Material Constraint Theory Part VI

III. ALLOCATION OF MATERIAL - “MANY TO MANY ALLOCATION”

Can we treat above problem as three dependent problems…..???
in two different manner …..

TYPE I : One to Many assignment :
Number of problems is equal to number of current operations.

TYPE II : Many to one assignment :
Number of problems is equal to number of successor operations
Let us analyze both the situation.

TYPE I : One to Many assignment : Number of problems is equal to number of current operations.

Problem one:

Problem two:
Problem three:
TYPE II: One to Many assignment: Number of problems is equal to number of successor operations.


Problem one:


Problem two:




Problem three:


The next question(s) comes in our mind is:

1] Whether the above division of problem is logical or not?

2] If yes: Which one is better…under what conditions….

3] Whether it is possible to embed both solution and generate a third possible solution.

4] Whether this solution can be generalized for [m, n] problem case.
It means m current operations and n immediate successor operations.

For each of the two problems, we have solution as derived earlier in this section and it could be interesting to find out impact of that solution on problem definition.

If one carefully observes the stated problems, can state following observations:

1] Though two problem defined seems to be different, but those can not be solved as two independent problems.

2] Interestingly the constraints equation are linear in nature. Therefore, it may be a good idea to try to formulate LP problem to solve the problem of material constraint.

3] For this purpose, we can divide total problem into two distinct problems:

a] Calculate total feasible inventory for each of the current operation(s) that can be assigned to all consuming operations, without violating stock policy.

b] Formulate a linear programming problem that can be solved by simplex
method, which provides an optimized solution for a given objective function.

4] This linear program can also be formulated for ‘one to many allocation’ case as well. This shall also provide an optimized solution for a given objective function.

5] The LP formulation can also solve the third case: many to one allocation problem.

6] One additional benefit of using LP is that is gives optimized solution in a very less time and it can handle big number of variables without any difficulty.

Now we will try to formulate LP for each of the case and solve it by simplex method.

It is easier to solve these LP problems using ‘Solver’ function from excel, with an assumption that problem is linear in nature and variables have non-negative values.


Generalized Linear Problem Solution:


Objective Function: Max / Min Σ Oi * Pi

Subject to: Σ Pi * Q i = Q

All Pi Σ 0 ,

Here quantities Qi are calculated by the algorithm described earlier and Oi is weight or penalty / cost for item i. The value Qi provides the total maximum feasible quantities of a particular item i, that can be assign for production, without violating stock policy.

Pi indicates the actual production quantity of item i.

Let us see again the same three classification of the problem domain:

A. Many to Many allocation problem

B. One to Many allocation problem

C. Many to One allocation problem