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# Formulas To Calculate Scrap, Rework And Inventories In sanitaryware Factory

Workers that spend many hours into the sanitaryware factories find difficult to explain to others departments, how the scrap and rework impact in the flow, however this knowledge is important in position as planners or bookkeeper, figures as scrap, lead time or inventories are important for them. Due to the complexity of the process and different plant layouts, in this job, only will be modeling a process of several stages in series with scrap, where the rework items are treated in a parallel line, so the number of parts decreases downstream due to the scrap. We believe that will be easy to translate this model to others organizations. At the end, you have a numerical example.

### 1. - DEFINITION OF VARIABLES

A temporal periodicity is assumed, in which, time is divided into segments. Each segment is called n.

**Capacity**: For all i = 1.... M in space time n, Bni is defined as the amount of net pieces that the workshop i can produce in period n.

**Orders received**: Dni is defined as the number of orders received at the workshop i at time n.

**Inventory: **defined I_{n}^{i}, i= 1,..M, as inventory at the start of each workshop at the beginning of time n. With the following conditions:

i= 2,. . .M I_{n}^{i}^{≥}

i=1 if I_{n}^{i}^{≥}

-I_{n}^{i} delays or pending

**Production**: is defined as P_{n}^{i }, i=1,. . .M as the workshop i production in period n. This production has been defined by a policy that took into account the capacity B_{n}^{i} and inventory I_{n}^{i+1}.

The process would work this way:

At the beginning of period n+1 the inventory in 1 is:

I_{n+1}^{1} = I_{n}^{1} – D_{n}^{1} + P_{n}^{1}

**Production** Pn1 is seen as demand for the Workshop 2, which generates demand for the workshop 3. So the whole process is determined by external demand.

**Demand**

Also E _{D}_{n}_{1}_{ }_{B}_{n}_{i}

**2.-FLOW**

**2.1. - SERIAL FLOW**

The main flow for the production cycle could be divided in: Casting, punching, first finishing, green drying, green finishing, white drying, glazing, trademark application, glazed drying, firing, packing.

We could say that the pieces are borned by filling a batch of molds with slip, but during the thickness phase, hardness and handling of the pieces out of the molds, we can have losses or scrap; “s”.

Definition of variables:

Si: waste or scrap of workshop i

si: so much per one of scrap of workshop i

So, each workshop has to process the processed parts in “m” minus all scrap pieces upstream.

For a number of produced pieces in m, the production in workshop i, will be:

And the first pass yield will be: number of units coming out divided by the number of units going into of workshop:

**2.2. - PARALLEL PROCESS**

We can have rework in some phases; these pieces are disaggregated and treated on one parallel line.

Definition of variables:

Rwp: pieces sent to rework (generated by workshop p)

Rwp’: pieces sent to rework generated by workshop p’

rwp: so much per one of rework workshop p:

rwp’: so much per one of rework workshop p’

Although we could have rework in each phase, for clarity, we are going to focus in one very usual; the firing process in sanitary ware manufacturing, so we give to p the number 2.

In workshop number 2, after the process, inspection disaggregates the pieces in three parts:

The good ones: for workshop 1.

To repair or rework: defined as Rw pieces, they have small defects that current technology is able to repair with reasonable costs. They will be sent to workshop 2’.

Waste or scrap s2: With the current techniques is not profitable (or impossible) to repair the piece and have the physical and chemical characteristics or appearance in order to be marketed within the established quality parameters. The piece will be destroyed, but usually their remains can be used as raw materials.

The pieces to repair go to workshop 2’, where the processed pieces are inspected and they will be newly disaggregated in:

The good ones: to workshop 1.

To repair Rw2’: Sent again to workshop 2’.

Waste s2’: to be destroyed

If the workshop m produces P^{m} units, the process 2 has to process:

From the process 2 the pieces will go to the conditioning workshop 1:

The processed pieces minus the disaggregated ones to rework or to waste:

The pieces from the process 2 disaggregated to the process 2’ will be: the processed by 2 minus the ones sent to conditioning, minus the scrap. They are defined as the process pieces multiplied by the percentage of rework.

The good pieces from process 2’ will be: the disaggregated to rework from 2 minus the scrap from the second fire.

But 2’ produces rework too, whose percentage average is identified as rw2’, and scrap named s2’. This rework pieces from 2’ produces rework and waste and so on.

The total pieces reprocessed by 2’ are:

Where Rw2’would be:

For a number of pieces P^{m }processed in m, the total pieces reprocessed by 2’ are:

And the pieces generated at the conditioning workshop 1 by 2’ will be, rework from 2 minus scrap from 2’

Knowing the proportions can be summarized as:

The total percentage of scrap of 2+2’ system will be the scrap pieces in the first fire plus the scrap of the second, divided by the pieces that have entered to the system, that is, the pieces processed in the first fire.

The total rework will be: all the pieces to rework divided for all pieces that have been processed.

**2.3.-PACKING**

We admit that the packing section has no rejection or rework, so it can be concluded that if we make P^{m} pieces in workshop M, the production in the final workshop will be P^{m} less all scrap pieces in the process:

And the waste of the system is the broken pieces in the whole process divided by the started parts.

And the yield or performance will be:

We have to take in consideration that the yield of the parallel process belongs to the set.

RTY, Rolled throughput yield, finish good without rework, it is the probability to produce a defect free unit. So we don´t have to take in consideration any rework process.

**3. LEAD TIME**

**3.1 TEMPORAL FLOW IN PARALLEL REWORK**

This workshop is going to produce an initial delay in the main flow, since a number of pieces are disaggregate and processed before coming back to the main flow. After a series of cycles, the main flow will have the losses caused by the parallel process, but the flow (as part of the intrinsic variation of the process) will be constant.

Although we have assumed only a process in parallel in one of the workshops, the conclusions would be extended if the process had several of them. As can be look throughout the chapter shall be observed initial delay and must be careful to take the scrap in the whole of the two branches.** **

The lead time that pieces (which have been disaggregated from the main line) will be in the inventory system will depend not only rework process time or “lead time” of the system itself, but the number of times these pieces have to re-enter the cycle of rework due to defects generated again, in addition scrap pieces will decrease the current inventory, increasing the rotation.

To evaluate the delay, it will be calculate the number of cycles required for that the contribution of the parallel line to the main one will be insignificant compared to the segregate from it.

Variables:

Q: Batch of pieces that go to the rework process for first time

rw2’: so much per one of rework in the parallel line

s2’: so much per one of waste from the parallel line

fpy2’: so much per one of units coming out of a process

j: number of cycles

x: so much per one of pieces coming out from the system in cycle j

P2’_{j:} number of good pieces coming out from rework line in cycle j

From the first cycle of the loop come out to the main line the pieces that have entered minus the pieces to rework newly, minus the waste pieces:

P2’_{1}= Q-Qs2'-Qrw2'= Q (1-s2'-rw2') = Qfpy2’

In the second cycle of the loop, we have to process Q*rw2’ pieces and come out to the main line:

P2’_{2}= Q rw2' fpy2’

In the third cycle come back to the main line:

P2’_{3}= Q rw2'^{2}fpy2’

That is, come out to the main line in the cycle j

P2’_{j}= Q rw2'^{(j-1)} fpy2’^{ }

The number of pieces that come out in each cycle is an exponential function which depends on the mantissa performance and the exponent of the number of cycles.

**3.1.1 - OUTPUT TRANSIENT DELAY**

But maybe we want to know the number of cycles to get a proportion or percentage “x” of those who have gone on which have entered the system:

So the come out to the main line in the cycle j, dividing by those who have entered the system:

Applying logarithms:

Clearing

So, if we want to know the number of cycles in order to come out the 98% of pieces from one batch and rw2'=0.2 and fpy 2'= 0.7, we need j= 3.2 cycles. However, generally, the time of cycles will not coincide with the period time that we use, for example we calculate the production per days, but we have a rework shuttle kiln with a curve of 16 hours.

**3.2. - LEAD TIME TOTAL PROCESS**

Definition of variables:

LT_{t }: Total lead time

LT_{i }: Lead time of workshop i

n: considered period

R_{t}: Manufacture turnover or total rotation

R_{i}: Manufacture turnover of workshop i

ρ_{i }: performance of workshop i

P^{1 }: Net production from process in period n

The following basic formulas are known also:

The total process time will be the sum of processing times of each of the sub process:

It is known that the parts that go to finished goods warehouse are:

P^{1 }=^{ }P^{m }ρ_{m }ρ_{m-1 . . . }^{ }ρ_{2+2’}

**Adjusted inventory: **is defined as the inventory in the section decreased by downstream losses based on performance. That is, the contribution of this inventory in the future is available pieces for the final user.* *

So:^{ }

Therefore:

But:

Rotation is a production system by stages, with waste and rework, will be: the produced pieces, divided by the sum of adjusted ongoing inventories.

**Total adjusted inventory**, inventory that will be good, is defined as:

Note, that we don't take in consideration the inventory of rework, because is sub-process of the main line, of course, we take its scrap.

The total processing time will be the considered period per total adjusted inventory divided by the net production of the period. If the period was the unit, the total processing time will be the total adjusted inventory divided by the net production:

The adjusted inventory will be the net production per time unit per process time.

Note that the adjusted inventory by rotation in any workshop will be equal to the good pieces from the warehouse.

**CASE DEVELOPMENT ADJUSTED INVENTORY APPLICATION**

For this case, we take the value for m=4: Casting, spray, first-second fire and packing.

**Conclusion:** We know the challenges of using the real inventory as a benchmark for planning with waste and rework. To properly assess the inventory at any given time, we define a new concept, adjusted inventory, which is the inventory minus downstream piece losses. Based on this new concept, we have redefined two classical terms: Rotation: Parts supplied by the system to the warehouse divided by the sum of the adjusted inventories. Lead time: adjusted total inventory divided by the net production. The temporal influence on the flow of parallel processing has been quantified: the number of pieces that go out in each cycle is an exponential function in which, the mantissa and exponent depend on the system performance and the number of cycles, respectively. This article is one small extract of my thesis doctoral "Planning and inventory management in a manufacturing environment with waste and stochastic reprocessing" managed by Dr. Javier Conde. Department of Management. Faculty of Economics and Business Administration. U.N.E.D,Madrid, Spain. You can contact me at [email protected]

##### José María Berges Asín

José María Berges Asín lives in Spain and has 27 years of experience in an international sanitary ware company, mainly in production area. His training is: Industrial Engineering, E.M.B.A. and PhD Business administration.

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