Transportation Model –

Formulations

The Transportation Model

A special class of LPPs that deals with transporting(=shipping) a commodity from sources (e.g. factories) to destinations (e.g. warehouses).

The objective is to determine the shipping schedule that minimizes the total shipping cost while satisfying supply and demand limits.

Assume that the shipping cost is proportional to the number of units shipped on a given route.

Transportation Problem

DesMoines

(100 unit

capacity)

Fort Lauderdale

(300 units capacity)

Cleveland

(200 units Demand)

Evansville

(300 units

capacity)

Albuquerque

(300 units

Demand)

Boston

(200 units

Demand)

Consider:

m sources 1,2, …, m and n destinations 1, 2, …, n.

cij the cost of shipping one unit from Source i to Destination j

Assume that the availability at source i is ai (i=1, 2, …, m) and the demand at the destination j is bj (j=1, 2, …, n). We make an important assumption: the problem is a balanced one. That is

n

j

j

m

i

i ba11

That is, total availability equals total demand.

a dummy source (if the total demand is more than the total supply) or a dummy destination (if the total supply is more than the total demand).

Let xij be the amount of commodity to be shipped from the source i to the destination j.

Thus the problem becomes the LPP

n

j

ijij

m

i

xcz11

Minimize

subject to

),...,2,1(

),...,2,1(

1

1

njbx

miax

j

m

i

ij

i

n

j

ij

0ijx

• Define problem

• Set up transportation table (matrix)

– Summarizes all data

• Develop initial solution

– Northwest corner/ Min Cost/ Vogel’s Methods

• Find optimal solution

– Transportation Simplex Method

Transportation Problem Solution Steps

Advantages of the transportation problem

• Advantages of the transportation problem and the minimum cost flow problem

– Integer solutions

– Very fast solution methods

– Extremely common in modeling

Transportation Table

Destination

Source Supply

Demand

1

2

:

m

a 1

a 2

:

a m

1 2 . . n

b 1 b 2 b n

Quantity demanded or

required

Example 5.1-1

MG Auto has three plants in Los Angeles, Detroit, and New Orleans, and two major distribution centers in Denver and Miami. The capacities of the three plants during the next quarter are 1000, 1500 and 1200 cars. The quarterly demands at the two distribution centers are 2300 and 1400 cars. The transportation cost per car from Los Angeles to Denver and Miami are $80 and $215 respectively. The corresponding figures from Detroit and New Orleans are 100, 108 and 102, 68 respectively.

mn decision variables xij and m+n constraints.

Since the sum of the first m constraints equals the sum of the last n constraints (because the problem is a balanced one), one of the constraints is redundant and we can show that the other m+n-1 constraints are LI. Thus any BFS will have only m+n-1 nonzero variables.

Though we can solve the above LPP by Simplex method, we solve it by a special algorithm called the transportation algorithm.

Example 5.1-2

MG Auto has three plants in Los Angeles, Detroit, and New Orleans, and two major distribution centers in Denver and Miami. The capacities of the three plants during the next quarter are 1000, 1300 and 1200 cars. The quarterly demands at the two distribution centers are 2300 and 1400 cars. The transportation cost per car from Los Angeles to Denver and Miami are $80 and $215 respectively. The corresponding figures from Detroit and New Orleans are 100, 108 and 102, 68 respectively.

Formulate the transportation Model.

the total demand = 3700 > 3500 (Total supply) dummy supply 3700-3500=200 units to make the problem a balanced one.

If a destination receives u units from the dummy source, it means that that destination gets u units less than what it demanded. We usually put the cost per unit of transporting from a dummy source as zero (unless some restrictions are there). Thus we get the transportation tableau

80 215 1000

100 108 1300

102 68 1200

0 0 200

2300 1400

Source

Destination

Denver Miami Supply

Demand

Los Angeles

Detroit

New Orleans

Dummy

We write inside the (i,j) cell the amount to be shipped from source i to destination j. A blank inside a cell indicates no amount was shipped.

Problem 5 Problem Set 5.1A Page 196

In the previous problem, penalty costs are levied at the rate of $200 and $300 for each undelivered car at Denver and Miami respectively. Additionally no deliveries are made from the Los Angeles plant to the Miami distribution center. Set up the transportation model.

The above imply that the "cost" of transporting a car from the dummy source to Denver and Miami are respectively 200 and 300. The second condition means we put a "high" transportation cost from Los Angeles to Miami. We thus get the tableau

80 M 1000

100 108 1300

102 68 1200

200 300 200

2300 1400

Source

Destination

Denver Miami Supply

Demand

Los Angeles

Detroit

New Orleans

Dummy

Note: M indicates a very "big" positive number. In TORA it is denoted by "infinity".

Problem 8 Problem Set 5.1A Page 196

Three refineries with daily capacities of 6,5, and 8 million gallons, respectively, supply three distribution areas with daily demands of 4,8, and 7 million gallons, respectively.Gasoline is distributed to the three distribution areas through a network of pipelines. The transportation cost is 10 cents per 1000 gallons per pipeline mile. The table below gives the mileage between the refineries and the distribution areas. Refinery 1 is not connected to the distribution area 3.

Distribution Area 1 2 3

Refinery

1 120 180 -

2 300 100 80

3 200 250 120

Construct the associated transportation model.

(Solution in the next slide)

12 18 M 6

30 10 8 5

20 25 12 8

4 8 7

Source

Destination

1 2 3 Supply

1

2

3

Demand

The problem is a balanced one. M indicates a very "big" positive number.

Refinery

Distribution Area

The total cost will be 1000* ij

i j

ij xc

3

1

3

1

Problem 10 Problem Set 5.1A Page 197

In the previous problem, suppose that the daily demand at area 3 drops to 4 million gallons. Surplus production at refineries 1 and 2 is diverted to other distribution areas by truck. The transportation cost per 100 gallons is $1.50 from refinery 1 and $2.20 from refinery 2. Refinery 3 can divert its surplus production to other chemical processes within the plant.

Formulate the problem as a transportation model.

We introduce a dummy destination. Solution follows.

12 18 M 15 6

30 10 8 22 5

20 25 12 0 8

4 8 4 3

Source

Destination

1 2 3 Dummy Supply 1 2 3

Demand

M indicates a very "big" positive number.

Refinery

Distribution Area

The total cost will be 1000* ij

i j

ij xc

3

1

3

1

• Define problem

• Set up transportation table (matrix)

– Summarizes all data

• Develop initial solution

– Northwest corner/ Min Cost/ Vogel’s Methods

• Find optimal solution

– Transportation Simplex Method

Transportation Problem Solution Steps

Determination of Starting Basic Feasible Solution

Determination of the starting Solution

There are three methods to determine a starting

BFS.

As mentioned earlier, any BFS will have only m+n-

1 basic variables (which may assume non-zero

=positive values) and the remaining variables will

all be non-basic and so have zero values.

In any transportation tableau, we only indicate the

values of basic variables. The cells corresponding

to non-basic variables will be blank.

Degenerate BFS

If in a cell we find a zero mentioned, it means that

that cell corresponds to a basic variable which

assumes a value of zero. In simplex language, we

say that we have a degenerate BFS.

NORTH-WEST Corner Method

The method starts at the north-west corner cell

(i.e. cell (1,1)).

Step 1. We allocate as much as possible to the

selected cell and adjust the associated amounts of

supply and demand by subtracting the allocated

amount.

Step 2. Cross out the row (column) with zero

supply (zero demand) to indicate that no further

assignments can be made to that row(column).

If both a row and a column are simultaneously

satisfied then

if exactly one row or column is left uncrossed

make the obvious allocations and stop. Else cross

out one only (either the row or the column) and

leave a zero supply(demand) in the uncrossed out

row(column).

Step 3. If no further allocation is to be made,

stop. Else move to the cell to the right (if a

column has just been crossed out) or to the cell

below if a row has just been crossed out. Go to

Step 1.

Consider the transportation tableau:

3 7 6 4 5

2 4 3 2 2

4 3 8 5 3

3 3 2 2

Destination

1 2 3 4 Supply

Source

1

2

3

Demand

3 2

2

1

1 1 1

1

1 2 2

Total shipping cost = 48

2. LEAST COST Method

In this method we start assigning as much as possible

to the cell with the least unit transportation cost (ties

are broken arbitrarily) and the associated amounts of

supply and demand are adjusted by subtracting the

allocated amount.

Cross out the row (column) with zero supply

(zero demand) to indicate that no further

assignments can be made to that row(column).

If both a row and a column are simultaneously

satisfied then

if exactly one row or column is left uncrossed

make the obvious allocations and stop. Else cross

out one only (either the row or the column) and

leave a zero supply(demand) in the uncrossed out

row(column).

Next look for the uncrossed out cell with the

smallest unit cost and repeat the process until no

further allocations are to be made.

Consider the transportation tableau:

3 7 6 4 5

2 4 3 2 2

4 3 8 5 3

3 3 2 2

Destination

1 2 3 4 Supply

Source

1

2

3

Demand

Total shipping cost = 36

2

1

1 4

3

0

2 2 0 2

3. Vogel’s approximation method (VAM)

Step 1. For each row (column) remaining under

consideration, determine a penalty by subtracting the

smallest unit cost in the row (column) from the next

smallest unit cost in the same row(column). ( If two unit

costs tie for being the smallest unit cost, then the

penalty is 0).

Step2. Identify the row or column with the largest

penalty. Break ties arbitrarily. Allocate as much as

possible to the cell with the least unit cost in the selected

row or column.(Again break the ties arbitrarily.) Adjust

the supply and demand and cross out the satisfied row or

column.

If both a row and a column are simultaneously

satisfied then

if exactly one row or column is left uncrossed

make the obvious allocations and stop. Else cross

out one only (either the row or the column) and

leave a zero supply(demand) in the uncrossed out

row(column). (But omit that row or column for

calculating future penalties).

Step 3. If all allocations are made, stop. Else go to

step 1.

3 7 6 4 5

2 4 3 2 2

4 3 8 5 3

3 3 2 2

Destination 1 2 3 4 Supply Row Penalties

S

o

u

r

c

e

1

2

3

Demand

Total shipping cost = 32

Column

Penalties

1

0

1

1 1 3 2

2

0

1

-

1

1 4 - 1

3

0

3 0 0 2

Iterative Computations

of the Transportation

Algorithm

Iterative computations of the Transportation

algorithm

After determining the starting BFS by any one of

the three methods discussed earlier

Step1: Use the Simplex optimality condition to

determine the entering variable as a current non-

basic variable that can improve the solution.

If the optimality condition is satisfied by all non-

basic variables, the current solution is optimal

and we stop. Otherwise we go to Step 2.

Step 2. Determine the leaving variable using

the Simplex feasibility condition. Change the

basis and go to Step 1.

The determination of the entering variable from

among the current non-basic variables is done

by the method of multipliers.

In the method of multipliers, we associate

with each row a dual variable (also called a

multiplier) ui and with each column we

associate a dual variable (also called a

multiplier) vj.

Note that each row corresponds to a

constraint and each column corresponds to a

constraint we recall from duality theory that

At any simplex iteration ,

Primal z-equation Left hand side Right hand side

coefficient of = of corresponding - of corresponding

variable xj dual constraint dual constraint

That is ijjiijij cvucz ""

(Verify this by taking m=3 and n=4 !)

Since there are m+n-1 basic variables and

since 0 ijij cz

ijji cvu

for all such basic variables, we have m+n-1

equations

to determine the m+n variables ji vu ,

We arbitrarily choose one of them and equate

to zero and determine the remaining m+n-1 of

them. Then we calculate ijjiijij cvucz

for all non-basic variables xij. Then the entering

variable is that one for which ijji cvu

is most positive.

We do all this on the transportation tableau itself

(and NOT separately) as the following example

shows.

3 7 6 4 5

2 4 3 2 2

4 3 8 5 3

3 3 2 2

Destination

Supply

S

o

u

r

c

e

Demand

3 2

1 1

1 2

Thus x32 enters the basis.

u1=0

v1=3 v2=7

u2= -3

v3=6

u3= 2

v4=3

Total Cost =48 Starting

Tableau

0 -1

-2 -2

1 6

Determining the leaving variable

• Construct a closed loop that starts and ends at

the entering variable cell.

•The loop consists of connected horizontal and vertical

segments only (no diagonals are allowed).

•Except for the entering variable cell, each vertex (or corner)

of the closed loop must correspond to a basic variable cell.

• The loop can cross itself and bypass one or more basic

variables.

• The amount to be allocated to the entering

variable cell is such that it satisfies all the demand

and supply restrictions and must be non-negative.

Usually

is the minimum of the amounts allocated to the

basic cells adjacent to the entering variable cell.

Having decided about the amount to be

allocated to the entering cell, for the supply and

demand limits to remain satisfied, we must

alternate between subtracting and adding the

amount at the successive corners of the loop.

In this process one of the basic variables will

drop to zero. In simplex language, we say it

leaves the basis.

We repeat this process till optimality is reached.

3 7 6 4 5

2 4 3 2 2

4 3 8 5 3

3 3 2 2

Destination

Supply

S

o

u

r

c

e

Demand

3 2

1 1

1 2

Thus x32 enters the basis.

u1=0

v1=3 v2=7

u2= -3

v3=6

u3= 2

v4=3

Total Cost =48 Starting Tableau

0 -1

-2 -2

1 6

Thus will become 1 and in the process both

the basic variables x22 and x33 will become

simultaneously zero. Since only one of them

should leave the basis we make x22 leave the

basis and keep x33 in the basis but with value

zero.

Thus the transportation cost reduces by 6 (as

x23 increases by 1) and we say one iteration is

over. The resulting new tableau is on the next

slide.

3 7 6 4 5

2 4 3 2 2

4 3 8 5 3

3 3 2 2

Destination

Supply

S

o

u

r

c

e

Demand

3 2

2

0 2

Thus x13 enters the basis.

u1=0

v1=3 v2=7

u2= -9

v3=12

u3= -4

v4=9

Total Cost =42 Start of Iteration 2

6 5

-8 -2

-5

-6

1

Thus will become 0 and x32 leaves the basis.

Again the BFS is degenerate . But the

transportation cost remains the same and we

say the second iteration is over. The resulting

new tableau is on the next slide.

3 7 6 4 5

2 4 3 2 2

4 3 8 5 3

3 3 2 2

Destination

Supply

S

o

u

r

c

e

Demand

3 2

2

2

Thus x14 enters the basis.

u1=0

v1=3 v2=7

u2= -3

v3=6

u3= -4

v4=9

Total Cost =42 Start of Iteration 3

-6

5

-2 4

-5

0

1

0

Sometimes it might be difficult to find the closed

loop from the entering cell by inspection. In that

case the following method can be used to find the

closed loop. We sketch the flowchart of the

sequence in which the variables ui and vj were

determined. For example in the above case the

flowchart is on the next slide. Now to find the loop

emanating from the non-basic cell (1,4), join u1 and

v4 by a dotted line (as shown). Then the closed

loop is:

(1,4) (1,2) (3,2) (3,4) (1,4)

u1=0

v1=3

v2=7

v3=6 u2= -9

u3= -

4 v4= 9

Thus the closed loop is

(1,4) (1,2) (2,3) (3,4)

(1,4)

Thus will become 2 and in the process both

the basic variables x12 and x32 will become

simultaneously zero.

Since only one of them should leave the basis

we make x32 leave the basis and keep x12 in

the basis but with value zero.

x32 becomes 3. Thus the transportation cost

reduces by 1*2+4*2=10 and we say third

iteration is over. The resulting new tableau is

on the next slide.

3 7 6 4 5

2 4 3 2 2

4 3 8 5 3

3 3 2 2

Destination

Supply

S

o

u

r

c

e

Demand

3 0

2

2

Thus this is the optimal tableau. Alt Opt solutions exist.

u1=0

v1=3 v2=7

u2= -

3

v3=6

u3= -4

v4=4

Total Cost =32 Start of Iteration 4

-6 -5

-2 -1

-5

0

3

0

Problem 4 Problem Set 5.3B Page 217

In the unbalanced transportation problem given

in the table below, if a unit from a source is not

shipped out (to any of the destinations) a

storage cost is incurred at the rate of $5, $4,

and $3 per unit for sources 1,2, and 3

respectively. Additionally all the supply at

source 2 must be shipped out completely to

make room for a new product. Use VAM to

determine the starting solution and determine

the optimum solution.

1 2 3

1 1 2 1

2 3 4 5

3 2 3 3

To balance the problem, we introduce a

dummy destination with transportation costs

20

40

30

30 20 20

$5, $M, $3 respectively.

(Solution in the next slide)

1

1 2 1 5 20

3 4 5 M 40

2 3 3 3 30

30 20 20 20

Destination 1 2 3 Dummy Supply Row Penalties

S

o

u

r

c

e

1

2

3

Demand

Total shipping cost = 240

Column

Penaltie

s

0

1

1

1 1 2 2

10

0

-

1

1 1 - M-3

20

30

10

20

0

10

10

-

1

1

1 1 - -

10

-

1

0

- 1 - -

1 2 1 5 20

3 4 5 M 40

2 3 3 3 30

30 20 20 20

Destination

Supply

S

o

u

r

c

e

Demand

3

0

10

20 0

Thus this is the optimal tableau. Alt Opt solutions exist.

u1=-2

v1=2 v2=3

u2= 1

v3=3

u3= 0

v4=3

Total Cost =240 Starting Tableau

-1

-4 -1

4-M

0

-1

10

20