Lecture No.21
Held on: Wednesday, October 11, 2000
Prepared by: Manish Ahuja & Gaurav Kumar
(Continued from last lecture for rectilinear case)
let X(1), … , X(m) are the x coordinates of the different plants
Let cost associated per unit distance to serve the facility
i be w(1),…………,w(m)
Cost becomes
C(x)=
At suppose
………………………..(i)
…………………………..(ii)
if (ii) is a strict inequality
Then optimal point is
if (ii) is an equality then
e
to solve the problem, define
Find first so that
Above inequality is iterated for increasing values of t, whenever above inequality is satisfied, i.e. twice the value of summation exceeds the value of C¢, that particular value of t is taken
Euclidian problem
this is analogous to the force balance which can be represented by the figure given below
at point we have optimal point
thus
General Plant Location allocation model
Suppose there are n users (customer locations) and m potential plant sites. There may or may not be an overlap.
Suppose
Ui , i = 1,2,…,m denote the capacity at plant i,
Dj , j = 1,2,…,n denote the demand at user j,
Cij denote the cost of transporting 1 unit from plant i to location j,
Fi denote the cost of constructing a plant at location i,
Xij amount transported from plant i to location j
Define:
Yi = 1 if plant is constructed at location i,
= 0 otherwise
Objective Function
The objective function is
Minimize
Subject to the following constraints:
It is an Integer Linear Programming problem.
Floor Layout: The Quadratic
Assignment Problem
Here we have m facilities, which have to be located in m locations
Let djl denote distance of location j from location l (from center to center)
wik denote cost per unit distance of travel from facility I to facility k
Xij = 1 if facility i is located at location j
0 otherwise
Objective Function
The objective function is less obvious in this case. Hence let us first define a configuration
Suppose that a facility i is located at j
and another facility k is located at l
Now we can write the objective function as follows:
Minimize
Subjected to following constraints
as seen
from the above is a zero – one
variable
Some special cases:
· If a machine is double the size the problem in this case can be solved by first assuming the machine to be split into two machines of equal size and then making the cost between the two machines very large.
· If fewer machines are present while the available space is more then dummy machines can be added with zero costs.
This method becomes quite cumbersome as the value of m increases and it is difficult to solve exactly for large m.
So we restore to heuristics, which finds good sub optimal solution. One such heuristic is
CRAFT : Computer Relative Allocation of Facilities Technique
Here the technique is to look at all pair wise exchanges in locations (i.e. nC2), take the location for which we have least cost. This is repeated for the new allocation.
The method is explained with the help of following example
Suppose we have 4 facilities to be located in 4 locations
The cost matrix is shown below
Facility 1 2 3 4
1 0 50 20 100
2 50 0 30 10
3 20 30 0 70
4 20 10 70 0
The distance matrix is shown below
A B C D
A 0 1 2 3
B 1 0 1 2
C 2 1 0 1
D 3 2 1 0
Let us assume that the facilities are allocated as follows as first step
A is allocated to 1
B is allocated to 3
C is allocated to 4
D is allocated to 2
The cost is =
= 510
First iteration:
Exchange pair 1-2 1-3 1-4 2-3 2-4 3-4
Cost change -100 -50 -80 -60 90 -100
For the exchanges 1-2 and 3-4 the reduction in the cost is same. Breaking ties arbitrarily we
perform the exchange 1-2 and get the following allocation
A is allocated to 2
B is allocated to 3
C is allocated to 4
D is allocated to 1
The associated cost = 410
Second iteration:
Exchange pair 1-2 1-3 1-4 2-3 2-4 3-4
Cost change 100 -40 10 30 100 100
Now we perform the exchange 1-3 for which there is cost reduction
This time no further iteration is required
However this method may not give an optimal solution.