llp and transportation problems solution
TRANSCRIPT
![Page 1: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/1.jpg)
Operations Research
![Page 2: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/2.jpg)
“”
CES Activity On LLP and Transportation Problems
Made By :- ADITYA ARORARoll No. :- 0141mba009Class :- MBA 3A
![Page 3: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/3.jpg)
Linear Programming
ProblemsGraphic Method
![Page 4: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/4.jpg)
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.
![Page 5: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/5.jpg)
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
![Page 6: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/6.jpg)
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
![Page 7: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/7.jpg)
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
Therefor, coordinates are (80,0)
x1 x2
0 4080 0
![Page 8: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/8.jpg)
By the help of eq. 2 we will find out the coordinates
3x1+ 2x2≤ 120
Let take x1 =0
0+ 2x2= 120
x2= 60
Therefor, coordinates are (0,60)
Let take x2 =0
3x1+ 0 = 120
x1= 40
Therefor, coordinates are (40,0)
x1 x2
0 6040 0
![Page 9: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/9.jpg)
0 10 20 30 40 50 60 70 80 900
10
20
30
40
50
60
70
A
B
C
FEASIBLE AREA
![Page 10: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/10.jpg)
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
![Page 11: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/11.jpg)
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
![Page 12: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/12.jpg)
Any QuestionsRelating To LLP
![Page 13: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/13.jpg)
Transportation Problem
![Page 14: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/14.jpg)
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.
![Page 15: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/15.jpg)
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
![Page 16: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/16.jpg)
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
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
The biggest
profit given in
the table
![Page 17: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/17.jpg)
PlantProduct
CapacityP1 P2 P3 Dummy
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
![Page 18: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/18.jpg)
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
![Page 19: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/19.jpg)
Step3: Using VAM method to solve
PlantProduct
Capacity PenaltiesP1 P2 P3 Dummy
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
![Page 20: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/20.jpg)
Step3: Using VAM method to solve
PlantProduct
Capacity PenaltiesP1 P2 P3 Dummy
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
![Page 21: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/21.jpg)
Step3: Using VAM method to solve
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
![Page 22: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/22.jpg)
Step3: Using VAM method to solve
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
![Page 23: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/23.jpg)
Step3: Using VAM method to solve
PlantProduct
CapacityP1 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
![Page 24: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/24.jpg)
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
![Page 25: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/25.jpg)
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
CapacityP1 P2 P3 Dummy
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
![Page 26: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/26.jpg)
Step5: Finding the value of Ui and Vj
The formula for finding them is Cij = Ui + Vj
Let us assume U1 = 0
PlantProduct
CapacityP1 P2 P3 Dummy
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
![Page 27: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/27.jpg)
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)
![Page 28: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/28.jpg)
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
![Page 29: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/29.jpg)
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
![Page 30: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/30.jpg)
• 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
![Page 31: LLP and Transportation problems solution](https://reader035.vdocuments.net/reader035/viewer/2022082209/58a6135a1a28ab57488b4ed3/html5/thumbnails/31.jpg)
Thank YouAny Questions