TRANSPORTATION PROBLEMS

Transportation problems are generally concerned with the distribution of a certain product from several sources (origin) to numerous facilities (demand points) at minimum cost. Let there be m origins, n destinations

Let xij be the no. of units transported from ith origin to jth destination

cij be the cost of shipping one unit from ith origin to jth destination.

Si and Dj be the supply capacity at origin and Demand at destination respectively.

Then the obj is to minimize the cost i.e.

Minimize z = Σ Σ cij xij

s.t.c

Supply constraints Σ xij = si v i=1,….,m

Demand constraints Σ xij = dj v j= 1,….,n

Assumption is that

Σ si = Σ dj

i=1

m

j=1

n

j=1

n

m

i=1

m

i=1 j=1

n

Generalized transportation table

 dn  d2d1Demand

smxmn  xm2xm1m

       

       

s2x2n  x22x212

s1x1n  x12x111

n  21Supply

Destination

Origin

c11 c12 c1n

c21 c22 c2n

cm1 cm2 cmn

.

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.

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.

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.

..

..

....

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A national truck rental firm, Yeh Haw trucking co., is planning for a heavy demand during the month of june. An inventory of its truck combined with projections for demand indicate that 3 areas will be short of the no. of trucks required to satisfy expected demand.

To prepare for the period of heavy demand, company officials wish to relocate trucks from areas which have surplus to the areas which have shortage at minimum cost. The cost of driving trucks between two cities as well as surplus and shortages are given in the table – next slide.

 80200140Demand (shortage

of trucks)

150   

3

160   

2

110   

1

321 

Supply (surplus of trucks)

Destination (Shortage area)Origin (Surplus

area)

50 100 100

200 300 200

100 200 300

420420

Initial Basic Feasible SolutionIt has (m+n) constraints with (mn) variables

i.e., in this problem we will have 3+3=6 constraints and 3*3=9 variables. But basic variables will be (m+n-1) i.e 6-1=5.

There are no. of ways of finding the initial basic feasible solution.

i) North-west corner ruleii) Minimum-cost method (Greedy algorithm)iii) Vogel’s approximation methodWe will apply each one of these for the

problem mentioned above.

• Northwest corner rule

 80200140Demand (shortage

of trucks)

150 80 70 

3

160  130 30

2

110   110

1

321 

Supply (surplus of trucks)

Destination (Shortage area)Origin (Surplus

area)

50

200

100

100

300

200

100

200

300

Total cost = 110*50+30*200+130*300+70*200+80*300 = rs.88,500

• Minimum-cost method (greedy algorithm)

 80200140Demand (shortage

of trucks)

150 120 30

3

160 80 80 

2

110  110

1

321 

Supply (surplus

of trucks)

Destination (Shortage area)Origin (Surplus

area)

50

200

100

100

300

200

100

200

300

Total cost = 110*50+30*100+80*300+120*200+80*200 = rs.72,500

• Vogel’s Approximation Method

 80200140

Demand (shortage of trucks)

150 101403

16080 80 2

110 1101

321 

Supply (surplus of

trucks)

Destination (Shortage area)

Origin (Surplus

area)

50 100 100

200 300 200

100 200 300

50

0

100

Penalty

50 100 100

0

100

100

100 100

100

100

100 100

1) Find the intial basic feasible solution usinga) North-west corner methodb) Row minima method; c) Least-cost Methodd) Column minima method e) VAM – Vogel's Approximation Method

2) Determine an initial basic feasible solution to the following transportation problem using a) North-west corner rule; b) Least cost method; c) Vogel's Approximation method VAM

DestinationI II III IV Supply

A11 13 17 14 250

B16 18 14 10 300

C21 24 13 10 400

Demand 200 225 275 250

Total cost = 110*100+90*300+10*200+80*200+140*100

= rs.70,000Check for no. of basic variables which in this

case have to be 3+3-1=5. As we can see from the table that 5 cells have been filled meaning we have 5 basic variables.

From here we move on to finding the optimum solution. We have mainly two methods to arrive at optimum solution:

1. Stepping-stone method2. MODI method

• Stepping-stone method

 80200140Demand (shortage

of trucks)

150 80 70 3

160  130 302

110   1101

321 

Supply (surplus of trucks)

Destination (Shortage area)Origin (Surplus

area)

50 100 100

200 300 200

100 200 300

+1

+1

-1

-1

Cell closed path Improvement index (Iij)1,2 1,2 – 1,1 – 2,1 – 2,2 – 1,2 -501,3 1,3 – 1,1 – 2,1 – 2,2 – 3,2 – 3,3 -1502,3 2,3 – 2,2 – 3,2 – 3,3 – 2,3 -2003,1 3,1 – 3,2 – 2,2 – 2,1 – 3,1 0

 80200140Demand (shortage

of trucks)

150  150 3

160 80 50 302

110   1101

321 

Supply (surplus of trucks)

Destination (Shortage area)Origin (Surplus

area)

50

200

100

100

300

200

100

200

300

cell closed path Iij

1,2 1,2 – 1,1 – 2,1 – 2,2 – 1,2 -50 1,3 1,3 – 1,1 – 2,1 – 2,3 – 1,3 50 3,1 3,1 – 3,2 – 2,2 – 2,1 – 3,1 0 3,3 3,3 – 2,3 – 2,2 – 3,2 – 3,3 200

 80200140Demand (shortage

of trucks)

150  150 3

160 80  802

110  50 601

321 

Supply (surplus of trucks)

Destination (Shortage area)Origin (Surplus

area)

50

200

100

100

300

200

100

200

300

Cell closed path Iij1,3 1,3 – 1,1 – 2,2 – 2,3 – 1,3 502,2 2,2 – 1,2 – 1,1 – 2,1 – 2,2 503,1 3,1 – 3,2 – 1,2 – 1,1 – 3,1 -503,3 3,3 – 2,3 – 2,1 – 1,1 – 1,2 – 3,2 – 3,3 150

 80200140Demand (shortage

of trucks)

150  90 603

160 80  802

110  110 1

321 

Supply (surplus of trucks)

Destination (Shortage area)Origin (Surplus

area)

50

200

100

100

300

200

100

200

300

Cell closed path Iij

1,3 1,3 – 1,2 – 2,2 – 2,3 – 1,3 1001,1 1,1 – 1,3 – 3,2 – 1,2 – 1,1 502,2 2,2 – 2,1 – 3,1 – 3,2 – 2,2 03,3 3,3 – 2,3 – 2,1 – 3,1 – 3,3 200Since all the improvement index values are >= 0Total cost = 110*100+80*200+60*100+90*200+80*200 = rs.67,000

MODI (Modified Distribution) method

Dual problem of the transportation problem is given by:

Maximize Z = Σ si ui + Σ dj vj

s.t.c ui + vj <= cij

ui, vj – unrestricted

If xij is nonbasic variable (xij=0), its corresponding dual slack variable is a basic variable and >0(unequal to 0) in the dual complimentary basic solution.

i=1

m

j=1

n

• If we use the symbol Iij for this dual slack variable and add it to the dual constraint we get Iij = cij – ui – vj

The value of Iij – dual slack variable is the marginal effect on the value of the primal objective function from introducing an additional unit of nonbasic xij.

ui are row indexes and vj are column indexes and the values are filled by assuming u1 = 0 and we know that

ui + vj = cij. So we will make use of cost coeff to get other values.

 80200140

Demand (shortage of trucks)

150 101403

16080 80 2

110 1101

321 

Supply (surplus of

trucks)

Destination (Shortage area)

Origin (Surplus

area)

50

200

100

100

300

200

100

200

300

u1 = 0

u2 = 200

V2 = 100v1 = 0 v3 = 0

u3 = 100

50 100

0

200

Since all the Iij values are +ve, it means that the optimal solution has been reached

• A CPU mfger wants to ship the products to warehouses in 5 different cities A,B,C,D&E from plants I, II, III. The quantity at plant, requirements at warehouses and shipping costs/unit shipped are given in the table.

Determine the amt that should be shipped from each plant to each warehouse to minimize the shipping costs.

cont…

2100030006000400050003000Demand

8000     III

4000     II

9000     I

supplyEDCBAPlant

 warehouses 

10

1 20 7 10 4

2

20

10

5

8

9

30 6

10

Assignment Model

• Five jobs are assigned to five people; each person will do one job only. The expected times (in hrs) required for each person to complete each job have been estimated & are shown in the following table. Use the Hungarian method to determine the optimal soln.

cont…

Person

Job 1 2 3 4 5

1 12 15 13 14 15

2 16 18 15 14 16

3 18 18 15 18 20

4 15 20 18 17 19

5 26 15 18 14 15

• A distributor has four sales territories which have to be assigned to four sales person. From the past sales experience the firm’s sales manager has estimated the annual sales vol (‘000s) for each sales representative in each sales territory. Find the territory assignments that maximize sales.

sales territorySales rep A B C DWashington 44 80 52 60Benson 60 56 40 72Fredricks 36 60 48 48Hodson 52 76 36 40

• A company produces certain product from three plants 1,2,3 which are distributed to distribution centres A,B,C,D,E. Capacities, Demand & cost of shipping/unit from various plants to destinations are given in table. Find the shipment plan to minimize costs

400400400Demand

3003

6002

4001

SupplyCBA

1.45 1.60

1.20

0.602.251.10

1.40

1.20 1.80

13001200

• A company produces certain product from three plants 1,2,3 and shipped to three warehouses A,B,C. Transportation costs/unit are shown in the table below. Solve the model to determine the minimum cost solution.

300400200Demand

1003

5002

3001

SupplyCBA

20 16

12

81010

24

18 10

900900

Premier consulting has three consultants C1, C2, C3 who have to handle 4 different clients ClientA, ClientB, ClientC, ClientD. The time availability with consultants and the time required by each client is given in the table. Hourly rates vary for the consultant-client combination and are based on several factors

The rates/hr for each consultant-client combo is given in the table. Find the optimal solution providing the hours each consultant should be scheduled for each client in order to maximize the firm’s billing. What is the schedule and what is the billing?

85

Client D

10075180Hrs required

140C3

160C2

160C1

Available hrs

Client CClient BClient A

460

440

100

120

155

135

125

150 140

115

115 100

120

130

Degenerate Problems• Solve the following transportation problem

• Goods have to be transported from S1, S2, S3 to destinations D1, D2, D3. The transportation costs per unit, capacities of the sources and demand are shown in the table. Determine a transportation schedule to minimize the cost.

D1 D2 D3 SupplyS1 8 5 6 120S2 15 10 12 80S3 3 9 10 80

Demand 150 80 50

• A product is manufactured by 4 factories. Unit production cost in them are Rs.2, Rs.3, Re.1 & Rs.5 respectively. Their production capacities and the demand from 4 stores and the transportation cost from each factory to each store is shown in the table below. Allocate to minimize the total cost.

Stores1 2 3 4 Supply

Factories

A 2 4 6 11 50B 10 8 7 5 70C 13 3 9 12 30D 4 6 8 3 50

Demand 25 35 105 20

Unbalanced Transportation Problems

• A company has factories at A,B,C & D which supply to warehouses at P,Q,R & S. The factory capacities are 230, 280, 180 resp. for regular production. If overtime production is utilized, the capacities can be increased to 300, 360 & 190 respectively. Increment unit costs are Rs.5, Rs.4 and Rs.6 resp. The current warehouse requirements are 165, 175, 205, 165 resp.Unit shipping costs in Rs. is shown in the table. Find the optimum distribution to minimize the cost.

P Q R SA 6 7 8 10B 4 10 7 6C 3 22 2 11

Transshipment

1Pune

2Nagpur

3Bhopal

4Bangalore

5Delhi

6Calcutta

7Hyderabad

8Chennai

2

3

3

1

5

2

6

6

3

6

4

4

200

150

350

300

600

400