -1 0 =0

0 0 0 1/6 4/6 1 4

s s s s x xbSol 432121i

0 1 0 0 1 1- 1

s s s s x xbSol 432121i

3 rd row (new):

0 1 0 1/6 10/6 0 5

s s s s x x bSol 432121i

Corresponding replacement row value:

4 th old row:

Old row element- Row element in key column* corresponding replacement value

2 0 4 =20 0 1 =01 0 4/6 =10 0 1/6 =00 0 0 =00 0 0 =01 0 0 =1

0 0 0 1/6 4/6 1 4

s s s s x xbSol 432121i

1 0 0 0 1 0 2

s s s s x xbSol 432121i

4 th row (new)

1 0 0 0 1 0 2

s s s s x x bSol 432121i

100 0 1020s

010 61 610050s

001 6168020s

000 61 64145x

sss s xxbSol B CB.V

000045c

4

3

2

1

432121i

j

Now compute again ΔJ

0 0 0 65 6

4- 0 j

100 0 1020s

010 61 610050s

001 6168020s

000 61 64145x

sss s xxbSol B CB.V

000045c

4

3

2

1

432121i

j

0 0 0 65 6

4- 0

j

100 0 1020s

010 61 610050s

001 6168020s

000 61 64145x

sss s xxbSol B CB.V

000045c

4

3

2

1

432121i

j

2sol/Ratio x

664

4

8

12

68

2

6 3

10

5

212

( ).

10 3/4- 1/8 001/20s

015/4- 3/8005/20s

003/4 81 103/24x

001/2- 1/4 0135x

sss sxxbSol B CB.V

000045c

4

3

2

1

432121i

j

0 0 1/2 6/8 0 0 Δ

10 3/4- 1/8 001/20s

01 5/4- 3/8005/20s

00 3/4 81 103/24x

00 1/2- 1/4 0135x

ss s sxxbSol B CB.V

000045c

j

4

3

2

1

432121i

j

21.21

*025

*023

*43)*(5SolC.BZ

isprofit maximum the and 23x3,by x given is and

attained. is solution optimum an hence 0 are s' Δall Since

21

j

• The Omega manufacturing company has discontinued the production of a certain unprofitable product lin. This act created considerable excess production capacity. Management is considering devoting this excess capacity to one or more of the three products; call them products 1,2 and 3. The available capacity on the machines that might limit output is summarized in the following table:

The number of machine hours required for each unit of the respective product is:

Machine Type Available Time(Machine Hrs per week)

Milling machine 500

Lathe 350

Grinder 150

Machine Type Product 1 Product 2 Product 3

Milling machine 9 3 5

Lathe 5 4 0

Grinder 3 0 2

The sales department indicates that the sales potential for product 3 is 20 units per week. The unit profit would be Rs. 50, Rs.20, and Rs.25 respectively on products 1,2 and 3. the objective is to determine how much of each product Omega should produce to maximize the profit.

• A firm produces three products A,B and C, each of which passes through three departments: Fabrication, Finishing and Packaging. Each unit of product A requires 3,4 and 2; a unit of product B requires 5, 4 and 4, while each unit of product C requires 2, 4 and 5 hours respectively in three departments. Everyday, 60 hours are available in the fabrication department, 72 hours in the finishing department and 100 hours in the packaging department. The unit contribution of product A is Rs.5, of product is Rs.10, and of product C is Rs.8.