chapter 1 lpp tps

7/30/2019 Chapter 1 Lpp Tps

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What is Operations Research?



Meaning Of OR:

According to T.L.Saaty “OR is the art of giving badresults to problems to which otherwise worseresults are given”.

According to Churchman “OR is the application of

scientific methods, techniques and tools toproblems involving the operations of systems soas to provide these in control of the applicationwith optimum solutions to the problem”.

According to H.M.Wagner “ OR is the scientificapproach to problem solving for executivemanagement”.



Scope of OR:

• In Agriculture

• In Finance

• In Industry• In Marketing

• In Production Management

• In Personnel Management• In Production Management



Methodology Of OR:

Define theProblem

Develop theModel

Obtain InputData

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 alinear function of the variables.

Step 2: Formulate the other conditions of the problem such as resources limitationsas linear equations or inequations intermsof 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 Brequires 2 hours of grinding and 5 hours of polishing. The manufacturer has 2 grinders and3 polishers. Each grinder works for 40 hours a

week and each polisher works for 60 hours aweek. Profit on model A is Rs.3 and on model Bis Rs.4. Whatever is produced in a week is soldin the market. How should the manufacturer

allocate his production capacity to the types of models so that he may make the maximum profitin week?



2. A manufacturing firm needs 5 component parts.

Due to inadequate resources, the firm is unable

to manufacture all its requirements. So themanagement 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. Therelevant 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



Where

M: Per unit milling time in hours

A: Per unit assembly time in hours

T: Per unit testing time in hoursTR: Total requirements in units

PP: Price per unit quoted in the market

PC: Per unit direct costs

Resources available are as follows:Milling hours: 300

Assembly hours: 160

Testing hours: 150

Formulate this as an LPP, taking the objective function asmaximization of saving by producing the componentsinternally.



3. The marketing department of Everest company

has collected information on the problem of

advertising for its product. This relates to theadvertising media available, the number of

families expected to be reached with each

alternative, cost per advertisement, the

maximum availability of each medium and theexpected exposure of each one. The information

is as given below:

Advertising Media

No.of families

expected tocover Cost perAd (Rs)

Maximum

availability (No.oftimes)

Expected

exposure(units)

TV(30 sec) 3000 8000 8 80

Radio(15 sec) 7000 3000 30 20

Sunday edition of a daily (1/4page) 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 asequation.

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

common region satisfying all theconstraints simultaneously.

Step 4: Determine the corner points of thefeasible region.

Step 5: Find the optimal solution by

substituting the corner points in theobjective 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

Contribution7 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 makeC = 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 or

tables

T > 0

C > 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



Painting

Constraint Line

2T + 1C = 1000

Intercepts

(T = 0, C = 1000)

(T = 500, C = 0)0 500 800 T

C

1000

600

0



0 100 500 800 T

C

1000

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(200,0)

B(500,0)

C(320,360)

D(200,450)E(450,100)



Z value at A(100,0) = 700

Z value at B(500,0) =3500Z 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 byT=320 & C=360 and max profit is Rs.4040



2. Solve graphically:

0x,0x18x2x

84x73x

60x45x Subject to

x146xZMinimize

21

21

21

21

21

and



3. Solve graphically

0x,0x

3xx

1xxSubject to

x23xZMinimize

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 minimizedconvert 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

5y2xMax Z

0y)(x,and

-7y-2x

265y-8xs.t

5y2xMax 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 isexpressed as the difference between two

variables which are non-negative

0,0where x x x x x



0y0,y0,x

7 yy-3x

7 )y-y(-3x

12y9-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 surplusvariables.

.constrainteachof

sidehandlefton thevariablesurplusasubstractthenof is

inequalityof typetheIf .constrainteachof sidehandlefttheonableslack variaaddthenof is inequalityof typetheIf



0y),(xand254y7x

842yx-

60y2xs.t

3y2xMax Z



0y),(xand254y7x

84s-2yx-

60sy2xs.t

3y2xMax Z

2

1



Simplex method of solving LPP



• Solve by simplex method

0)x,(xand

2x

1xx- 62xx

244x6x .x45xMax Z

21

2

21

21

21

21

t s



Step 1: Convert the LPP into standard form.



)0s0,,0,0,0x,0(xand2s x

1 xx- 6 2xx

244x6x .x45xMax Z

432121

42

321

221

121

21

sss

s

s

st s



)0s0,,0,0,0x,0(xand2s x

1 xx-

6 2xx

244x6x .

.0.0.0.0x45xMax Z

432121

42

321

221

121

432121

sss

s

s

st s

ssss



Step 2: Prepare simplex table



j j

j

s).(zwhere

computeNow

or x BC

c z

j

j j



5

(I.V)



/x Ratios(Sol 1 )

6



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 thecorresponding values in the new table.

1st row (new)

0 0 0 1/6 4/6 1 4

s s s s xxbSol 432121i



For each row other than the key row,

valuerowtreplacemeningCorrespond

columnkeytheinelementRow

-elementrowOld

elementrowNew



Corresponding replacement row value:

2 nd old row:

Old row element- Row element in key column* corresponding replacementvalue

6 1 4 =2

1 1 1 =0

2 1 4/6 =8/6

0 1 1/6 =-1/6

1 1 0 =1

0 1 0 =0

0 1 0 =0

0 0 0 1/6 4/6 1 4


0 0 1 0 2 1 6




2 nd row (new):

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

s s s s x x bSol432121i




3 rd old row:


1 -1 4 =5

-1 -1 1 =0

1 -1 4/6 =10/6

0 -1 1/6 =1/6

0 -1 0 =0

1 -1 0 =1

0 -1 0 =0

0 0 0 1/6 4/6 1 4


0 1 0 0 1 1- 1




3 rd row (new):

0 1 0 1/6 10/6 0 5





4 th old row:


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

0 0 1/6 =00 0 0 =00 0 0 =01 0 0 =1

0 0 0 1/6 4/6 1 4


1 0 0 0 1 0 2




4 th row (new)

1 0 0 0 1 0 2




Now compute again ΔJ



0 0 0 6

5 6

4- 0 j



0 0 0 6

5 6

4 - 0 j

sol/ Ratio

6 6 4

4

8

12

6 8

2

3

( ) .



• The Omega manufacturing company has



• 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 and3. The available capacity on the machines

that might limit output is summarized in the

following table:



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

Machine Type Available Time(Machine Hrs perweek)

Milling machine 500Lathe 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 products1,2 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



• 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 threedepartments. Everyday, 60 hours are

available in the fabrication department, 72

hours in the finishing department and 100hours in the packaging department. The unit

contribution of product A is Rs.5, of product

chapter 1 lpp tps

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