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**Managing Inventory under Risks**

Leadtime and reorder point Uncertainty and its impact Safety stock and service level The lot-size reorder point system Managing system inventory

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**Leadtime and Reorder Point**

Q Usage rate R Inventory level Average inventory = Q/2 Reorder point Time Receive order Place order Receive order Place order Receive order Leadtime

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When to Order? ROP (reorder point): inventory level that triggers a new order ROP = LR (1) Example: R = 20 units/day Q*= 200 units L = leadtime with certainty μ = LR = leadtime demand L (days) ROP 2 40 7 140 14 280 22 440

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**Motorola Hong Kong Revisited**

It takes the supplier 3 full working days to deliver the material to Motorola Consumption rate is 90 kg/day At what inventory level should Mr. Chan place an order?

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**Uncertainty and Its Impact**

Sandy is in charge of inventory control and ordering at Broadway Electronics. The average demand for their best-selling battery is on average 1,000 units per week with a standard deviation of 250 units With a one-week delivery leadtime from the supplier, Sandy needs to decide when to order, i.e., with how many boxes of batteries left on-hand, she should place an order for another batch of new stock What is the difference between Mr. Chan’s task at Motorola and this one?

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**Forecast and Leadtime Demand**

Often we forecast demands and make stocking decisions accordingly trying to satisfy arriving customers from on-hand stock Often, forecasting for a whole year is easier than for a week Leadtime demands usually can not be treated as deterministic

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**Inventory Decision Under Risk**

Inventory on hand When you place an order, you expect the remaining stock to cover all the leadtime demands Any order now or later can only satisfy demands after the leadtime L When to order? ROP1? order ROP1 ROP2 L L

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ROP under Uncertainty When DL is uncertain, it often makes sense to order a little earlier, i.e., at an inventory level higher than the mean ROP = + IS (2) IS = safety stock or extra inventory IS = zβ ×s (3) zβ = safety factor

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**Random Leadtime Demand**

Random Variable Mean std Demand Leadtime Leadtime demand (DL) R L = LR

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**Safety Stock Leadtime Inventory on hand L Time t L order order order**

ROP mean demand during supply lead time safety stock Time t L Leadtime

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**Some Relations ROP safety stock safety stock safety factor**

safety factor service level Given demand distribution, there is a one-to-one relationship, so we also have ROP Is zβ β

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**Safety Stock and Service Level**

Service level is a measure of the degree of stockout protection provided by a given amount of safety inventory Cycle service level: the probability that all demands in the leadtime are satisfied immediately SL = Prob.( LT Demand ≤ ROP) =β

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**Service Level under Normal Demands**

Service Level: SL = ? (The area of the shaded part under the curve) Is= ROP – µ = 200 Mean: µ = 1,000 ROP = 1,200 SL = Pr (LD ROP) = probability of meeting all demand (no stocking out in a cycle)

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**Compute Cycle Service Level**

Given Is and σ Use normal table, we find β from zβ Use excel: SL= NORMDIST(ROP, ,σ,True) (5) (4)

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**Example 7.3 (MBPF) zβ = or SL = NORMDIST(24,000,20,000, 5,000, True)**

ROP = 24,000, µ = 20,000, σ = 5,000 zβ = β = or SL = NORMDIST(24,000,20,000, 5,000, True) NT 9-EX1

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**Compute Safety Stock Given β, we obtain zβ from the normal table**

Use (3), we obtain the safety stock Use (2), we obtain ROP Given β, we can also have zβ = NORMSINV (β) (6) ROP = NORMINV(β, µ, σ ) (7)

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**Example 7.4 (MBPF) µ = 20,000, σ = 5,000 β = 85% 90% 95% 99% zβ =**

β = 85% 90% 95% 99% zβ = ROP = NT

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**Price of High Service Level**

NORMSINV ( 0.999)·200 NORMSINV ( 0.99)·200 Safety Stock NORMSINV ( 0.97)·200 NORMSINV ( 0.95)·200 NORMSINV ( 0.90)·200 NORMSINV ( 0.85)·200 Service Level Suppose I said I want 100% service level and it turns out it costs me $8.1b and the company only has $2m to invest. What should I do? Look at the curve. More towards perfection, more costly. Go to the computer example! 9-EX2

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Example, Broadway Sandy orders a 2-week supply whenever the inventory level drops to 1,250 units. What is the service level provided with this ROP ? If Sandy wants to provide an 95% service level to the store, what should be the reorder point and safety stock ? Average weekly demand µ = 1,000 Demand SD = 250 Reorder point ROP = 1,250

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**The Service Level Safety stock Is = Safety factor zβ = Service level**

By normal table β = By excel SL= NORMDIST (1250, 1000, 250, True) NT 9-EX1

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**Safety Stock for Target SL**

For 95% service rate By the normal table z0.95 = ROP = Is = By excel ROP =NORMINV (0.95, 1000, 250) NT 9-EX1

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**Lot Size-Reorder Point System**

Having determined the reorder point, we also need to determine the order quantity Note that we can forecast the annual demand more accurately and hence treat it as deterministic Then, the order quantity can be obtained using the standard EOQ

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**The Average Inventory Let the order quantity be Q**

The average inventory level = (Q+Is+ Is)/2 = Q/2 +Is The holding cost = HQ/2+HIs The ordering cost = S(R/Q) The optimal inventory cost = HQ* + HIs Inventory on hand Q +Is order ROP mean demand during supply lead time safety stock Time t Leadtime

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**Example, Broadway R=52000/year (52 weeks) H=$1/unit/year S=$200/order**

Lot-size Reorder point Order quantity Q* = For 95% service rate Is = 250zβ = Inventory cost = Sandy’s current policy µ= 1000, Q = 2000 ROP = 1,250, SL =84% Holding cost = Ordering cost Inventory cost 9-EX1

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**Managing System Inventory**

There are different stocking points with inventories and at each stocking point, there are inventories for different functions Total average inventory includes three parts: Cycle + Safety + Pipeline inventories Total Average Inventory = Q/2 + Is + RL (8)

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Pipeline Inventory If you own the goods in transit from the supplier to you (FOB or pay when order), you have a pipeline inventory Average pipeline inventory equals the demand rate times the transit time or leadtime by Little’s Law Pipeline inventory = RL

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**Sandy’s Current System Inventory**

Q=2,000, L =1 week, R = 1,000/week ROP = 1250, Safety stock = Is = 250 Total system average inventory: not own pipeline I = 2000/2+250 = 1250 owns pipeline I = 2000/ = 2250

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**Managing Safety Stock Levers to reduce safety stock**

- Reduce demand variability - Reduce delivery leadtime - Reduce variability in delivery leadtime - Risk pooling

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**Demand Aggregation By probability theory**

Var(D1 + …+ Dn) = Var(D1) + …+ Var(Dn) = nσ2 As a result, the standard deviation of the aggregated demand is (9)

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**The Square Root Rule Again**

We call (9) the square root rule: For BMW Guangdong Monthly demand at each outlet is normal with mean 25 and standard deviation 5 Replenishment leadtime is 2 months. The service level used at each outlet is 0.90 The SD of the leadtime demand at each outlet of our dealer problem The leadtime demand uncertainty level of the aggregated inventory system

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**Cost of Safety Stock at Each Outlet**

The safety stock level at each outlet Is = z0.9σ = 1.285×7.07 = 9.08 The monthly holding cost of the safety stock TC(Is) = H×Is = 4,000x9.08 = 36,340RMB/month

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**Saving in Safety Stock from Pooling**

System-wide safety stock holding cost without pooling 4×C(Is) = 4 ×36,340=145,360 RMB/month System-wide safety stock holding cost with pooling C(Isa ) = 2 ×36,340=72,680 RMB/month Annual saving of 12x(145,360-72,680) = 872,160 RMB!!

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**BMW’s System Inventory**

With SL = 0.9: L = 2, Q = 36 (using EOQ), R=100/month z0.9 =1.285, Is =(1.285)(14.4)= 18.5 ROP = 2x =218.5 Total system average inventory: not own pipeline I = 36/ = 36.5 owns pipeline I = 36/ = 236.5

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**Takeaways (1) ROP = + IS = RL + zβσ**

Leadtime demand usually must be treated as random, and hence creates risks for inventory decision We use safety stock to hedge the risk and satisfy a desired service level Together with the EOQ ordering quantity, the lot-size reorder point system provide an effective way to manage inventory under risk Reorder point under normal leadtime demand ROP = + IS = RL + zβσ

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**Takeaways (2) For given target SL For given ROP**

ROP = + zβσ = NORMINV(SL, ,σ) For given ROP SL = Pr(DL ROP) = NORMDIST(ROP, , σ, True) Safety stock pooling (of n identical locations) Total system average inventory = Q/2 + Is not own pipeline = Q/2 + Is+RL owns pipeline

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