Math Models of OR: The Transportation Problemeaton.math.rpi.edu/.../15B_transportationbeamer.pdfThe...

Math Models of OR:The Transportation Problem

John E. Mitchell

Department of Mathematical SciencesRPI, Troy, NY 12180 USA

October 2018

Mitchell The Transportation Problem 1 / 20

The Transportation Problem

Outline

1 The Transportation Problem

2 The transportation tableau

3 Finding an initial basic feasible solution

The transportation problem

Network flow problems are defined on graphs, with material flowingalong edges from node to node. A transportation problem is a networkflow problem with a special structure:

There is a collection of m sources, with source i having supply aiavailable.There is a collection of n destinations, with destination j havingdemand bj .The only edges connect a source directly to a destination.Each edge from source i to destination j has a per-unit shippingcost cij .The total supply is equal to the total demand:

∑mi=1 ai =

∑nj=1 bj .

(We will discuss relaxing this assumption later.)

∑mi=1 ai =

∑nj=1 bj .

∑mi=1 ai =

∑nj=1 bj .

∑mi=1 ai =

∑nj=1 bj .

∑mi=1 ai =

∑nj=1 bj .

An example

d1c11 = 2

c12 = 4

c13 = 3

c21 = 1

c22 = 5

c23 = 2

c31 = 1

c32 = 1

c33 = 620

supply sources destinations demand

Linear optimization problem representation

A transportation problem can be written as a linear optimizationproblem:

minx∈IRm×n∑m

i=1∑n

j=1 cijxij

subject to∑n

j=1 xij = ai i = 1, . . . ,m (supply constraints)∑mi=1 xij = bj j = 1, . . . ,n (demand constraints)

xij ≥ 0 i = 1, . . . ,m, j = 1, . . . ,n

This is in standard form, so we could use the simplex algorithm tosolve it.

However, the problem can be large, with the number of variables equalto the product mn.

Tableau for our example problem

x11 x12 x13 x21 x22 x23 x31 x32 x330 2 4 3 1 5 2 1 1 6

20 1 1 1 0 0 0 0 0 020 0 0 0 1 1 1 0 0 020 0 0 0 0 0 0 1 1 110 1 0 0 1 0 0 1 0 030 0 1 0 0 1 0 0 1 020 0 0 1 0 0 1 0 0 1

Notice that each row has exactly two “ones”, with all other entriesequal to zero. Notice also that the sum of the first three rows isidentical to the sum of the last three rows, so the constraints arelinearly dependent. This means one of the constraints could bediscarded: if we have the right amount of material leaving all thesources and the right amount going in to all but one of the destinations,then we must have the right amount entering the final destination.

The constraints are linearly dependent

Algebraically, we can find a nontrivial linear combination of theconstraints that gives the trivial constraint 0 = 0, showing theconstraints are linearly dependent:

We have∑n

j=1 xij = ai for i = 1, . . . ,mand

∑mi=1 xij = bj for j = 1, . . . ,n

=⇒∑m

(∑nj=1 xij

)−∑n

j=1(∑m

i=1 xij)=∑m

i=1 ai −∑n

j=1 bj

=⇒ 0 = 0 by assumption that total supplyand total demand are equal

The transportation tableau

Outline

Simplex, but written differently

Since a transportation problem is a linear optimization problem, we aregoing to solve it using the simplex algorithm.

It won’t look like the simplex algorithm.

We exploit the structure of the problem to use a more compact type oftableau, and to calculate the pivots more efficiently.

The basic feasible solutions have a special structure in the network.

Find a feasible solution

First, we exploit the structure of the problem to find a feasible solution.

We ship from the first source with available supply to the firstdestination that has unmet demand.

We update the available supply and unmet demand and repeat, untilall material has been shipped.

The feasible solution

d1x11 = 10

d1x11 = 10x12 = 10

0 10 20

Finding the feasible solution (continued)

d1x11 = 10x12 = 10

0 10 20

d1x11 = 10x12 = 10x22 = 20

0 10 20

30 20 0

d1x11 = 10x12 = 10x22 = 20

x33 = 200 20

0 10 20

30 20 0

Transportation tableauThe transportation tableau has one row for each source and onecolumn for each destination.

The cost cij is placed in cell (i , j); these entries are updated to thereduced costs as the algorithm proceeds.

The flows xij are stored as superscripts in the corresponding cells.

Thus, the solution we found can be stored in the followingtransportation tableau:

supply

demandb1 = 10 b2 = 30 b3 = 20

a1 = 20 210 410 3a2 = 20 1 520 2a3 = 20 1 1 620

Finding an initial basic feasible solution

Outline

Basic feasible solutionsThe simplex algorithm works with basic feasible solutions. The solutionwe found earlier is a feasible solution, but is it a basic feasiblesolution? Not quite, but it can be modified easily to give a basicfeasible solution.

DefinitionA basic feasible solution for a transportation problem with m sourcesand n destinations satisfies the following properties:

it is feasible.it contains exactly m + n − 1 basic variables.the basic variables correspond to a spanning tree in the network;that is, for any two nodes u and v in the network, there is exactlyone path from u to v that traverses only the arcs corresponding tobasic variables, perhaps traversing some arcs in the reversedirection.

Getting a basic feasible solution

The feasible solution we found earlier has positive values of xij for 4edges.

d1x11 = 10

x12 = 10

x22 = 20

x33 = 2020

We have m + n − 1 = 3 + 3− 1 = 5. So we don’t have a basic feasiblesolution.

Also, there is no path from s1 to d3 (for example) that only uses thearcs with positive xij .

Adding degenerate basic variablesThis can be remedied by declaring x23 a basic variable, with flowx23 = 0.

d1x11 = 10

x12 = 10

x22 = 20

x23 = 0

x33 = 2020

We now have a (unique) path from s1 to d3, namelys1 → d2 → s2 → d3.

The Northwest Corner Rule

Basic feasible solutions can be constructed directly using a slightmodification of the earlier approach. This is known as the NorthwestCorner Rule:

1 If there are fewer than m + n− 1 basic variables then pick the costentry in the upper left (northwest) corner of the tableau and shipas much as possible along that arc so that supply is used up ordemand is used up.

2 If the assignment just made uses up the demand for the columnthen eliminate that column from further consideration and return toStep 1.

3 If the assignment just made uses up the supply for the row theneliminate that row from further consideration and return to Step 1.

If use up supply and meet demand simultaneously

Note that if an assignment uses up both the supply and the demandthen we only eliminate the column.

This means that on the next iteration the top-most source will haveavailable supply of zero, so the next arc added to the set of basicvariables will have xij = 0.

The NW corner rule for the example

10 30 2020 2 4 320 1 5 220 1 1 6

10 30 2020 210 4 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 520 220 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 620

10 30 2020 2 4 320 1 5 220 1 1 6

10 30 2020 210 4 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 520 220 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 620

10 30 2020 2 4 320 1 5 220 1 1 6

10 30 2020 210 4 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 520 220 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 620

10 30 2020 2 4 320 1 5 220 1 1 6

10 30 2020 210 4 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 520 220 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 620

10 30 2020 2 4 320 1 5 220 1 1 6

10 30 2020 210 4 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 520 220 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 620

10 30 2020 2 4 320 1 5 220 1 1 6

10 30 2020 210 4 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 5 220 1 1 6

10 30 2020 210 410 320 1 520 220 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 6

10 30 2020 210 410 320 1 520 20

20 1 1 620

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