Transportation Problem

Content..

• Generic Model• Initial Solutions• VAM• Least Cost Method• Modi Method

Transportation Problem• This problem involves the shipment of a

homogeneous product from a number of supply locations to a number of demand locations.

• Problem: given needs at the demand locations, how should we take limited supply at supply locations and move the goods. Further suppose we wish to minimize cost.

Basic Concept

• Objective:Minimize cost

• Variables:Quantity of goods shipped from eachSupply point to each demand point

• Restrictions:Non negative shipmentsSupply availability at a supply pointDemand need at a demand point

Formulation of Problem• Suppose there are m sources and n destinations• Let ai be the number of supply units available at source i (i=1,2,3 …

m).• Let bj be the number of demand units required at destination j (j=

1, 2, 3…n) • The total number of variables is m x n.• The total number of constraints is m + n.• Let cij represent unit transportation cost for transporting unit from

source i to destination j.• We have determine number of units to be transported from i

source to j destination so that total cost of transportation is minimum.

Formulation of Problem cont…• Let xij be the number of units shipped from source i

to destination j.• Let ai be the row i total supply

• Let bj be the column j total demand Min SScijxij i j

subject to S xij < ai for each source i j

S xij > bj for each destination j i

xij > 0 for all i and j

Balanced Transportation Problem

• The total supply from all the sources equals the total demand in all the destinations.

In case of balanced transportation problem, all constraints are expressed as equations i.e. there is no inequality.

Example 1:

Plant Distribution Centres Supply (million litres)

1 2 3 4

1 2 3 11 7 6

2 1 0 6 1 1

3 5 8 15 9 10Demand 7 5 3 2 17

A dairy firm has three plants and four distribution centers. The daily production capacity and demand at each distribution centre is given in the table below. The cost shipping of one million litres of milk from plant to distribution centre is also given in hundreds of rupees.

Formulation: Step 1

Here xij are the decision variables i.e. the quantities of milk to be shipped from different plants to different distribution centres.

When i=1,2,3 and j= 1, 2, 3,4 and xij ≥ 0

Formulation: Step 2

• Minimize the objective function i.e.

– Minimize Z = 2x11 + 3x12 + 11x13 + 7x14 + x21 + 0x22 + 6x23 + x24 + 5x31 + 8x32 + 15x33 + 9x34

• In general we can write:

Min SScijxij i j

Formulation: Step 3

• Here constraints are the supply of milk is limited. So,

• Here constraints are the demand of milk

Solving Transportation Problem• The transportation problem is solved in two phases:

– Phase I — Obtaining an initial feasible solution– Phase II — Moving toward optimality

• In Phase I, – North-West Corner Rule– The Least-Cost Method– Row Minima Method– Column Minima Method– Vogel’s Approximation Method (VAM)

• In Phase II, – The Stepping Stone Method, – The MODI method ( Modified Distribution Method)

North-West Corner Rule

1. Assign largest possible allocation to the cell in the upper left-hand corner of the tableau

2. Repeat step 1 until all allocations have been assigned3. Stop. Initial tableau is obtained4. The proposed solution is feasible solution since all

demand and supply constraints are fully satisfied.5. Assignment has been done without considering cost,

solution may not be the best or optimal.6. The value of xij can be taken as zero when they are not

allocated in any cell.7. A cell that gets an allocation is called a basic cell.

Allocation of supply using NWC rule

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3 11 7 6

11

0 6 1 1

5 85

153

92

10

71

5 3 2 17

Example 2: Transportation Tableau

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Northwest corner

150

Step 1

Max (150,200)

Step 2

150

50

-- -- 0

150

--

--

-- 0

0

125

Allocation of supply using NWC rule

Row Minima

• Allocate as mush as possible in the lowest cost cell of the first row so that either the capacity of the first plant is exhausted or the requirement at jth distribution center is satisfies or both.

• In case of tie among the cost, select arbitrarily.

Least cost Method

Here, we use the following steps:Steps:

Step 1 Find the cell that has the least costStep 2: Assign as much as allocation to this cellStep 3: Block those cells that cannot be allocatedStep 4: Repeat above steps until all allocation have been

assigned.

• In case tie among the cost, select the cell where allocation of more number of units can be made.

Example 1:Allocation of supply using Least Cost Method

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3 11 7 6

1 01

6 1 1

51

84

153

92

10 9 5 2

71

54

3 2 17

22

Example 2: Allocation of supply using Least Cost Method

Step 1:

The min cost, so allocate as much resource as possible here

200

Step 3

Step 2:

23

Step 3:

200

--

--

0

75

Second iteration, step 4

Example 2: Allocation of supply using Least Cost Method cnt…

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Example 2: Allocation of supply using Least Cost Method cnt…The initial solution

• Stop. The above tableau is an initial tableau because all allocations have been assigned

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Vogel’s approximation methodOperational steps:Step 1: for each column and row, determine its penalty cost by subtracting their two of their least cost Step 2: select row/column that has the highest penalty cost

in step 1Step 3: assign as much as allocation to the

selected row/column that has the least costStep 4: Block those cells that cannot be further allocatedStep 5: Repeat above steps until all allocations have been

assignedNote: In case the two least costs are equal, the penalty should

be taken as zero.

Vogel’s approximation methodIf Then

Case 1 Tie among the highest penalties Select the row /column having minimum cost

Case 2 Tie among minimum cost Select the cell which can have max. allocation

Case 3 Tie among maximum allocation Select cell arbitrarily for allocation

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subtracting their two of their least cost

Step 1

(8-6)

(11-7)

(5-4)

(6-4) (8-5) (11-10)

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Steps 2 & 3

Highest penaltycost

Step 2:

Step 3: this has the least cost

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Step 4

--- ---

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Step 5Second Iteration

--- ---

---

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3rd Iteration of VAM

---

--- ---

---

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Initial tableau for VAM

Some definitions…

• Feasible Solution: A set of non-negative values of xij that satisfy the constraints of transportation problem. The demand and supply constraints are satisfied.

• Basic Feasible Solution: A feasible solution that contains no more than m+n-1 non-negative allocations, satisfy demand and supply constraints, non-negative constraints and constraints are independent and do not form loop.

Some definitions…

• Optimal Solutions: A feasible solution which makes the cost of transportation is minimum. The solution need not to be basic feasible solution.

• Degenerate Basic Feasible Solution: A basic feasible solution that contains less than m+n-1 non-negative allocations.

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Finding Optimal Solution:

Test for Optimality

• After finding initial solution, test for optimality is done.

• The two methods can be used to test optimality:– Stepping Stone Method– Modified distribution Method (MODI Method) or

u-v method.

Stepping-stone method1. Determine an initial basic feasible solution using any three method discussed

above.2. Select a non-basic cell (unoccupied cell) and allocate +1 unit to this cell. 3. Trace a closed loop so that the turning points of the loop is in the allocated cells.4. Calculate net change in the transportation cost. 5. Carry out all above four steps for all unallocated cells.6. Negative value in net change in transportation cost implies that the cost can be

further reduced.7. In case net change is positive or zero, the solution in the question is optimal.8. Select the cell giving highest negative net cost change and determine maximum

number of units that can be assigned to that cell. Also addition and subtraction have to made to other cells in the loop to satisfy the demand and supply constraints.

9. Again go to step 2 and follow subsequent steps till net change in all unoccupied cells is positive or zero.

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Stepping-stone methodLet consider the following initial tableau from the Min Cost algorithm

These are basicvariables

There areNon-basic variables

Question: How can we introducea non-basic variable into basic variable?

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Introducea non-basic variable into basic variables

• Here, we can select any non-basic variable as an entry and then using the “+ and –” steps to form a closed loop as follows:

let consider this nonbasic variable

Then we have

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Stepping stone

+

- +

-

The above saying that, we add min value of all –ve cells into cell that has “+” sign, and subtractsthe same value to the “-ve” cellsThus, max –ve is min (200,25) = 25, and we add 25 to cell A1 and A3, and subtract it from B1 and A3

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Stepping stone

The above tableaus give min cost = 25*6 + 120*10 + 175*11 175*4 + 100* 5 = $4525

We can repeat this process to all possible non-basic cells in that abovetableau until one has the min cost! NOT a Good solution method

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Getting optimal solution using MODI method

• In this method, net cost change is evaluated for all unoccupied cells and only one closed path is traced for the most negative value of net cost change.

• There is considerable time saving over the Stepping –Stone Method.

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Modified distributed method (MODI)

• It is a modified version of stepping stone method• MODI has two important elements:

1. It determines if a tableau is the optimal one2. It tells you which non-basic variable should be firstly

considered as an entry variable3. It makes use of stepping-stone to get its answer of next

iteration

– How it works?

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Procedure (MODI)Step 0: let ui, vj , cij variables represent rows, columns, and cost in the transportation tableau, respectivelyStep 1: (a) form a set of equations that uses to represent all basic variables

ui + vj = cij

(b) solve variables by assign one variable = 0

Step2: (a) form a set of equations use to represent non-basic variable (or empty cell) as such

cij – ui – vj = kij

(b) solve variables by using step 1b informationStep 3: Select the cell that has the most –ve value in 2bStep 4: Use stepping-stone method to allocate resource to cell in step 3Step 5: Repeat the above steps until all cells in 2a has no negative value

Example

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MODIConsider to this initial tableau:

Step 0: let ui, v , cij variables represent rows, columns, and cost in the transportation tableau, respectively

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Step 0

C3A

Step 1: (a) form a set of equations that uses to represent all basic variables

ui + vj = cij

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ui + vj = cij

(b) solve variables by assign one variable = 0

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Set one variable = 0

Because weadded an non-basic variable

Step2: (a & b)

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Step2: (a & b)

Note this may look difficult and complicated, however, we can add theseV=values into the above tableau as well

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Step2: (a & b), alternative

-1

6-0-7

-1 +2

+5

Step 3: Select the cell that has the most –ve value in 2b

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Step3

-1

-1 +2

+5

Select either one, (Why?)These cells mean, introduce it will reduce the min z to -1 cost unit

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Step 4: Use stepping-stone method

+

- +

-

From here we have ….

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Step 4: Use stepping-stone method

Step 5: we repeat steps 1-4 again for the above tableau, we have

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Step 5

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Step 5 cont

All positivesSTOP