Material Constraint Theory Part III

(III) PROBLEM ANALYSIS AND SOLUTION ALLOCATION OF MATERIAL - “MANY TO ONE ALLOCATION”


Production quantity of Op 10, 20 & 30 is being consumed by Op40. Total quantity produced at current operation is not sufficient to produced planned quantity by op40. System maintains inventory for items produced by Op10, 20 & 30, namely item A, B & C.

Problem is to find out a feasible allocation of inventory of items A, B & C from stock & WIP to produce Item D through Op40, based on objective function.

Constraints:

1. WIP inventory of items produced by Operation 10, 20 & 30 at a particular time ‘t’.Stock Policy.
2. Consumption ratio of each item produced by Operation 10, 20 & 30 for item to be produced by Operation 40.
3. Planned production quantity of Item, (to be produced by Operation 40).
4. Scheduled receipts of items produced by operation 10, 20 & 30.

Assumptions:

1. There is no other demand on items ‘A’ , ‘B’ and ‘C’ (produced by operations 10, 20 and 30 respectively, or otherwise are subcontracted), at particular time ‘ t ‘.
2. Next scheduled receipt for items ‘A’, ‘B’ and ‘C’ are known.
3. Demand on items ‘A’ , ‘B’ and ‘C’ by other MO’s, for a finite horizon defined is known.
4. The inventory system follows base stock policy {[s, S] policy}.

(IV) PROBLEM ILLUSTRATION – MANY TO ONE ALLOCATION


Look ahead horizon = T = max {Ti, Tj, Tk}

We should know for Look-ahead period [time t to time T]:


With all above calculations and information, we may proceed to algorithm, which provides a feasible solution for finite inventory allocation problem.

It is proved to be a good practice to assign available finite material to all consuming operation using the look-ahead concept. This is natural as it tries to visualize forthcoming problem(s) and tries to minimize its effect at least locally or sometimes even globally.

In the next part let us try to build a simple heuristic, which will give us a good feasible solution for the stated problem.