Introduction to Operations Research

The Transportation Problem Simplex Method Introduction to Operations Research

Transportation Problem Remember our initial transportation problem with the associated cost to send a unit from the factory to each of the four warehouses

This created a nice, large matrix Transportation Problem

There are important properties in the transportation problem that allow us to handle large problems without having to solve a very large B-1. Requirements assumption: • each source has a fixed (not dynamic) set of supplies • each destination has a fix need of supplies Feasible solution property: • if ∑ si = ∑ dj for all i, j (the supplies meets the demands and vice versa) the TP has a feasible solution Cost assumption: • the cost to distribute is linearly determined by the number of units delivered Integer solution property: • if the si and dj are integer values, every CPF solutions will consists of integer values Fundamental assumptions

In the regular simplex, whenever we have constraints with = sign we used artificial variables to jump start an initial solution, then we move on to get rid of them and then go from point to point until we find an optimal solution, by selecting the entering variable (the most negative value) from row 0 of a non basic variable. If we can calculate the initial row (Z) easily, we would not need to follow the whole Gauss Jordan elimination process. Transportation Problem

Using the revised simplex method we obtained the values for the initial row as follows: CbijB-1A - Cij, where Cij is the initial cost value It just happens that this computation is obtained simply from the initial cost values and the values of the current dual variables (restrictions) for both the supply Ui and destination Vj, Z = Vi + Uj - Cij Transportation Problem

For the initial row of Z, we can now concentrate on the equation Cij - Vi - Uj = 0 • To start, we will assign the value of 0 to one of the Uiand solve for the rest of the m+n-1 restrictions. Let u1 = 0 Or choose that Ui that appears more often = 0 • From there we select the entering variable. Since we are under minimization, we will compute Cij - Vi - Ujand select the one with the most negative value non basic variable to enter the basis. Transportation Problem

We use a better format to keep this data in a table fashion. Transportation Problem

Transportation EXAMPLE Rather than using the LP Simplex with 9 constraints and 9 artificial variables, we use the TP tableau

How do I get and initial CPF feasible distribution? • How do I check for optimality? • How do I select the entering variables? • How do I select the leaving variable? Transportation EXAMPLE

Initial cpf Remember that if ∑ si = ∑ dj the TP has a feasible solution. The simplest approach is using the Northwest Corner Method : Start depleting each source by filling the demands from the top left corner destination and move right till the source is finished. Move to the next source to continue filling demands fro left to right and top to bottom until the problem is completed.

Now, compute the objective function by multiplying the Cij’sby the amount delivered by the all basic variable Xij’s Initial cpf This is an initial solution, yet not an optimal. But we got there without the use of artificial variables.

Optimal Solution? Remember Cij - Ui - Vj Determine the dual variables values from the m+n restriction equations Calculate coefficients on row zero for non basic variables If there is at least one where Cij - Ui - Vj < 0 that is the entering variable Let U1 = 0 and do it by hand or use Gauss Jordan elimination C1,1 = u1 + v1 → 16 = 0 + v1 v1 = 16

Anyone of these is a candidate Optimal Solution? where Cij - Ui- Vj< 0 Look for the largest negative value to represent the entering variable

Loop closes nicely ! -1010 +10 -1050 - 10 +10 = 20 Becomes the leaving variable Note there are only 8 basic variables a CPF solution Leaving variable? Chain reaction of events in a loop Take from other cells and compensate properly Only take the smallest quantity in the loop + 10

Now compute the objective function Cpf solution We get a better distribution (minimize the cost)

Optimal Solution (TP1)? Compute new Ui and Vj Calculate new Cij - Ui- Vj Choose entering variable

-1050 - 10 +10 = 20 Leaving variable? Chain reaction of events in a loop Take from other cells and compensate properly Only take the smallest quantity in the loop + 10

Now compute the objective function Cpf solution We get a better distribution (minimize the cost)

Redistribute quantities Compute the objective function Cpf solution We get a better distribution (minimize the cost) $2,590

Optimal Solution (TP6)? Compute new Ui and Vj Calculate new Cij - Ui- Vj Choose entering variable Using this a basic variable will not improve the minimization since the value is already 0 (degenerate)

Transportation problem recap • The process was slow and tedious because we started from a very limited initial CPF, i.e. we used no previous knowledge in the distribution. We will see later better methods to obtain an knowledge-based initial solution • Vogel’s Algorithm • Russell’s Approximation • But before we do that lets recap the TP solving method • Set up the initial tableau from the cost table • Obtain a initial CPF (Northwest, Vogel’s, Russell’s) • Test for optimality • Using Cij - Ui - Vj = 0 for basic variables determine U’s and V’s • For non basic variables compute Cij - Ui - Vj where this is < 0 select that as the entering variable • Find a loop in the tableau to determine the leaving variable • Compensate and return to 3 until there are no more non basic variables where Cij - Ui - Vj < 0

Introduction to Operations Research