Transportation Modeling

Problem

04/10/23 1

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:

04/10/23 2

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

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

04/10/23 5

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

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

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

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

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

04/10/23 11

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

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

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

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

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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04/10/23 18

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.

04/10/23 19

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

04/10/23 26

• 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

Otsukaresamadeshita!!!

Thank You for listening!!!

04/10/23 27