Inventory control is concerned with minimizing the total cost of inventory.
The three main factors in inventory control decision making process are:
The cost of holding the stock (e.g., based on the interest rate).
The cost of placing an order (e.g., for row material stocks) or the set-up cost of production.
The cost of shortage , i.e., what is lost if the stock is i nsufficient to meet all demand.
The third element is the most difficult to measure and is often handled by establishing a " service level " policy, e. g, certain percentage of demand will be met from stock without delay.
The ABC Classification The ABC classification system is to grouping items according to annual sales volume, in an attempt to identify the small number of items that will account for most of the sales volume and that are the most important ones to control for effective inventory management.
There is a trade-off between lot size and inventory level.
Frequent orders (small lot size): higher ordering cost and lower holding cost.
Fewer orders (large lot size): lower ordering cost and higher holding cost.
28.
EOQ Inventory Order Cycle Demand rate 0 Time Lead time Lead time Order Placed Order Placed Order Received Order Received Inventory Level Reorder point, R Order qty, Q As Q increases, average inventory level increases, but number of orders placed decreases ave = Q/2
30.
Answer to Inventory Management Questions for EOQ Model
Keeping track of inventory
Implied that we track continuously
How much to order?
Solve for when the derivative of total cost with respect to Q = 0: -SD/Q^2 + iC/2 = 0
Q = sqrt ( 2SD/iC)
When to order?
Order when inventory falls to the “Reorder Point-level” R so we will just sell the last item as the new order comes in:
R = DL
31.
The EOQ Model Q = Number of pieces per order Q* = Optimal number of pieces per order (EOQ) D = Annual demand in units for the Inventory item S = Setup or ordering cost for each order H = Holding or carrying cost per unit per year
32.
An EOQ Example Determine optimal number of needles to order D = 1,000 units S = $10 per order H = $.50 per unit per year Q* = 2 DS H Q* = 2(1,000)(10) 0.50 = 40,000 = 200 units
33.
An EOQ Example Determine optimal number of needles to order D = 1,000 units Q* = 200 units S = $10 per order H = $.50 per unit per year = N = = Expected number of orders Demand Order quantity
34.
An EOQ Example Determine optimal number of needles to order D = 1,000 units Q* = 200 units S = $10 per order N = 5 orders per year H = $.50 per unit per year = T = Expected time between orders Number of working days per year N
35.
An EOQ Example Determine optimal number of needles to order D = 1,000 units Q* = 200 units S = $10 per order N = 5 orders per year H = $.50 per unit per year T = 50 days Total annual cost = Setup cost + Holding cost TC = S + H D Q * Q * 2
= d x L ROP = Lead time for a new order in days Demand per day d = D Number of working days in a year
37.
Reorder Point: Example Demand = 10,000 kg /year Store open 311 days/year Daily demand = 10,000 / 311 = 32.154 kg /day Lead time = L = 10 days R = dL = (32.154)(10) = 321.54 kg = 322 kg
buffer added to on hand inventory during lead time
Stockout
an inventory shortage
Service level
probability that the inventory available during lead time will meet demand
39.
Variable Demand with a Reorder Point Reorder point, R Q LT Time LT Inventory level 0
40.
Reorder Point with a Safety Stock Reorder point, R Q LT Time LT Inventory level 0 Safety Stock
41.
Reorder Point With Variable Demand R = dL + z d L where d = average daily demand L = lead time d = the standard deviation of daily demand z = number of standard deviations corresponding to the service level probability (service factor) z d L = safety stock
42.
Reorder Point for a Service Level Probability of meeting demand during lead time = service level Probability of a stockout R Safety stock d L Demand z d L
44.
Reorder Point for Variable Demand The carpet store wants a reorder point with a 95% service level and a 5% stockout probability For a 95% service level, z = 1.6 4 d = 30 m per day L = 10 days d = 5 m per day R = dL + z d L = 30(10) + (1.6 4 )(5)( 10) = 32 5.9 m Safety stock = z d L = (1.6 4 )(5)( 10) = 2 5 . 9 m