Material Constraint Theory Part IV

(V) HEURISTIC – MANY TO ONE ALLOCATION

1] For given lot size of item L, calculate actual requirement of immediate Predecessor items.
q[i] = DL * x , q[j] = DL * y , q[k] = DL * k ,

2] Calculate shortages for each predecessor item.
Sq[i] = q[i] – Qi’’ , Sq[j] = q[j] – Qj’’ , Sq[k] = q[k] – Qk’’ ,

3] Calculate the feasible quantity of each item that can be assign at particular time t, from store stock, based on stock policy.
Fq[i] = max { (Qi’ + Q[Ti] – Di – s[i] ), 0 }
Fq[j] = max { (Qj’ + Q[Tj] – Dj – s[j] ), 0 }
Fq[k] = max { (Qk’ + Q[Tk]+Q[Tk+1] – Dk – s[k] ), 0 }

4] Calculate the total quantity available at time t, to manufacture item L.
TQ[I] = Qi’’ + min { Sq[i] , Fq[i] } ,
TQ[j] = Qj’’ + min { Sq[j] , Fq[j] } ,
TQ[k] = Qk’’ + min { Sq[k] , Fq[k] } ,

5] Calculate the actual production of item L possible from total quantity available of items i, j and k.
PQ[L-i] = TQ[i] / x ,
PQ[L-j] = TQ[j] / y ,
PQ[L-k] = TQ[k] / z ,

Net quantity of item L, which can be produced, shall be:NQ[L] = min { PQ[L-i], PQ[L-j] , PQ[L-k] }

6] Calculate actual quantity of predecessor items required to produce NQ[L].
NQ[i] = NQ[L] / x ,
NQ[j] = NQ[L] / y ,
NQ[k] = NQ[L] / z ,

7] Calculate the quantity required to be assigned from the stock, for each predecessor items.
SQ[i] = NQ[i] – Qi’’ ,
SQ[j] = NQ[j] – Qj’’ ,
SQ[k] = NQ[k] – Qk’’ ,

8] Update store stock by deducting the assigned quantity for each predecessor item.
Qi = Qi’ – SQ[i],
Qj = Qj’ – SQ[j],
Qk = Qk’ – SQ[k],

9] Update the lot size to be produced for item L, from DL to NQ[L] .
Exit.

The above algorithm works on the principal:

Produce feasible quantity of current item under consideration, while minimizing the probability of shortages for all other items that needs to be produced over defined finite horizon, and consumes at least one item from those required for the current item.

Off course, alterations are possible in the above algorithm based on various conditions. For example:

A] Change in stock policy.
B] Current item is very important. One should try to maximize production quantity of current item, irrespective of any resulting shortages for any other item in the future and / or stock policy.
C] Compare the priorities of the items conflicting for the consumption of same items over finite horizon, and try to minimize probability of shortage for most important item.

The algorithm is flexible enough to consider any of above recommendation without any major changes. Mainly the calculations for step three shall change, based on changed condition.

II. ALLOCATION OF MATERIAL - “ONE TO MANY ALLOCATION”


Constraints:

1] WIP inventory of item A produced by Operation 10 at a particular time t.
2] Stock Policy.
3] Consumption ratio of item A produced by Operation 10 for items to be produced by Operation 20, 30 and 40.
4] Planned production quantity of Items. (to be produced by Operation 20, 30 & 40).
5] Stock on hand of item A at time ‘t’, and scheduled receipts of item A produced by operation 10.

Assumptions:

a] There is no other demand on item ‘A’ than by operations 20, 30 & 40 (produced by operations 10, or otherwise is sub contracted), at particular time ‘ t ‘ .
b] Next scheduled receipt for item ‘A’ is known over a finite horizon T.
c] Demand on item ‘A’ by other MO’s, for a finite horizon T is known.
d] The inventory system follows base stock policy { [s, S] policy }.

(VI) PROBLEM ILLUSTRATION –ONE TO MANY ALLOCATION

Lot sizes of operation 20, 30 & 40 are Q20, Q30 and Q40. Consumption ratio for operations 20, 30 & 40 are r[20], r[30] and r[40] respectively. We should know for Look-ahead period [time t to time T]:



As now the downstream consuming operations [may be called succ_op] are many, one has to carefully consider demand on succ_op, over finite forward horizon T also to derive a particular solution.

The total analysis can be divided into two different set of problems. Lets look at it the next part: