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TRANSPORTATION PROBLEM: FINDING THE OPTIMAL SOLUTION
LECTURE NINE
Lecture Objectives
At the end of the lesson, the student should be able to: Find the Optimal Solution to a Transportation problem Solve and find Optimal Solution to special cases of transportation problems Introduction Transportation problem is essentially a class of allocation problems. It’s used to assign quantities of a single commodity from various sources to certain destinations. The objective is to identify the transportation routes, which result in either minimum cost or maximum benefits.
The general structure of the problem
A product is to be transported from M sources S 1 , S 2 , S 3 …..Sm to N destinations d 1 , d 2 ,d 3 …..dn.
The following quantities must be identified
𝑎𝑖 the quantity available at each source Si. for example, a 2 -S2. Bj the quantity required at each destination dj for example, bj-dj Cij is the unit cost of transportation from every source to every destination.
Let Xij represent the amount transported from every source to every destination. The Xij are referred to as the decision variables. The structure may best be presented inform of a table as follows.
d 1 d 2 d 3 dn supply S 1 X 11 X 12 X 13 X1n S 2 X 21 X 22 X 23 X2n A 2 S 3 X 31 X 32 X 33 X3n A^3 Sn Xm1 Xm2 Xm3 Xmn An Demand B 1 B 2 B 3 Bn
C 11 C 12 C 13
C 21 C 22 C 23
C1n C2n C 31 C 32
Cm 1 Cm2 Cm
C 33 C3n
Cmn
The objective function is denoted as:
min 𝐶 = ∑.
𝑛
𝑖=
𝑚
𝑗=
In the circumstances that the sum of aj is equal to the sum of bj , that is , ∑𝑎𝑖=∑𝑏𝑗 .The transport problem
in said to be balanced otherwise it is unbalanced.
The above linear programming problem will have as many constraints as the total number of sources and destinations.
The ordinary simplex procedure will be cumbersome for use and hence a special simplex process referred to as the transportation process is used.
This procedure only applies if the problem is balanced. If the problem is unbalanced, the deficits have to be included to balance it.
This is done by adding a dummy source or a dummy destination whichever is less.
Given a balanced transportation problem the basic solution will always have, M+N-1 non-zero variables out of the MXN decision variables.
SOLUTION PROCESS
There are two broad steps
I. Finding the initial basic feasible solutions (IBFS). II. Finding the optimal solution (test for optimality)
FIND THE INITIAL BASIC FEASIBLE SOLUTION (IBFS)
There are three method of finding the IBFS. The methods are
a) Northwest corner method b) Least cost method c) Vogel’s Approximation method (VAM)
Demand^50 55
Least cost method
The procedure is very similar to the N.W corner method only that the maximum assignments are made based on the least costs and not the direction
NB: In case of a tie in the minimum cost, the allocation is done arbitrarily.
Vogel’s Application method (VAM)
- Calculate the difference between the lowest and the next highest cost for each row and column. This difference is referred to as a penalty
- Choose the row or column with the largest penalty and make the maximum possible allocation to the cell with the lowest cost
- Omit/exclude from subsequent consideration the satisfied row or column
- Repeat the procedure
A. The unbalanced problem
In case the supply isn’t equal to demand the problem is unbalanced. The simplex process above only applies for a balanced problem. Therefore in case the problem is unbalanced a dummy source or destination is added to balance the problem.
D1 D2 D3 SUPPLY
S1 80
S2 70
DEMAND 70 70 60
The supply is short by 30units. A dummy source is added to take care of the excess so that the costs or benefits to a dummy are all zero.
D 1 D 2 D 3 SUPPLY
S 1 80
S 2 70
Sd 30
DEMAND 70 70 60
B. Degeneracy
In case the number of allocated cells is less than M+N -1 the problem is said to degenerate
Degeneracy may occur at two stages. During the initial allocation if a row and a column are satisfied simultaneously. During the optimization process, in case the minimum losing cells are more than 1.
