Operations Final case

7/29/2019 Operations Final case

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Group - 4

Aakash B. Pathak (PB1101)

Abhishek Mandloi (PA1101)

Anamika Singh (PA1117)Kuntal Chatterjee (PA1118)

Rohit Sharma (PA1129)

Shalinee Pujari(PA1133)

Sumit Kumar(PB1117)


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Introduction

Linear programming (LP, orlinear

optimization) is a mathematical method for

determining a way to achieve the best outcome

(such as maximum profit or lowest cost) in a

given mathematical model for some list of

requirements represented as linear

relationships. Linear programming is a specificcase of mathematical programming

(mathematical optimization)


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Terminology

Objective Function(Zmax/min

):The solution of

a linear-programming problem reduces to

finding the optimum value (largest or smallest,

depending on the problem) of the linear

expression called the objective function.

For example, in our problem the objective function is:

Z(min)=125X1+121X2+118X3+132X4+124X5+119X6+130X7+126X8+129X9+123X10+ 125X11+129X12+

0.X13+0.X14+0.X15+ 0.X16


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Constraint: In mathematics, a constraint is a condition

that a solution to an optimization problem is required by

the problem itself to satisfy. There are two types of

constraints: equality constraints and inequality

constraints. The set of candidate solutions that satisfy

all constraints is called thefeasible set.

Decision Variable: in our equation, X1, X2, X3,.,

X16 are the decision variables.


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W1 W2 W3 W4 Supply

P1 125 121 118 132 400

P2 124 119 130 126 300

P3 129 123 125 129 500

P4 0 0 0 0 600

Demand 200 300 700 600

Q.1Formulate the primal

model


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Primal Solution

Objective Function

Z(min)=125X1+121X2

+118X3+132X4+124X5

+119X6+130X7+126X8+129X9+123X10+

125X11+129X12+

0.X13+0.X14+0.X15+

0.X16

Where, Xi 0.

i = 1,2,3,16

Subject to Constraint1. Supply Constraints:

X1+X2+X3+X4=400

X5+X6+X7+X8=300

X9+x10+X11+X12=500 X13+X14+X15+X16=600

2. Demand Constraints:

X1+X5+X9+X13=200

X2+X6+X10+X14=300

X3+X7+X11+X15=700

X4+X8+X12+X16=600


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W1 W2 W3 W4 Supply

P1 125 121 118 132 400

P2 124 119 130 126 300

P3 129 123 125 129 500

P4 0 0 0 0 600

Demand 200 300 700 600

Q.2Solve the LP using

Solver


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=C4*C14+D4*D14+E4*E14+F4*F14

+G4*G14+H4*H14+I4*I14+J4*J14+

K4*K14+L4*L14+M4*M14+N4*N14

+O4*O14+P4*P14+Q4*Q14+R4*R14

This one is the formula for the highlighted cell


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=C5*$C$14+D5*$D$14+E5*$E$14+

F5*$F$14+G5*$G$14+H5*$H$14+I5

*$I$14+J5*$J$14+K5*$K$14+L5*$

L$14+M5*$M$14+N5*$N$14+O5*$

O$14+P5*$P$14+Q5*$Q$14+R5*$R

$14

This one is the formula for the highlighted cell


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Procedure to Solve using solver

STEP BY STEPFOLLOW THE POINTER


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THIS IS THE OPTIMALSOLUTION


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Here, Zmin = 1,46,000


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Q.3

Write the Dual Model For the

primal & report the Solver

solution


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Dual Solution

Objective function

Z(max)=400U1+300U2+500U3+600U4+200V1+300V2+

700V3+ 600V4

Where the dual variables Ui& Vj are unrestricted in

sign. i = 1,2,3,4

j = 1,2,3,4

Subject to constraint

U1+V1125

U1+V2121

U1+V3118

U1+V4132

U2+V1124

U2+V2119

U2+V3130

U2+V4126

U3+V1129

U3+V2123

U3+V3125 U3+V4129

U4+V10

U4+V20

U4+V30

U4+V40


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=C4*C22+D4*D22+E4*E22+F4*

F22+G4*G22+H4*H22+I4*I22+J

4*J22


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=C6*$C$22+D6*$D$22+E6*$E$

22+F6*$F$22+G6*$G$22+H6*$

H$22+I6*$I$22+J6*$J$22


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STEP BY STEPFOLLOW THE POINTER


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THIS IS THE OPTIMAL

SOLUTION


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Here, Zmin value from dual model is 1,46,000.

Also, Zmin value from primal model is 1,46,000

It indicates that our both model is equal to each other.


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W1 W2 W3 W4 Supply

P1 125 121 118 132 400

P2 124 119 130 126 300

P3 129 123 125 129 500

P4 0 0 0 0 600Demand 200 300 700 600

Q.4

Obtain the optimal solution by

LCM & UV method.


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D1 D2 D3 D4 Supply

S1 125 121 118 132 400

S2 124 119 130 126 300

S3 129 123 125 129 500

S4 0 0 0 0 600Demand 200 300 700 600

Solution of Transportation Problem Using Least Cost Method

TOTAL no. of supply constraints : 4

TOTAL no. of demand constraints : 4Problem Table is


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D1 D2 D3 D4 Supply

S1 125 121 118 132 400

S2 124 119 130 126 300

S3 129 123 125 129 500

S4 0(200) 0 0 0 400

Demand 0 300 700 600

Allocate At [4][1] = 200

Step = 1


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D1 D2 D3 D4 Supply

S1 125 121 118 132 400

S2 124 119 130 126 300

S3 129 123 125 129 500

S4 0(200) 0(300) 0 0 100

Demand 0 0 700 600


Step = 2


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D1 D2 D3 D4 Supply

S1 125 121 118 132 400

S2 124 119 130 126 300

S3 129 123 125 129 500

S4 0(200) 0(300) 0(100) 0 0

Demand 0 0 600 600


Step = 3


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D1 D2 D3 D4 Supply

S1 125 121 118(400) 132 0

S2 124 119 130 126 300

S3 129 123 125 129 500

S4 0(200) 0(300) 0(100) 0 0

Demand 0 0 200 600


Step = 4


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D1 D2 D3 D4 Supply

S1 125 121 118(400) 132 0

S2 124 119 130 126 300

S3 129 123 125(200) 129 300

S4 0(200) 0(300) 0(100) 0 0

Demand 0 0 0 600


Step = 5


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D1 D2 D3 D4 Supply

S1 125 121 118(400) 132 0

S2 124 119 130 126(300) 0

S3 129 123 125(200) 129 300

S4 0(200) 0(300) 0(100) 0 0

Demand 0 0 0 300


Step = 6


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D1 D2 D3 D4 Supply

S1 125 121 118(400) 132 0

S2 124 119 130 126(300) 0

S3 129 123 125(200) 129(300) 0

S4 0(200) 0(300) 0(100) 0 0

Demand 0 0 0 0


Step = 7


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D1 D2 D3 D4 Supply

S1 125 121 118(400) 132 400

S2 124 119 130 126(300) 300

S3 129 123 125(200) 129(300) 500

S4 0(200) 0(300) 0(100) 0 600

Demand 200 300 700 600

Final Allocation Table is


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Here m + n - 1 = 4 + 4 - 1 = 7 & No. of basiccell=7

Hence, (m+n-1) = No. of basic cell

The solution is feasible.

Total Transportation cost = 118 400 + 126

300 + 125 200 + 129 300 + 0 200 + 0

300 + 0 100 + = 148700

-----------------------------------------------------------

The least total transportation cost = 148700

--------------------------------------------------------


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D1 D2 D3 D4 Supply Ui

S1 125 121 118(400) 132 400 118S2 124 119 130 126(300) 300 122

S3 129 123 125(200) 129(300) 500 125

S4 0(200) 0(300) 0(100) 0 600 0

Demand 200 300 700 600Vj 0 0 0 4

Find Ui and VjFormula: Ui + Vj = COSTij

UV / MODI / PENALTY method


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S1 125[7] 121[3] 118(400) 132[10] 400 118

S2 124[2] 119[-3] 130[8] 126(300) 300 122

S3 129[4] 123[-2] 125(200) 129(300) 500 125

S4 0(200) 0(300) 0(100) 0[-4] 600 0

Demand 200 300 700 600

Vj 0 0 0 4

Calculate penalties(Pij) for all non-basic

cell.

FORMULA: Pij = COSTij (Ui + Vj)


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S1 125[7] 121[3] 118(400) 132[10] 400 118

S2 124[2] 119[-3] 130[8] 126(300) 300 122

S3 129[4] 123[-2] 125(200)(+)129(300)

(-)500 125

S4 0(200) 0(300) 0(100)(-) 0[-4](+) 600 0

Demand 200 300 700 600

Vj 0 0 0 4

Here, minimum penalty cell is considered as Home

cell. Cell S4D4 is home cell.


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S0 125 121 118(400) 132 400 122

S1 124 119 130 126(300) 300 126

S2 129 123 125(300) 129(200) 500 129

S3 0(200) 0(300) 0 0(100) 600 0

Demand 200 300 700 600

Vj 0 0 -4 0

Minimum allocation among all -VE Once = 100.

Substract 100 from all (-) and Add it to all (+).

Find Ui and Vj


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S1 125[3] 121[-1] 118(400) 132[10] 400 122

S2 124[-2] 119[-7] 130[8] 126(300) 300 126

S3 129[0] 123[-6] 125(300) 129(200) 500 129

S4 0(200) 0(300) 0[4] 0(100) 600 0

Demand 200 300 700 600

Vj 0 0 -4 0

Modified Table 1:

Calculate penalties(Pij) for all non-basic cell.



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S1 125[3] 121[-1] 118(400) 132[10] 400 122

S2 124[-2] 119[-7](+) 130[8] 126(300)(-) 300 126

S3 129[0] 123[-6] 125(300) 129(200) 500 129

S4 0(200) 0(300)(-) 0[4] 0(100)(+) 600 0

Demand 200 300 700 600

Vj 0 0 -4 0

Here, minimum penalty cell is considered

as Home cell. Cell S2D2is home cell.

Minimum among all -VE Once = 300.


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S0 125 121 118(400) 132 400 -7

S1 124 119(300) 130 126 300 0

S2 129 123 125(300) 129(200) 500 0

S3 0(200) 0 0 0(400) 600 -129

Demand 200 300 700 600

Vj 129 0 125 129

Minimum among all VE Once 300.

Substract 300 from all (-) and Add it to all (+).

Find Ui and Vj

Modified Table 2:


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S1 125[3] 121[128] 118(400) 132[10] 400 -7

S2 124[-5] 119(300) 130[5] 126[-3] 300 0

S3 129[0] 123[123] 125(300) 129(200) 500 0

S4 0(200) 0[129] 0[4] 0(400) 600 -129

Demand 200 300 700 600

Vj 129 0 125 129

Calculate penalties(Pij) for all non-basic

cell.



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S1 125[3] 121[128] 118(400) 132[10] 400 -7

S2 124[-5](+) 119(300) 130[5] 126[-3] 300 0

S3 129[0] 123[123] 125(300) 129(200) 500 0

S4 0(200) 0[129] 0[4] 0(400) 600 -129

Demand 200 300 700 600

Vj 129 0 125 129

Here, minimum penalty cell is considered as

Home cell. Cell S2D1is home cell.


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S0 125 121 118(400) 132 400 -2

S1 124(-1) 119(300) 130 126 300 0

S2 129 123 125(300) 129(200) 500 5

S3 0(200) 0 0 0(400) 600 -124

Demand 200 300 700 600

Vj 124 119 120 124

Modified Table 3:


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S1 125[3] 121[4] 118(400) 132[10] 400 -2

S2 124(-1) 119(300) 130[10] 126[2] 300 0

S3 129[0] 123[-1] 125(300) 129(200) 500 5

S4 0(200) 0[5] 0[4] 0(400) 600 -124

Demand 200 300 700 600

Vj 124 119 120 124

Modified Table 3:




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S1 125[3] 121[4] 118(400) 132[10] 400 -2

S2 124(-1) 119(300) 130[10] 126[2] 300 0

S3 129[0] 123[-1](+) 125(300) 129(200) 500 5

S4 0(200) 0[5] 0[4] 0(400) 600 -124

Demand 200 300 700 600

Vj 124 119 120 124


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S0 125 121 118(400) 132 400 -7

S1 124(-1) 119(300) 130 126 300 -4

S2 129 123(-1) 125(300) 129(200) 500 0

S3 0(200) 0 0 0(400) 600 -128

Demand 200 300 700 600

Vj 128 123 125 128

Minimum among all -VE Once = -1.

Substract -1 from all (-) and Add it to all (+) Find Ui and Vj


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S1 125[4] 121[5] 118(400) 132[11] 400 -7

S2 124(-1) 119(300) 130[9] 126[2] 300 -4

S3 129[1] 123(-1) 125(300) 129(200) 500 0

S4 0(200) 0[5] 0[3] 0(400) 600 -128

Demand 200 300 700 600

Vj 128 123 125 128




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D1 D2 D3 D4 Supply

S1 125 121 118(400) 132 400S2 124(-1) 119(300) 130 126 300

S3 129 123(-1) 125(300) 129(200) 500

S4 0(200) 0 0 0(400) 600

Demand 200 300 700 600

Since all Pij0. Hence,the allocation is optimum.


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Here m + n - 1 = 4 + 4 - 1 = 7 & No. of basic cell=7Hence, (m+n-1) = No. of basic cell

The solution is feasible.

Total Transportation cost = (118 400) + (124 -1) +

(119 300) + (123 -1) + (125 300) + (129 200) +

(0 200 )+ (0 400 )+ = 145953

-----------------------------------------------------------

The minimized total transportation cost = 145953-----------------------------------------------------------


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W1 W2 W3 W4 Supply

P1 125 121 118 132 400

P2 124 119 130 126 300

P3 129 123 125 129 500

P4 0 0 0 0 600

Demand 200 300 700 600

Q.5

What will be the impact on the

planning if the route P2-W3 andthe route P1-W4 are blocked for

maintenance ? X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16

Z 125 121 118 0 124 119 0 126 129 123 125 129 0 0 0 0

C1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0


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C1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0

C2 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0

C3 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0

C4 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1

C5 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0

C6 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0

C7 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0

C8 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1

DV

Z 0

LHS RHS

0 400

0 3000 500

0 600

0 200

0 300

0 700

0 600

=C4*C14+D4*D14+E4*E1

4+F4*F14+G4*G14+H4*H14+I4*I14+J4*J14+K4*K1

4+L4*L14+M4*M14+N4*

N14+O4*O14+P4*P14+Q4

*Q14+R4*R14


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STEP BY STEP

FOLLOW THE POINTER


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X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16

Z 125 121 118 0 124 119 0 126 129 123 125 129 0 0 0 0

C1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0


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C2 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0

C3 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0

C4 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1

C5 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0

C6 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0

C7 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0

C8 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1

DV 0 0 0 400 0 0 300 0 0 300 200 0 200 0 200 200

Z 61900

LHS RHS

400 400

300 300

500 500

600 600

200 200

300 300

700 700

600 600

THIS IS THE OPTIMAL

SOLUTION


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If we blocked the route P2-W3 &

the route P1-W4, then then the

minimum cost Z=61,900

Operations Final case

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