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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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Procedure to Solve using solver
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
Allocate At [4][2] = 300
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
Allocate At [4][3] = 100
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
Allocate At [1][3] = 400
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
Allocate At [3][3] = 200
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
Allocate At [2][4] = 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
Allocate At [3][4] = 300
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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D1 D2 D3 D4 Supply Ui
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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D1 D2 D3 D4 Supply Ui
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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D0 D1 D2 D3 Supply Ui
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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D1 D2 D3 D4 Supply Ui
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.
FORMULA: Pij = COSTij (Ui + Vj)
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D1 D2 D3 D4 Supply Ui
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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D0 D1 D2 D3 Supply Ui
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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D1 D2 D3 D4 Supply Ui
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.
FORMULA: Pij = COSTij (Ui + Vj)
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D1 D2 D3 D4 Supply Ui
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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D0 D1 D2 D3 Supply Ui
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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D1 D2 D3 D4 Supply Ui
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:
Calculate penalties(Pij) for all non-basic cell.
FORMULA: Pij = COSTij (Ui + Vj)
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D1 D2 D3 D4 Supply Ui
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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D0 D1 D2 D3 Supply Ui
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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D1 D2 D3 D4 Supply Ui
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
Calculate penalties(Pij) for all non-basic cell.
FORMULA: Pij = COSTij (Ui + Vj)
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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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Procedure to Solve using solver
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