Reorder point

Operations ManagementReorder Point

DON PAUL

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Reorder Point (ROP)Definition:

• The Reorder Point (ROP) is the level of inventory which triggers an action

to replenish that particular inventory stock.

• It is normally calculated as the forecast usage during the

replenishment lead time plus safety stock.

• In the EOQ (Economic Order Quantity)model, it was assumed that there is

no time lag between ordering and procuring of materials. Therefore the

reorder point for replenishing the stocks occurs at that level when the

inventory level drops to zero and because instant delivery by suppliers, the

stock level bounce back.

• Reorder point is a technique to determine when to order; it does not

address how much to order when an order is made.

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Cont.…

• In real life situations one never encounters a zero lead time. There is

always a time lag from the date of placing an order for material and the

date on which materials are received. As a result the reorder point is

always higher than zero, and if the firm places the order when the

inventory reaches the reorder point, the new goods will arrive before the

firm runs out of goods to sell. The decision on how much stock to hold is

generally referred to as the order point problem.

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General Expression

• The two factors that determine the appropriate order point are the

delivery time stock which is the Inventory needed during the lead time (i.e.,

the difference between the order date and the receipt of the inventory

ordered) and the safety stock which is the minimum level of inventory that is

held as a protection against shortages due to fluctuations in demand.

Reorder Point (R) = Normal consumption during lead-time + Safety Stock

ROP : Constant demand and lead time

• For basic EOQ model with const. demand and const. lead time to receive an order

is equal to the amount demanded during lead time.

Reorder Point (R) = Normal consumption during lead-time

R = d x LTWhere d = demand rate per period(units per day or week)

LT = lead time in days or weeks

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Problem 1

An ePaint Internet store is open 311 days per year. If annual

demand is 10,000 gallons of Ironcoat paint and the lead time to

receive an order is 10 days, determine the reorder point for

paint.

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SolutionDemand (d) = 10,000 gallons/yearStore open 311 days/year

Daily demand = 10,000 / 311 = 32.154 gallons/day

Lead time = LT = 10 days

R = d x LT = (32.154)(10) = 321.54 gallons

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ROP : Variable demand and constant lead timeIf the inventory level might be depleted at a faster rate during lead time,

Variable demand with a Reorder point

ROP : Variable demand and constant lead timeWhen demand is uncertain, a safety stock of inventory is frequently added to

the expected demand during lead time.

Reorder point with a Safety Stock

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Reorderpoint, R

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LT Time LT

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0Safety Stock

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Service Level

• The probability that the inventory available during lead time will meet demand.

• Stockout means an inventory shortage.• A service level of 90% means there is 0.90 probability that

demand will be met during the lead time and the probability that a stockout will occur is 10%.

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ROP : Variable demand and const. lead time Expression

• R = dLT + z d

where d = average daily demandLT = lead timed = standard deviation of daily demandZ = number of standard deviations corresponding to the service level probabilityz d = safety stockThe term d in this formula for the reorder point is the square root of the sum of the daily variances during lead time:

Variance = (daily variance) x (number of days of lead time)

= Standard deviation =

= d

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Problem 2For ePaint internet store, assume that daily demand for Ironcoat

paint is normally distributed with an average daily demand of 30

gallons and a standard deviation of 5 gallons of paint per day.

The lead time for receiving a new order of paint is 10 days.

Determine the reorder point and safety stock if the store wants a

service level of 95% with the probability of a stock out equal to

5%.

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Solution

d = 30 gallons per dayLT = 10 daysd = 5 gallons per day

For a 95% service level, z = +1.65(from table Area under the standardized normal curve)

Safety stock= z d

= (1.65)(5)( )

= 26.1 gallons

R= dLT + z d

= 30(10) + 26.1

= 326.1 gallons

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ROP : Const. demand and Variable lead time ExpressionIf only lead time is variable, then dLT = dLT

• R = dLT + zdLT

where d = demand rateLT = average lead timeLT = standard deviation of lead timeZ = number of standard deviations corresponding to the service level probability

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Problem 3A motel uses approximately 600 bars of soap each day and this

tends to be fairly constant. Lead time for soap delivery is

normally distributed with a mean of six days and a standard

deviation of two days. A service level of 90 percent is desire.

a) Find the ROP.

b) How many days of supply are on hand at the ROP?

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Solutiond = 600 bars per day SL = 90%, so z = 1.28 (from table Area under the standardized normal curve)

LT = 6 days LT = 2 daysa) R = dLT + zdLT = 600x(6) + 1.28x(600)x(2) = 5136 bars of

soapb) No of Days = R/d =5136/600 = 8.56 days

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ROP : Variable demand and Variable lead time ExpressionIf both demand and lead time are variable, then dLT =

• R =

where d = average demand rateLT = average lead timeLT = standard deviation of lead timed = standard deviation of demand rateZ = number of standard deviations corresponding to the service level probability

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Problem 4The motel replaces broken glasses at a rate of 25 per day. In the

past, this quantity has tended to vary normally and have a

standard deviation of 3 glasses per day. Glasses are ordered from

a Cleveland supplier. Lead time is normally distributed with an

average of 10 days and a standard deviation of 2 days. What ROP

should be used to achieve a service level of 95 percent ?

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Solutiond = 25 glasses per day SL = 95%, so z = +1.65 (from table Area under the standardized normal curve)

LT = 10 days LT = 2 days d = 3 glasses per dayR =

= 25(10) + 1.6= 334 glasses

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Problem 5A restaurant uses an average of 50 jars of a special sauce each

week. Weekly usage of sauce has a standard deviation of 3 jars.

The manager is willing to accept no more than a 10 percent

risk of stockout during lead time, which is two weeks. Assume

the distribution of usage is normal.

a. Which of the above formulas is appropriate for this

situation? Why?

b. Determine the value of z.

c. Determine the ROP.

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Solution

d = 50 jars per week LT = 2 weeksd = 3 jars per week

Acceptable risk = 10 percent, so service level is 0.90

a. Because only demand is variable(i.e has a standard deviation),

formula R = dLT + z d is appropriate

b. From table for service level 0.9000, z = +1.28

c. R = dLT + z d

= 50(2) + 1.28(3)

= 105.43

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How much to order: Fixed-order-interval model

• Orders are placed at fixed time intervals

• If demand is variable, the order size will tend to vary from

cycle to cycle

Two types :

Fixed-quantity ordering

Fixed-interval ordering

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Determining the amount to order• If both the demand rate and lead time are constant, the fixed-interval

model and the fixed-quantity model function identically.

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Fixed-order-interval model : Expression

OI = Order interval (length of time between orders)A = Amount on hand at reorder time

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Problem 6Given the following information, determine the amount to order.

d = 30 units per day Desired service level = 99 percent

d = 3 units per day

Amount on hand at reorder time = 71 units

LT = 2 days

OI = 7 days

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Solutionz = 2.33 for 99 percent service level

Amount to order = d(OI + LT) + z d - A

=30(7+2) + 2.33(3) - 71 = 220 units

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Problem 7Given the following information, determine the amount to order.

d = 10 units per day

d = 2 units per day

A = 43 units

Q = 171 units

LT = 4 days

OI = 12 days• Determine the risk of a stockout ata. The end of the initial lead time.b. The end of the second lead time.

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Solutiona. R= dLT + z d 43 =10x4 + z(2)(2)

z =0.75 So from the table service level is 0.7734The risk = 1-0.7734 = 0.2266, which is fairly high.

b. Amount to order = d(OI + LT) + z d - A

171=10(4+12) + z x (2) - 43Solving z = +6.75

This value is way out in the right tail of the normal distribution, making the service level virtually 100 percent and thus, the risk of a stockout at this point is essentially equal to zero.

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Single–Period Model

• Single-period model (sometimes referred to as the newsboy

problem) is used to handle ordering of perishables (fresh

fruits, vegetables, seafood, cut flowers) and items that have a

limited useful life (newspapers, magazines, spare parts for

specialized equipment).

• Focuses on two costs: Shortage cost

Excess cost

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Cont.…

• Shortage cost may include a charge for loss of customer

goodwill as well as the opportunity cost of lost sales. Generally,

shortage cost is simply unrealized profit per unit.

C shortage (Cs) = Revenue per unit - Cost per unit

• Excess cost pertains to items left over at the end of the period.

In effect, excess cost is the difference between purchase cost

and salvage value.

C excess (Ce) = Original cost per unit - Salvage value per unit

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Continuous stocking levels• The stocking level equalizes the cost weights.• The service level is the probability that demand will not

exceed the stocking level, and computation of the service level is the key to determining the optimal stocking level(So).

Service level (SL)= Cs / ( Cs + Ce)

Where Cs = shortage cost per unit

Ce = excess cost per unit

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Problem 8Sweet cider is delivered weekly to Cindy’s Cider Bar. Demand

varies uniformly between 300 liters and 500 liters per week.

Cindy pays 20 cents per liter for the cider and charges 80 cents

per liter for it. Unsold cider has no salvage value and cannot be

carried over into the next week due to spoilage. Find the optimal

stocking level and its stockout risk for that quantity.

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SolutionCe = Cost per unit - Salvage value per unit

=$0.20 - $0=$0.20 per unit

Cs = Revenue per – Cost per unit=$0.80 - $0.20=$0.60 per unit

SL = Cs / ( Cs + Ce) = 0.60/ ( 0.60 + 0.20) = 0.75Thus, the optimal stocking level must satisfy demand 75 percent of the time. For the uniform distribution, this will be at a point equal to the minimum demand plus 75 percent of the difference between maximum and minimum demands

So = 300 + .75(500 - 300) = 450 liters

The stockout risk = 1-0.75 = 0.25

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Problem 9Cindy’s Cider Bar also sells a blend of cherry juice and apple

cider. Demand for the blend is approximately normal, with a

mean of 200 liters per week and a standard deviation of 10 liters

per week. Cs= 60 cents per liter and Ce= 20 cents per liter. Find

the optimal stocking level for the apple-cherry blend.

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Solution

SL = Cs / ( Cs + Ce) = 0.60/ ( 0.60 + 0.20) = 0.75

This indicates that 75 percent of the area under the normal curve must be to

the left of the stocking level.

The value of z is between 0.67 and 0.68, say, +0.675, will satisfy this. The

optimal stocking level is So = mean + z

So = 200 liters+ 0.675(10 liters) = 206.75 liters

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Summary

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