Rarely moving materials have very limited nos of issues over a long period of time. If
the rarely moving material is critical and unpredictable in consumption then it
is necessary to stock, but how much to stock?
Let's analyze the consumption of below mentioned material
Material code : SKU10
Procurement lead time : 6 Months
Issue Date--------------Request Qty----------Issued Qty
10/12/2000-----------------40------------------------40
17/09/2003-----------------37------------------------37
24/12/2004-----------------52------------------------50
16/05/2006-----------------45------------------------42
25/12/2007-----------------35------------------------32
So over a period of 7 years, there have been 5 nos of issues for the material and each issue has different qty.
Before going into further detail let us explore few concepts
Service level = Nos of reservations or Requests served/Total Nos of Reservations or requests
Fill Rate = Total Quantity issued against a request/Total requested Qty of a particular request
Let us analyses again the issue pattern of material SKU10
Demand dated 24/12/2004 was having service level of 100% but fill rate is not 100%,
But It is 50/52 x100 = 96.15%
Now it can be seen that issue pattern of the material SKU10 is not normally distributed but discretely distributed.
Discrete portability distribution can be applied to find the probability of issue of this material during its lead time.
There are many discrete probability distribution such as
- Bernoulli distribution
- Geometric distribution
- Binomial Distribution
- Poisson Distribution-Which is limiting case of binomial distribution
Here Poisson distribution is enough to help us in resolving our query
P (λ, k) = λke−λ / k!
For the above issue pattern
λ = ((5-1)/7)/2 = 0.28
Where λ is average rate of occurrence of event (i.e. issue of material) over a period (Lead time of half year)
k is Number of occurrences of an event (i.e. Nos of issues)
Probability that there shall be no issue during lead time P(0.28,0) = 0.280e−0.28 / 0! = 0.752
Probability that there shall be exactly during lead time P(0.28,1) = 0.214324
Probability that there shall be one or less than one issue = P(0.28,0) + P(0.28,1) = 0.966
So for maintaining a service level of 96.6% during lead time we assume that there shall probably be 1 No of request during leading time.
But what could be the request quantity against this request?
To answer this question we should first decide what should be the fill rate.
Let us assume desired fill rate for this material is 95%
Norm inverse for probability of 95% with standard deviation of above requested
quantities and mean value = 51.85381 ~52(After rounding off)
It can be calculated as below.
Expected Request Quantity = Mean Value of requested quantity + k* Standard
Deviation of the requested quantity
k = Factor for fill rate
Now for 95% of fill rate and 96.6% of service level qty to be kept as safety
stock = Nos of request during Lead time x Qty per request = 1x52 =52
So 52 Nos shall be the Re-order point qty for the fill rate of 95% and service level of 96.6%.
If the stock falls below 52, procurement proposal should be generated to maintain the desired service level and fill rate.
Related Words: Service Levels, Fill Rate, Poisson Distribution
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