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In the following a problem from Nahmias, Production and Operatons Analysis (sixth Edition, page 279). The particularity is that the reorder point s is greater than the max level of Inventories on Hand (EOQ+safety stocks).

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This is a re-elaborated version of the Original (Nahmias). A shop supply automotive components and in particular the one considered in the following: an oil filter. The cost of this component is 1.5 €/item. Order Setup cost A is 100 €/order (see Silver et al., 1998). Warehouse carrying costs are 28% €/€-y.

Monthly demand follow a normal distribution with average demand D=280 items/month and standard deviation σ=77 items/months.

Order Lead time is 5 months.

Unsatisfied demand is backordered with a backorder cost of 12.8 €/item.

Find the following:

 

a) the optimal level of order quantity and the reorder point

b) the yearly average carrying costs, setup costs and backorder, on the hypothesis to follow an optimal policy

c) Evaluate the cost of the uncertainty of this process, by comparing the cost obtained in  point b) with the one in the hypothesis of no variance.

d) What happen in the case of LT=6 months but A=50€?

e) What happen if LT reduce to 4 months but a standard deviation of 1 month?

 


 Solution to point a)

In order to minimize the Total Replenishment Costs in Continuous Review (s,Q) we should find together Q (the EOQ) and s= (the reorder point, remember s=xL+k•σl).

Considering that when there is a stock out, missed items are backordered, the total replenishment cost (see Silver et al.) is:

(1)

Where  B2,the penalty for each item in shortage, is equal to:

In TRC(k,Q),  Gu(k) is:

and in particular Gu(k) derivate is:

In order to find the minimum we impose then following conditions:

 

To minimum conditions then:

(2)

and

(3)

We can approximate the solution by neglecting the backorder cost in (2) and calculating separately the classical EOQ on one side and the k from (3) on the other side.

items/order

 

k=2.246, then

s=1786.7 items/order

 

To find the exact solution, we could satisfy both conditions (2) and (3) for k and Q by iterating (by using excel for instance). After few iterations we can find following results:

 

Qexact=1326 items/order

sexact=1783.5 items

 

 It is possible to note that differences between exact solution and the approximate one are very small:

 

Δ(Qexact-EOQ)=61.46=4.6%

 

Δ(sexact-s)=-3.26=-0.18%


It is noteworthy that the reorder point s is above the max level of stocks.

We can show this graphycally:

this results it is strange but correct. the reorder point, in fact, refers to the inventory position and not to the Stocks on Hand.

 

 


Solution to point b)

 

Carrying costs= 28%*1.5*(1265/2+2.246*77)=428.05 €/y

Setup Costs= A*D*12/Q=100*280*12/1265=265.61 €/y

StockOut Costs=B2*v*Gu(k)*σLT*D*12/Q=25.68 €/y


Solution to point c)

Costs in the case of variance equal to zero:

 

TRC(novariance, Q)=531.26 €/y

TRC(k,Q)=719.34 €/y

The costs of uncertainty is then 719.34-531.26=188.08 €/y=+35.4%


 

 Solution to point d)

by increasing the lead time to 6 months and reducing the setup costs to 50 €/order and determining EOQ and s optimal we obtain:

EOQLT6A50=894.4 items/order

sLT6A50=1823.6 items

 

if we calculate the exact solution via iteration:

EOQLT6A50=960.2 items/order (+6.85%)

sLT6A50=1843.3 items (+1.07%)

so costs are:

carrying costsLT6A50=365.75 (differences -62.30)

setup costsLT6A50=187.83 (differences -77.78)

stockout costsLT6A50=28.62 (differences 2.18)

total costs of candidate supply (LT=6 and A=50) =582.20 €/y

Differerence between candidate and present = -137.9 €/y=-19.15%

so the supply of point d is more convenient that the present one.

 


Solution to point e)

IF we consider a variable lead time tLT with average E(tLT)=4 months and standard deviation σtLT=1 months, we should consider the following expression for the evaluation of the LT standard deviation σLT:

 

 Then by calculating the order quantity Q and reorder point s (exact both) it is possible to evaluate the costs. But it is clear that we have a reduction in the variance, so solution of point e is better than present solution. There is no need to do other calculations.

 

Note: you can find here an excel file and a word file I used for this document.