Safety Stock
What must be the level of safety stock? In a simple condition where merely the usage rate is variable and the maximum usage rate can be given, the safety stock needed to search for total protection against stock out as:
(Maximum usage rate - Average usage rate) × Lead time
As both the lead time and usage rate vary, that is frequently the case and the range of variation is broad complete protection against stock out may need an excessively large safety stock. For illustration, if the lead time different in between 60 days and 180 days along with an average value of 90 days and the usage rate varies in between 75 units and 125 units per day along with an average value of 100 units daily, a safety stock of 13,500 units is needed for finish protection against stock-out. It has been worked out given as:
Maximum possible usage
|
-
|
Normal or Average usage
|
Maximum daily usage
X Maximum lead time
125 × 180
|
-
×
-
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Average or Normal daily usage
Average lead time
100 × 90 = 13,500
|
Because inventory-carrying costs are proportional to the point of inventories carried, this rarely having sense to search for total protection against stock-out. In view of the trade-off between stock-out cost and inventory carrying cost, the optimal level of safety stock is generally much less than the level of safety stock needed to attain total protection against stock-out.
A manufacturing company will need 50,000 units of a product throughout the next year. The cost of processing a demand is Rs.20 and the carrying cost per units is 50 paise per year. Lead-time of a demand is 5 days and the company will maintain a safety stock of two days usage.
You are needed to compute:
a. Economic Order Quantity;
b. Re-order Point;
c. Minimum Inventory;
d. Maximum inventory and
e. Average Inventory. Here assume 250 days in a year.
Solution
a. Economic Order Quantity
EOQ = √(2RO/C)
Here R = Annual Requirements or Usage
O = Ordering cost per order
C = Carrying cost per unit per year
EOQ = √ (2 × 50,000 × Rs.20/Re .0.5)
= √40,00,000
= 2,000 Units
b. Re-order Point
R.O.P. = (L× UR) + S
Where; L = Lead Time;
U = Usage Rate (50,000 ÷ 250) = 200 (units per day);
S = Safety Stock
R.O.P. = (5 × 200) + (2 × 200)
= 1,000 + 400
= 1,400 Units
c. Minimum Inventory
It is the Safety Stock kept by the company that is 400 units. Minimum inventory will be zero, if there is no safety stock.
d. Maximum Inventory
Maximum Inventory = EOQ + Safety Stock
= 2,000 + 400
= 2,400 units.
e. Average Inventory
= (Maximum Inventory + Minimum Inventory)/2
= (2,400 + 400)/2
= 1,400 Units