llp and transportation problems solution

Operations Research

“”

CES Activity On LLP and Transportation Problems

Made By :- ADITYA ARORARoll No. :- 0141mba009Class :- MBA 3A

Linear Programming

ProblemsGraphic Method

Product Platinum Required/unit (gms)

Gold Required/unit (gms)

Profit/Unit (Rs.)

A 2 3 500

B 4 2 600

Q1) A company manufactures two types of parts which use precious metals platinum &

gold. Due to shortage of these precious metals, the government regulates the amount that

may be used per day. The relevant data with respect to supply, requirements and profits are

summarized in the table as follows:

Daily allotment of platinum & gold are 160 gm and 120 gm respectively. How should the

company divide the supply of scarce precious metals. Formulate as a LPP.

Solution Q1)

After reading the question we know this is a maximization question as we have a limited

supply of Gold and Platinum for 2 product which required the item to produce the product

Let us assume product A is x1 and product B is x2

Therefor,

Zmax = 500 x1 + 600x2

For the equation we know that Gold and Platinum daily allotment is 120gm and 160gm .

Now, for equation-1 we will take all the requirement of platinum per unit of all product witch will be

multiplied with the units to be allotted to a product (i.e., x1 and x2) and then add them with each other witch

will be less then or equal to zero(0)

i.e., 2x1+ 4x2≤ 160

Now, for equation-2 we will take all the requirement of gold per unit of all product witch will be multiplied

with the units to be allotted to a product (i.e., x1 and x2) and then add them with each other witch will be less

then or equal to zero(0)

i.e., 3x1+ 2x2≤ 120

Also a non-negativity restriction for preventing negative values i.e., x1 ,x2 ≥ 0

By the help of eq. 1 we will find out the coordinates

2x1+ 4x2≤ 160

Let take x1 =0

0+ 4x2= 160

x2= 40

Therefor, coordinates are (0,40)

Let take x2 =0

2x1+ 0 = 160

x1= 80


x1 x2

0 4080 0

By the help of eq. 2 we will find out the coordinates

3x1+ 2x2≤ 120

Let take x1 =0

0+ 2x2= 120

x2= 60


Let take x2 =0

3x1+ 0 = 120

x1= 40


x1 x2

0 6040 0

0 10 20 30 40 50 60 70 80 900

10

20

30

40

50

60

70

A

B

C

FEASIBLE AREA

From the graph we got the fishable area OABC and thehe coordinate are

We will use equation of zmax to find the quantity per product

Zmax = 500 x1 + 600x2

• For O

Zmax = 0

• For A

Zmax = 0x1 + (600*40)

Zmax = 24000

• For B

Zmax = (500*20) + (600*30)

Zmax = 28000

x1 x2

O 0 0

A 0 40

B 20 30

C 40 0

• For C

Zmax = 500*40 + 0x2

Zmax = 20000

As the value of coordinate C is the maximum, therefor, the producer should use

• For product A = 20 units of Platinum and 20 unit of Gold and

• For product B = 30 units of Platinum and 30 units of Gold

Therefor, making total of 140 units of Platinum and 120units of Gold

Any QuestionsRelating To LLP

Transportation Problem

Q2) A company is producing three product P1, P2 and P3 at two of its plant situated in cities A and B. The

company plans to start a new plant in city C or in city D. the unit profits from the various plants are

listed in the table, along with the demand for various products and capacity available in each of the

plants. PlantProduct

CapacityP1 P2 P3

A 35 24 20 600B 30 28 25 1000C 20 25 37 800D 24 32 28 800

Demand 500 800 600 1900 3200

The company would set up the new plant on the bases of maximising aggregate profits from the three

cities plants. Using the transportation method determine in witch city would the plant be set up and

what would the corresponding profit be.

Solution Q2)

Step1: To check if the demand and supply are balanced

As the demand and supply are not equal we will introduce dummy column in the table to get

Supply = Demand

PlantProduct

CapacityP1 P2 P3 Dummy

A 35 24 20 0 600

B 30 28 25 0 1000

C 20 25 37 0 800

D 24 32 28 0 800

Demand 500 800 600 1300 3200

Step2: To see if the given matrix is minimisation matrix o not

As we can see in the question it was clearly mention that the matrix given is the profit from the product

from different plant, Hence, it’s a maximisation matrix so to convert it in to minimisation matrix we

have to select the biggest profit and subtract it from every other profit to get the new matrix

PlantProduct


A 35 24 20 0 600

B 30 28 25 0 1000

C 20 25 37 0 800

D 24 32 28 0 800

Demand 500 800 600 1300 3200

The biggest

profit given in

the table

PlantProduct


A 2 13 17 37 600

B 7 9 12 37 1000

C 17 12 0 37 800

D 13 5 9 37 800

Demand 500 800 600 1300 3200

Step3: Using VAM method to solve

PlantProduct

Capacity PenaltiesP1 P2 P3 Dummy

A 2 13 17 37 600 11

B 7 9 12 37 1000 2

C 17 12 0(600) 37 800 200 12

D 13 5 9 37 800 4

Demand 500 800 600 1300

3200Penalties

5 4 9 0


PlantProduct


A 2(500) 13 17 37 600 100 11 11

B 7 9 12 37 1000 2 2

C 17 12 0(600) 37 800 200 12 5

D 13 5 9 37 800 4 8

Demand 500 800 600 1300

3200Penalties

5 4 9 05 4 X 0


PlantProduct


A 2(500) 13 17 37 600 100 11 11 24

B 7 9 12 37 1000 2 2 28

C 17 12 0(600) 37 800 200 12 5 25

D 13 5(800) 9 37 800 4 8 32

Demand 500 800 600 1300

3200Penalties

5 4 9 05 4 X 0X 4 X 0


PlantProduct

Capacity Penalties

P1 P2 P3 Dummy

A 2(500) 13 17 37 600 100 11 11 24 37

B 7 9 12 37(1000) 1000 2 2 28 37

C 17 12 0(600) 37 800 200 12 5 25 37

D 13 5(800) 9 37 800 4 8 32 X

Demand 500 800 600 1300 300

3200Penalties

5 4 9 05 4 X 0X 4 X 0

X X X 0


PlantProduct

Capacity Penalties

P1 P2 P3 Dummy

A 2(500) 13 17 37(100) 600 100 11 11 24 37

B 7 9 12 37(1000) 1000 2 2 28 37

C 17 12 0(600) 37(200) 800 200 12 5 25 37

D 13 5(800) 9 37 800 4 8 32 X

Demand 500 800 600 1300 300

3200Penalties

5 4 9 05 4 X 0X 4 X 0

X X X 0


PlantProduct


A 2(500) 13 17 37(100) 600

B 7 9 12 37(1000) 1000

C 17 12 0(600) 37(200) 800

D 13 5(800) 9 37 800

Demand 500 800 600 1300 3200

Step4(a) : Feasibility test to check if the matrix is feasible or not

The test id done by using formula No. of occupied cells =

m+n-1

Where, m =

no. of rows and n = no. of columns

No. of occupied cells = 4+4-1=7

But there are only 6 occupied cells. Hence the matrix is not feasible.Plant

ProductCapacity

P1 P2 P3 Dummy

A 2(500) 13 17 37(100) 600

B 7 9 12 37(1000) 1000

C 17 12 0(600) 37(200) 800

D 13 5(800) 9 37 800

Demand 500 800 600 1300 3200

Step4(b) : Introducing EPSILON (e)

Epsilon is used to make a unoccupied cell into occupied cells and its vale is near to zero or nil

It is placed on lowest value in the matrix by keeping in mind 2 things

i) It should be introduced at least coast unoccupied cell

ii) It should not form a loop with other occupied sell. If it form a loop at the least cost then move to the

second least cost .

PlantProduct


A 2(500) 13 17 37(100) 600

B 7 9(e) 12 37(1000) 1000

C 17 12 0(600) 37(200) 800

D 13 5(800) 9 37 800

Demand 500 800 600 1300 3200

Step5: Finding the value of Ui and Vj

The formula for finding them is Cij = Ui + Vj

Let us assume U1 = 0

PlantProduct


A 2(500) 13 17 37(100) 600 U1=0

B 7 9(e) 12 37(1000) 1000 U2=0

C 17 12 0(600) 37(200) 800 U3=0

D 13 5(800) 9 37 800 U4=-4

Demand500 800 600 1300

3200V1=2 V2=9 V3=0 V4=37

Step6: Optimal Test

Formula for this test is

Unoccupied cells = Ui + Vj - Cij

• C12(13) = U1 + V2 – C13 = 0+9-13=(-4)

• C13(17) = U1 + V3 – C13 = 0+0-17=(-17)

• C21(7) = U2 + V1 – C21 = 0+2-7=(-5)

• C23(12) = U2 + V3 – C23 = 0+0-12=(-12)

• C31(17) = U3 + V1 – C31 = 0+2-17=(-15)

• C32(12) = U3 + V2 – C32 = 0+9-12=(-3)

• C41 (13) = U4 + V1 – C41 = (-4)+2-13=(-15)

Step6: Optimal Test

Formula for this test is

Unoccupied cells = Ui + Vj - Cij

• C43 (9) = U4 + V3 – C43 = (-4)+0-9=(-13)

• C44 (37) = U4 + V4 – C44 = (-4)+37-37=(-4)

As all the values of optimal cost is negative or zero, Therefor, the new shold be set up in

If we take cities A,B and C the profit will be Total profit (ABC) = (35*500)+(0*100)+(28*e)+(0*1000)+(37*600)+(0*200)= 17500+0+0+0+22200+0 =39700

If we take cities A,B and D the profit will be Total profit (ABD) = (35*500)+(0*100)+(28*e)+(0*1000)+(32*800)= 17500+0+0+0+25600 =43100

• As the Profit earned is more in cities ABD then in cities ABC i.e., 43100 and 39700 respectively

• Therefor, the new plant should be setup in city D rather then in City C

Thank YouAny Questions

llp and transportation problems solution

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