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In the following excercise a problem concerning Safety  Stock is considered. Safety stocks are determined for several items with a fixed budget.

 

Several models are used: Cycle Service Level, Fill Rate, Time between Stockouts etc. Afterwards variable Lead Times are considered and their impact on Inventory Performance are evaluated. Finally the fact that stock are committed is introduced and implications in safety stocks are discussed.  For further informations see Silver et alii, Inventory Management and Production Planning and Scheduling, 1998.

 Let us consider the following items that follow a Fixed Quantity, Continuos Review (s,Q) Inventory Policy:

 

Table 1 - Excercise Data 
Items i Demand Di [items/year] Standard Deviations σ[items/year] Value v €/unit Lead Times [weeks]
1  1200   360   2   2
2  3000   1300  1.5   2
3  2500   300   1.8  2
4  3800   500  1  2
5  600   300  6  4
6  400   50   3  8

A, the order setup cost is 120 €/order, the carrying cost r is 0.25 €/€/year and the total maximum value of safety stocks is 1000 €.  

Find the reorder point si , the Expected Total Stocksout Occasion per Year (ETSOPY), the Expected Total Shortage in Value per Year (ETSVPY), in the following cases: 

  1. The same safety stock factor k for all the items;
  2. The same Cycle Service Level P1 for all the items;
  3. The same costs for the items in shortage B2 [% of item values/item];
  4. The same Time Between Stock Outs; 
  5. the same Fill Rate, P2, complete backlog;
  6. The same costs for item in shortage for unit of time B3 [% of item vales/items-unit of time];
  7. The same costs for stock out incidents B1;
  8. Minimum Expected Total Stockout Occasions per Year ETSOPY;
  9. Minimum Expected Total Shortage in Value per Year ETSVPY;

10. Considering then only the results in case 2, (same P1) determine the  variation in P1 if Lead Time is not fixed but probabilistic. The average value of LT is the same as the one in Table 1, while standard deviation is 1 week for each item.

11. Considering now only case 2 (same P1) and only item 1. We can suppose that when an order is received inventory are immediatly committed, but items are physically picked from the warehouse one day after. So there is a distinction in between stocks are committed and when they are effettively used. What are the Safety Stocks reduction for item 1 if we include this day items waits between commitment and effective use? 


Concerning case 1, the same k, it is then possible to write the Safety Stocks in value as:

where the value of viσi are:

Table 2 Determination of viσi
Item i viσi
1 144
2 390
3 108
3 100
5 509
6 60
Total 1311

 

Considering that the Total costs of safety stocks must be 1000 € it follows:

 

 so reorder points si are:

Table 3 Reorder points si in the case 1) same k
Item i si
1 103
2 318
3 146
4 228
5 113
6 79

To calculate the ETSOPY we considering  the following formula:

 (1)

While for ETVSPY we use the:

(2)

While:

Case ETSOPY [stockout occasions/year] ETSVPY [€/year]
1) The same k 2.44 328.2

Considering that the same P1 or (CSL Cycle Service Level), means the same k for all the items, then case 2) is exactly the same of case 1:

Table 4 Reorder points si in the case 2) the same P1
Item i si
1 103
2 318
3 146
4 228
5 113
6 79

While:

Case ETSOPY [stockout occasions/year] ETSVPY [€/year]
2) The same P1 2.44 328.2

 


Considering the same B2, it follows that all the items i must satisfiy the following: 

 

(3)

 

So we start from a value of B2 (let say 0.8) and we determine kby using the (3) for each items i and then the total Safety costs with:

 

(4) 

 

 Then we can iterate by decreasing B2 if SS>1000 or increasing B2 if SS<1000

The  value of B2 should converges to B2=0.5845

 

Table 5 Reorder points si in the case 3) the same B2
Item i si
1 92
2 341
3 151
4 230
5 113
6 70


While:

Case ETSOPY [stockout occasions/year] ETSVPY [€/year]
3) The same B2 2.57 323.6

 


Considering that same B2 implies the same TBS, it follows that the solution is exactly the same: 

 

 

Table 6 Reorder points si in the case 4) the same TBS
Item i si
1 92
2 341
3 151
4 230
5 113
6 70


While:

Case ETSOPY [stockout occasions/year] ETSVPY [€/year]
4) The same TBS 2.57 323.6

 

 


From the hypothesis of same P2 it follows that all the items i must satisfiy the following: 

 

(5)

 

So we start from a value of P2 (let say 98%) and we determine kby using the (5)for each items i and then evaluating the total Safety costs with (4):

 

 Then we can iterate by decreasing P2 if SS>1000 or increasing P2 if SS<1000

The  value of P2 should converges to P2=98.11%

 

Table 7 Reorder points si in the case 5) the same P2
Item i si
1 83.8
2 352
3 104
4 160
5 141
6 67

While:

 

Case ETSOPY [stockout occasions/year] ETSVPY [€/year]
5) The same P2 3.61 379

 


For the case of same B3 we can use the equivalence to the case of P2 so results are:

 

Table 8 Reorder points si in the case 6) the same B3
Item i si
1 83.8
2 352
3 104
4 160
5 141
6 67

While:

 

Case ETSOPY [stockout occasions/year] ETSVPY [€/year]
6) The same B3 3.61 379

 

 


From the hypothesis of same B1 it follows that all the items i must satisfiy the following: 

 

(6)

 

So we start from a value of B1 (let say 150) and we determine kby using the (5)for each items i and then evaluating the total Safety costs with (4):

 

 Then we can iterate by decreasing B1 if SS>1000 or increasing B1 if SS<1000

The  value of B1 should converge to B1=156.95

 

Table 9 Reorder points si in the case 7) the same B1
Item i si
1 150
2 331
3 207
4 331
5 48
6 99

 


 While:

Case ETSOPY [stockout occasions/year] ETSVPY [€/year]
7) The same B1 1.74 508

 

 


Since we reach the minimum ETSOPY with the same B1 results are exactly the one in case 7) the same B1

 

 

Table 10 Reorder points si in the case 8) Minimum ETSOPY
Item i si
1 150
2 331
3 207
4 331
5 48
6 99

 


 While:

Case ETSOPY [stockout occasions/year] ETSVPY [€/year]
8) Minimum ETSOPY 1.74 508

 

 


Considering that same B2 implies the Minimum ETVSPY, it follows that the solution of case 9) is exactly the same of cases 3) and 4): 

 

 

Table 11 Reorder points si in the case 9) the Minimum ETVSPY
Item i si
1 92
2 341
3 151
4 230
5 113
6 70


While:

Case ETSOPY [stockout occasions/year] ETSVPY [€/year]
9) minimum ETSVPY 2.57 323.6

 

 


 

if we compare the solutions we have the following

 

Table 12 Solution comparisons
Item i

 

sin cases:

1) same k,

2) same P1

 

 

sin cases:

3) same B2,

4) same TBS,

9)minimum ETVSPY

 

 

sin cases:

5) same P2,

6) same B3

 

 

sin cases:

7) same B1,

8) minimum ETSOPY

1 103 92 83.8 ->min   max->150
2 318->min  341 352  331
3 146 151 104->min  max->207
4 228 230  160->min  max->331
5 113 113 max>141  48->min
6 79 70

67->min  max->99

 

One can note that B1 tends to decrease the safety stocks when the vi and σi goes up, (Silver, page 262) so if we consider Table 2, we can see why in B1 such lower value in case 5, where we have an high value of item vi and an high value of standard deviation σi.

Finally if we compare the ETSOPY and the ETSVPY we obtain:

 

Cases ETSOPY [stockout occasions/year] ETSVPY [€/year]

1) same k, 2) same P1

2.44 328.2

3) same B2, 4) same TBS,  9) minimum ETVSPY

2.57 323.6

5) same P2, 6) same B3

3.61 379

7) same B1, 8) minimum ETSOPY

1.74 508

 

We can nothe how the minimum is reached for case 9) and 8), as expected. But, in particular, with the minimum of ETSOPY we reach the maximum of ETSVPY.

 


 

When we have variable lead times, the standard deviation increases following the formula:

(7)

In order to answer the questions we shoud find k' , the Stock safety Factor with LT variable, via the following formula:

 

(8)

in this manner it is possible to determine for each item i the probability difference as  where:

 

 

Table 13 probability with variable Lead Times
item i k k' pu≥(k') pu(k) Δp
1 0.76 0.72 23% 22% 1%
2 0.76 0.74 23% 22% 23%
3 0.76 0.59 28% 22% 4%
4 0.76 0.60 27% 22% 4%
5 0.76 0.76 23% 22% -5%
6 0.76 0.71 24% 22% -3%

When an Order arrives, stock are first "committed" and then used in the manufacturing process. Usually the time Inventory are committed is not considered. But in the case it is not negligle it should be included. In the following the time an item is only committed will denoted as Committed Time, CT

Considering a standard Inventory cycle, the average behaviour can be shown in the following figure:

It is possible to note the Inventory Position behaviour during time, also the Inventory on Hand. Reorder point s and Lead Time LT are shown.

Inventory on Hand represents what is really present in the warehouse during the time, while Inventory Position what is in the warehouse and the order that are arriving.

If there is an interval of time between the moment an item is committed and the moment is used, named Committed Time CT, the situation changes, as shown in the following figure:

 Now when an item is committed it still belongs to the Inventory on hand, while it does not belong anymore to the Inventory Position.

So differently from the classical case, Inventory Position and Inventory on Hand are never equal. The effect is that Inventory On Hand increases by an amount proportional to the Committed Time. 

Safety Stocks with Committed Time SSCT must cover only the time interval LT-CT and not all LT. So ΔSS, the safety stocks reduction can be evaluated. Considering how standard deviation change with time (by the root square root) and that for item 1 in case P1 safety stocks are equal to 54.9 then:

 

 

 so the reduction is equal to 14.68 units.