Material Constraint Theory - Last Part

SOLUTION BY LINEAR PROGRAMMING

MANY TO MANY ALLOCATION PROBLEM





We can try to formulate the linear problem for this solution with different objective functions.

For current example, we can consider following three objective functions:
a. Maximize Profit and/or Priority
b. Maximize Total production Quantity
c. Minimize Total Delay [Cost / Time]


Problem Definition: • Parts to be produced are P1, P2 and P3. These are produced by operation 40, 50 and 60 respectively. • Illustration above shows the part requirement to produce each unit of these parts. For example, to produce one unit of P1, it requires one unit of Q10 and two units of Q20. • The available stock of Q10, Q20 and Q30 is 20, 35 and 29 respectively. • The need is to determine the quantity of parts P1, P2 and P3 to be produced in order to satisfy the stock constraint and achieve the objective function defined.

(I) MAXIMIZE PROFIT / PRIORITY




(II) MAXIMIZE TOTAL PRODUCTION QUANTITY



(III) MINIMIZE TOTAL DELAY [TIME/COST]





ONE TO MANY ALLOCATION PROBLEM

(I) MAXIMIZE PROFIT / PRIORITY



(II) MAXIMIZE TOTAL PRODUCTION QUANTITY

(III) MINIMIZE TOTAL DELAY [TIME/COST]

MANY TO ONE ALLOCATION PROBLEM


Let us take only one case here as this is simpler problem environment compare to other two. We can consider here the problem of maximizing total production quantity.


All above section gives a good overview of the problem related to stock limitation / constraints. It is logical to understand the problem first, domain requirements second and then apply the most practical solution, even I sometime(s) it is not the optimal solution.

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

Material Constraint Theory Part V

(VI) PROBLEM ILLUSTRATION – ONE TO MANY ALLOCATION - Contd...

Problem A] Find out total feasible quantity of item ‘A’, which is available for production at particular time ‘t’. Consider here, WIP and Store stock as well. Say it is Q[A].

If Q[A] Σ D[ t ], Then there is no problem for inventory allocation, as sufficient quantity is available in stock to process succ_op, without violating constraints defined. Else, problem B] will be applicable.


Problem B] Find out inventory allocation rule, which assigns Q[A] to succ_op in an effective manner, without violating constraints and satisfying the objective function, as defined.

One can think of following parameters, basis on which derivation of allocation rule may be possible, mostly related to demand:
1] Current lot sizes of each of succ_op,
2] Demand on each of succ_op, over finite horizon T, {D20 / D30 / D40}
3] Weight of each item in succ_op, (weight may be defined as delay cost per unit or importance)
4] Propagated end item demand on each of succ_op.
5] Consumption ratio for each of succ_op.

Let us try to derive the effect of all stated parameters on allocation problem. For the Illustration, let us solve a simple problem, as given below:



Here
s = 5, S = 18, D10 = 8, QT10 = 12, Q’10 = 14. Total demand on op10 = 16+10+21 = 47
Qi = Lot size [planned production quantity of current operation i ],
Di = demand of items produced by succ_op i, over finite horizon T.

Problem A: Find out feasible quantity that can be used for production of operations 20, 30 & 40, without violating stock policy.
The algorithm as used earlier can also be applied here. Therefore,
Feasible quantity that can be assigned from stock can be calculated as:

Fq[ A ] = 14 + 10 – 8 – 5 = 11

Feasible quantity that can be assign from stock = 11
Total quantity available for allocation = 11 + 22 = 33

Problem B: Find out a good solution for allocation of available inventory for succ_op, based on criteria defined.

I] Allocate inventory proportionate to lot size quantity:
[Minimize variance]

Ratio of allocation --- op20 : op30 : op40 : : 16 : 10 : 21
Total amount available: 33



II] Allocate inventory proportionate to demand:
[Minimize shortage]

Ratio of allocation --- op20 : op30 : op40 : : 16 : 9 : 18
Total amount available: 33


III] Allocate inventory proportionate to penalty/delay cost:
[Minimize delay cost ]

Ratio of allocation --- op20 : op30 : op40 : : 4 : 3 : 2
Total amount available: 33



Though above problems do not represent common / generalized results, but one can observe that allocation of inventory based on delay cost shall yield a result with reduces total penalty or delay cost.

One has to identify the suitable objective function and then the allocation rule can be derived for the same.

Actually, after determining the feasible quantity that can be assigned at particular time t, one can determine numerous methods to allocate available inventory to all downstream consuming operations.

Different individual parameters can be identified as stated above, or one can derive combination of this parameter to find out the proportionate allocation to a particular operation.

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:

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.

Material Constraint Theory Part II

MATERIAL CONSTRAINT THEORY

(I) MATERIAL CONSTRAINT ON “MANY TO ONE ALLOCATION”

The problem defined could be treated as the case of - Allocation of on-hand inventory to a particular MO and calculate feasible quantity which can be produced [of current Item]. In this case, at an operation level, the producing part consumes multiple WIP/RM and generates a case where all required parts are not available in required quantities.

In other words, the problem can be identified as:

a) Assigning available on-hand inventory of all required components for manufacturing current MO based on stock policy.
b) Calculate the feasible quantity of current item I than can be produced.

Following illustration is helpful to understand the stated problem.

(II) MATERIAL CONSTRAINT ON “ONE TO MANY ALLOCATION”

The problem defined could be treated as the case of - Allocation of on-hand inventory of a particular part / item to multiple MO/operation and calculate feasible quantity which can be produced [of current Item]. In this case, at an operation level, a particular part/WIP is being consumed by multiple downstream operations and generates a case where a required part is not available in sufficient quantities to produce sum of all next level parts demand.

In other words, the problem can be identified as:

a) Assign available inventory to multiple consuming MO, based on some objective function and stock policy.
b) Calculate the Production quantity for each of the successor MO.
Following illustration is helpful to understand the stated problem.

(III) MATERIAL CONSTRAINT ON “MANY TO MANY ALLOCATION”


The problem defined could be treated as the case of - Allocation of on-hand inventory of a particular part / item to multiple MO/operation and calculate feasible quantity which can be produced [of current Item]. In this case, at an operation level, a particular part/WIP is being consumed by multiple downstream operations and generates a case where a required part is not available in sufficient quantities to produce sum of all next level parts demand.

In other words, the problem can be identified as:

a) Assign available inventory to multiple consuming MO, based on some objective function and stock policy.
b)Calculate the Production quantity for each of the successor MO.

Following illustration is helpful to understand the stated problem.



Next part, we shall try to enumerate in detail each problem independently and try to find out feasible solution(s) possible without violating constraints defined.

Please note, there are number of feasible solution possible, depending on how one see a particular problem and how the required objective function is defined.

Material Constraint Theory - Part I

OVERVIEW

Typical manufacturing environment is said to be governed by three “M’s”, namely:.

• Men
• Machine
• Material

There has been a continuous practice to observe better usage of resources, mainly men and machine. The focus on material was not there except in the recent past, when the competition to get higher business share and still one needs to provide quality products in reasonable price.

It is interesting to note here that all these three resources have different behavior and characteristics. A man has created value primarily because of his skills, a machine because of capacity and processing capabilities and material as base resource [raw material].

The good question that could be asked here is – “what is the importance of material in getting the best or required results?”

If one reads the resources carefully, it is simple to establish that:

Men – has skills, which is function of time and skills are not readily available in the open environment. It requires training and inputs. This means, even if you have shortage of men or skills, it is difficult to bridge the gap in shot time span.

Machine – has build in capacity and processing capability. It generates faster, better and accurate results as compare to a manual process. It is also a fact that in case of capacity shortage, one can’t build the capacity overnight.

Material – forms the base or the raw material for the product to be developed. All the value additions throughout the process is added to material / raw material to create the desired result. It is interesting to note that in case of material shortage, one can fine additional capacity in the open environment to push through larger volumes of the material through the system. Nevertheless, it is a fact that the factor ‘material’ is one of the important points, which affects the process, output or desired results in more than 50% of the cases. Therefore, let’s try to understand how one can manage better in case of ‘material shortage’. This paper will try to explain different scenarios and situations and also put forward the possible action plan to manage the case to achieve the good value out of it.

(I) MATERIAL CONSTRAINT THEORY

In a real life situation, not only the capacity is a constraint, but the material is also a constraint. It means when one starts scheduling from today working forward, has to consider demand, the material available to work with and capacity available in terms of resources. In the other words, launch the work if there is a match of demand, materials and resource.

CASE EXAMPLE:

The manufacturing scope of the Items C1, C2 and C3 is outside the shop. This means either these components are subcontracted or may be are bought outs. Therefore, the due dates of these components are generated by MRP, providing date of scheduled receipts.
The facility has stock on-hand for each of the item.

The manufacturing lot size for item I has been generated by MRP, based on end item demand and stock on-hand. Scheduler takes the same lot size quantity to be produced and generates required manufacturing order.

Normally MRP runs on the base stock policy {(S, s) policy} and shows us what we want to happen and not what actually will happen. It assumes no constraint, and therefore generates a plan, which may not be feasible.

The Problem is:

When the scheduler tries to schedule the Manufacturing Order / Operation for item I, that time it do not know whether the right quantity of all components are available or not. There could be two situations: is there sufficient quantity is available in the stock?

If YES: it proceeds without any problem, and planning is not affected by the material quantity.

If NO : Problem starts, if scheduler assumes that all required quantity is available and schedules the MO/Operation for lot size given. At the end of the MO/operation, it will assume that all the quantity is produced and similar quantity propagates further in the chain of operations till the last operation.
As shown, if the actual receipt of components is less than the scheduled receipt, than quantity to be produced for item I shall be less than planned quantity.

Further to that all downstream operations shall suffer because of material shortage at operation for item I. This problem becomes more complicated as system has on-hand stock for components at time, and one should decide whether to use the stock or not.

THE NEED:
The need is to find out an effective solution that will:

1] Consider material as a constraint and finite resource.
2] Assign the available material to all consuming operation(s), in order to satisfy predefined objective function.
3] It should utilize stock on-hand, as and when necessary to achieve the objective function to the possible extent.
4] Assign the actually produced quantity to downstream operation in an effective manner.

Information Required for Problem analysis:

1] On-hand stock of all components required for processing current MO. [may be confirmed receipts quantities of components on or before time t]
2] Stock Policy [for example Base stock policy (s,S) ].
3] Ratio of consumption of all components required for processing current MO.
4] Objective of the assignment. [for allocating available inventory to all consuming operations].
5] Product demand for those products, which consumes current item.

(II) PROBLEM DEFINITION
The material constraint problem can be classified in distinct three types of definitions:

1) Allocation of material on “Many to One” basis
2) Allocation of material on “One to Many” basis
3) Allocation of material on “Many to Many” basis

In the next post, let us try to understand the base nature of each of the problems identified.