transportation problem
DESCRIPTION
TRANSCRIPT
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Transportation Modeling
Problem
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THE PROBLEM
The Epsilon Computers Co. sells desktop computers to universities along University belt, and ship them from three distribution warehouses. The firm is able to supply the following numbers of desktop computers to the universities by the beginning of the academic year:
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Distribution warehouse
Supply
Sta. MesaTaft AveDivisoria
15020050
total 400
Universities have ordered desktop computers that must be delivered and installed by the beginning of the academic year:
University/College Demand
CEUFEUUE
10080220
total 400
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The shipping cost per desktop computer from each distributor to each university are as follows:
From
To
A (CEU
)
B (FEU
)
C (UE)
1(Sta. Mesa)2 (Taft Ave)3 (Divisoria)
7106
5123
9104
With cost minimization as criterion, Epsilon Company wants to determine how many desktop computers should be shipped from each warehouse to each university. Compare alternatives using,a.Northwest Corner Rule (NCR)b. Least Cost Method (LCM)c. Stepping Stone Methodd. Vogel’s Approximation Method (VAM)04/10/23 4
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FromFrom To To A (CEU)A (CEU) B (FEU) C (UE)B (FEU) C (UE) SupplySupply
1 150(STA. MESA) X1a X1b X1c
2 200 (TAFT AVE) X2a X2b X2c
3 50(DIVISORIA) X3a X3b X3c
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
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C= cost of shipment of all Xij= no. of computerd delivered
i= origin ; j= destinationObjective Function:
Minimize: C= 7X1a + 5X1b + 9X1c + 10X2a + 12X2b + 10X2c + 6X3a + 3X3b + 14 X3c
Constraints: X1a +X1b + X1c= 150 X2a + X2b + X2c = 200X3a + X3b + X3c = 50X1a + X2a + X3a= 100X1b + X2b + X3b= 80X1c+ X2c + x3c= 220
Xij ≥ 004/10/23 6
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150
(STA. MESA) 100 50
2 200
(TAFT AVE) 30 170 3 50 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
C= 700 + 250 + 360 + 1700 + 700 = P 371004/10/23 7
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 100 30 20 150(STA. MESA)
2 200 200 (TAFT AVE)
3 50 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
LEGEND:
1 2 3 4 5 C = P 319004/10/23 8
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150
(STA. MESA) 100 50
2 200
(TAFT AVE) 30 170 3 50 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
Table derived through Northwest Corner Rule
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Compute for Improvement Indices:
Improvement Index- the increase/decrease in total cost that would result from reallocating one unit to an unused square.
Unused Square
Closed Path Improvement
Indices
X1c +X1c-X1b+X2b-X2c+9-5+12-10=
+6
X2a +X2a-X1a+X1b-X2b +10-7+5-12= -4
X3a+X3a-X1a+X1b-
X2b+X2c-X3c+6-7+5-12+10-
14= -12
X3b +X3b-X2b+X2c-X3c+3-12+10-14=
-1304/10/23 10
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150
(STA. MESA) 100 50
2 200
(TAFT AVE) 30 170 3 50 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150
(STA. MESA) 100 50
2 200
(TAFT AVE) 200 3 30 20 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
Improved Transportation Tableau @ C= P 3,320 04/10/23 12
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NOTE: Test the solution for improvement by computing again the improvement indices of the improved transportation tableau. If all indices are positive then an optimal solution has been obtained, otherwise, repeat the same steps for SSM.
Unused Square
Closed Path Improvement
Indices
X1c +X1c-X1b+X3b-X3c +9-5+3-14= -7
X2a+X2a-X1a+X1b-X3b+X3c-X2b
+10-7+5-3+14-10= +9
X2b +X2b-X2c+X3c-X3b+12-10+14-3=
+13
X3a +X3a-X1a+X1b-X3b +6-7+5-3= +104/10/23 13
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150
(STA. MESA) 100 50
2 200
(TAFT AVE) 200 3 30 20 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
Previously improved transportation tableau04/10/23 14
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150
(STA. MESA) 100 30 20
2 200
(TAFT AVE) 200 3 50 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
New improved Transportation Tableau @ C= P 318004/10/23 15
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Unused Square
Closed Path Improvement
Indices
X2a +X2a-X1a+X1c-X2c+10-7+9-10=
+2
X2b +X2b-X1b+X1c-X2c+12-5+9-10=
+6
X3a +X3a-X1a+X1b-X3b +6-7+5-3= +1
X3c +X3c-X3b+X1b-X1c +14-3+5-9= +7Since all improvement indices are all positive, optimal solution is obtained @ C= P 3,180.
Again, computing for improvement indices:
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This method is an algorithm that finds an initial feasible solution to transportation problem by considering the “penalty cost” of not choosing cheapest available route.
Opportunity Cost- the cost of opportunities that are sacrificed in order to take a specific action.
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1. For each row with an available supply and each column with an unfilled demand, calculate an opportunity/penalty cost by subtracting the smallest cost per unit entry from the second smallest entry for a minimization problem. While, for maximization problem, opportunity cost is calculated by getting the difference between the highest and the second highest entry.
2. Identify the row or column with largest opportunity(difference) cost.
3. Allocate maximum amount possible to the available route with the lowest cost for minimization or highest revenue for maximization in the row/column selected from step 2.
4.Reduce appropriate supply and demand by the amount allocated in step 3.
5.Remove any rows with zero available supply and columns with unfilled demand for further consideration.
6.Return to step 1.
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150(STA. MESA) X1a X1b X1c
2 200 (TAFT AVE) X2a X2b X2c
3 50(DIVISORIA) X3a X3b X3c
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
Original Transportation Tableau
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Row/ Column Second Lowest Cost
Lowest Cost Opportunity Cost
1 7 5 2
2 10 10 0
3 6 3 3
A 7 6 1
B 5 3 2
C 10 9 1
OPPORTUNITY COST FOR THE FIRST ALLOCATION
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150(STA. MESA)
2 200
(TAFT AVE) 3 50 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
First allocation
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OPPORTUNITY COST FOR THE SECOND ALLOCATION
Row/ Column Second Lowest Cost
Lowest Cost Opportunity Cost
1 7 5 2
2 10 10 0
A 10 7 1
B 12 5 7
C 10 9 1
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150(STA. MESA) 30
2 200
(TAFT AVE) 3 50 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
Second allocation
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OPPORTUNITY COST FOR THE THIRD ALLOCATION
Row/ Column Second Lowest Cost
Lowest Cost Opportunity Cost
1 9 7 2
2 10 10 0
A 10 7 1
C 10 9 1
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FromFrom To To A (CEU) B (FEU) C (UE)A (CEU) B (FEU) C (UE) Supply Supply
1 150(STA. MESA) 100 30 20
2 200 200
(TAFT AVE) 3 50 50(DIVISORIA)
DemandDemand 100 100 80 80 220 220 400 400
77 9955
10101010 1212
66 33 1414
Third allocation C= P 3180
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• DEGENERATE PROBLEMS
>> (column + row) – 1 , SSM not applicable
>> put zero (0) to any unused square and proceed with SSM steps
• UNBALANCED PROBLEMS
>> demand ≠ supply>> add dummy column/row
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Otsukaresamadeshita!!!
Thank You for listening!!!
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