Consider the following example: Adani Power Limited, which is a electric power producing company in India, has three electric power plants that supply the needs of four cities. Each power plant can supply the following numbers of kWhr of electricity: Plant 1 – 35 million; Plant 2 – 50 million; Plant 3 – 40 million.
Problem
Formulation -Example
Consider the following example: Adani Power
Limited, which is a electric power producing company in India, has three
electric power plants that supply the needs of four cities. Each power plant
can supply the following numbers of kWhr of electricity: Plant 1 – 35 million;
Plant 2 – 50 million; Plant 3 – 40 million.
The peak power demands in
these cities, which occur at the same time, are as follows (in kWhr): City 1 –
45 million; City 2 – 20 million; City 3 – 30 million; City 4 – 30 million. The
costs of sending 1 million kWhr of electricity from plant to city depend on the
distance the electricity must travel (see Table).
Formulate
this problem to minimize the cost of meeting each city’s peak power demand.
Solution We begin by defining a variable for each decision
that Adani Power has to make. We define, (for i = 1, 2, 3 and j
= 1, 2, 3, 4) Xij = number of (million) kWhr
produced at plant i and sent to city j Then, the objective of the Adani Power is to
minimize the total transmission cost; this is achieved by allocating
appropriately by linking the plant and city through the lowest cost of
transmission. As we
have discussed in the Mathematical Modelling of Problems, in the Unit-2, we
derive the following objective function; Min Z = 8X11 + 6X12 + 10X13 + 9X14 + 9X21 + 12X22 + 13X23 + 7X24 + 14X31
+ 9X32 + 16X33 + 5X34 This objective can be realized
subject to the following constraints. X11 + X12 + X13 + X14 ≤ 35 | ------ | (Supply
constraint from Plant-1) |
X21 + X22 + X23 + X24 ≤ 50 | ------ | (Supply
constraint from Plant-2) |
X31 + X32 + X33 + X34 ≤ 40 | ------ | (Supply
constraint from Plant-3) |
X11 | + X21 | + X31 ≥ 45 ----- (Demand
constraint at the city-1) |
X12 | + X22 | + X32 ≥ 20 | ----- | (Demand
constraint at the city-2) |
X13 + X23 + X33 ≥ 30 | ----- | (Demand
constraint at the city-3) |
X14 + X24 + X34 ≥ 30 | ----- | (Demand
constraint at the city-4) |
And obviously, Xij ≥ 0 | | |
This is a
special case of LP problem. It can be solved by the simplex algorithm, but
specialized algorithms are much more efficient.
In
general, a transportation problem is specified by the following parameters: 1. A set of ‘m’ supply points from which a good is
shipped. Supply point ‘i’ can
supply, at most, Si units
(in the above situation, m =3, S1
= 35, S2 = 50, S3 = 40) 2. A set of ‘n’ demand points to which the good is
shipped. Demand point ‘j’ must
receive at least dj units of the
shipped goods (you can see above, n = 4; d1 = 45, d2 = 20, d3 = 30, d4 = 30). 3. Each unit
produced at supply point ‘i’ and
shipped to demand point ‘j’ incurs a
variable cost cij. (In the above
case, for example, c12 = 6). Let Xij = number of units shipped from supply point i to demand point j. Thus, a general formulation is:
As we mentioned earlier, if ∑ Si = ∑ dj;
that is supply equal demand, it is said to be balanced transportation problem.
Adani issue is a balanced transportation problem. Thus, for a balanced
transportation problem,
Solution of balanced transportation problem is
simpler, therefore, it is desirable to formulate a transportation problem as a
balanced transportation problem. Thus, a transportation problem is specified by the
supply, the demand, and the shipping costs, so the relevant data can be
summarized in a transportation tableau. The cell in row i and column j
corresponds to the variable Xij.
If Xij is a basic variable, its value is
normally placed in the middle corner of the ijth cell of the tableau. The costs
are normally shown in the upper right corner. The following tableau is for the
Adani Power problem. (The optimal solution values are also given, which we are
going to disucss shortly).
Deriving the solution for a transportation problem
is done in two stages; in the first stage, we try to obtain an initial basic
feasible solution; in the stage-2, we try to check, whether the solution
obtained in stage-1 is optimal or not; if not optimum, we modify the
allocations made [known as MODI Method / u-v method]. There are many methods for finding such a starting
Basic Feasible Solution [BFS]. The easier ones are the northwest-corner method, the
column
minima method, the row-minima method and the matrix minima method (or least
cost method).
Tags : Operations Management - Transportation / Assignment & Inventory Management
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