This presentation explains about the Operations Management concept Reorder point, different cases with examples, fixed order interval model, single period model etc.

1. Operations Management
Reorder Point
DON PAUL

2. 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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3. 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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4. 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
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5. 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 LT
Where d = demand rate per period(units per day or week)
LT = lead time in days or weeks
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6. 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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7. Solution
Demand (d) = 10,000 gallons/year
Store 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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8. ROP : Variable demand and constant lead time
If the inventory level might be depleted at a faster rate during lead time,
Variable demand with a Reorder point
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9. ROP : Variable demand and constant lead time
When 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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Reorder
point, R
Q
LT
Time
LT
Inventorylevel
0
Safety Stock

10. Service Level
10
• 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%.

11. ROP : Variable demand and const. lead time Expression
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• R = dLT + z d 𝑳𝑻
where d = average daily demand
LT = lead time
d = standard deviation of daily demand
Z = number of standard deviations corresponding to the service level
probability
z d 𝐿𝑇 = safety stock
The term d 𝐿𝑇 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)
= d
2 𝐿𝑇
Standard deviation = d2 𝐿𝑇
= d 𝐿𝑇

12. Problem 2
For 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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13. Solution
13
d = 30 gallons per day
LT = 10 days
d = 5 gallons per day
For a 95% service level, z = +1.65(from table Area under the standardized
normal curve)
Safety stock= z d 𝐿𝑇
= (1.65)(5)( 10 )
= 26.1 gallons
R= dLT + z d 𝐿𝑇
= 30(10) + 26.1
= 326.1 gallons

14. ROP : Const. demand and Variable lead time Expression
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If only lead time is variable, then dLT = dLT
• R = dLT + zdLT
where d = demand rate
LT = average lead time
LT = standard deviation of lead time
Z = number of standard deviations corresponding to the service level
probability

15. Problem 3
A 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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16. Solution
d = 600 bars per day
SL = 90%, so z = 1.28 (from table Area under the standardized normal curve)
LT = 6 days
LT = 2 days
a) R = dLT + zdLT = 600x(6) + 1.28x(600)x(2) = 5136 bars of
soap
b) No of Days = R/d =5136/600 = 8.56 days
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17. ROP : Variable demand and Variable lead time Expression
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If both demand and lead time are variable, then dLT =
• R =
where d = average demand rate
LT = average lead time
LT = standard deviation of lead time
d = standard deviation of demand rate
Z = number of standard deviations corresponding to the service level
probability

18. Problem 4
The 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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19. Solution
d = 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 day
R =
= 25(10) + 1.6 10 32 + (252)(22)
= 334 glasses
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20. Problem 5
A 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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21. Solution
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d = 50 jars per week
LT = 2 weeks
d = 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 𝐿𝑇 is appropriate
b. From table for service level 0.9000, z = +1.28
c. R = dLT + z d 𝐿𝑇
= 50(2) + 1.28(3) 2
= 105.43

22. 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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23. 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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24. Fixed-order-interval model : Expression
OI = Order interval (length of time between orders)
A = Amount on hand at reorder time
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25. Problem 6
Given 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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26. Solution
z = 2.33 for 99 percent service level
Amount to order = d(OI + LT) + z d 𝑂𝐼 + 𝐿𝑇 - A
=30(7+2) + 2.33(3) 7 + 2 - 71 = 220 units
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27. Problem 7
Given 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 at
a. The end of the initial lead time.
b. The end of the second lead time.
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28. Solution
a. R= dLT + z d 𝐿𝑇
43 =10x4 + z(2)(2)
z =0.75 So from the table service level is 0.7734
The risk = 1-0.7734 = 0.2266, which is fairly high.
b. Amount to order = d(OI + LT) + z d 𝑂𝐼 + 𝐿𝑇 - A
171=10(4+12) + z x (2) 12 + 4 - 43
Solving 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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29. 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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30. 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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31. 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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32. Problem 8
Sweet 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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33. Solution
Ce = 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) = 0.75
Thus, 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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34. Problem 9
Cindy’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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35. 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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36. Summary
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