a) if average demand per day is 20 units and lead time is 10 days. assuming zero safety stock. ss=0,...

a) If average demand per day is 20 units and lead time is 10 days. Assuming zero safety stock. ss=0, compute ROP.

b) If average demand per day is 20 units and lead time is 10 days. Assuming 70 units of safety stock. ss=70, compute ROP?

c) If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss.

d) If average demand per day is 20 units and standard deviation of daily demand is 5, and lead time is 16 days. Compute ROP at 90% service level. Compute ss.

e) If demand per day is 20 units and lead time is 16 days and standard deviation of lead time is 4 days. Compute ROP at 90% service level. Compute ss.

ROP Problems

ROP (Re-Order Point) in periodic inventory control is the beginning of each period.

ROP in perpetual inventory control is the inventory level equal to the average demand during the lead time plus a safety stock to cover demand variability.

Lead Time is the time interval from placing an order until receiving the order.

If there were no variations in demand (it was constant) then safety stock (ss) is zero. In the case of ss=0, ROP is when inventory on hand is equal to the average demand during the lead time.

Demand Forecast, and Lead Times

a) If average demand per day is 20 units and lead time is 10 days. Assuming zero safety stock. ss = 0Compute ROP.

ROP = demand during lead time + ssROP = demand during lead timeDemand during lead time = (lead time) × (demand per unit of time)Demand during lead time = 20 × 10 = 200

ROP; Fixed d, Fixed LT, Zero SS

b) If average demand per day is 20 units and lead time is 10 days. Assuming 70 units of safety stock.ss = 70Compute ROP.

ROP = demand during lead time + ssDemand during lead time = 200ss = 70ROP = 200+70 = 270

ROP; Fixed d, Fixed LT, Positive SS

c) If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss

ROP; total demand during lead time is variable

Risk of astockout

Service level

Probability ofno stockout

Safety stock

0 z

Quantity

z-scale

Safety Stock and ROP

Each Normal variable x is associated with a standard Normal Variable z

Averagedemand

There is a table for z which tells us a) Given any probability of not exceeding z. What is the value of z b) Given any value for z. What is the probability of not exceeding z

x is Normal (Average x , Standard Deviation x) z is Normal (0,1)

ROP

Common z Values

Risk Service level z value0.1 0.9 1.280.05 0.95 1.650.01 0.99 2.33

Risk of astockout

Service level


Safety stock

0 z

Quantity

z-scale

Averagedemand

ROP

Relationship between z and Normal Variable x

Risk of astockout

Service level


Safety stock

0 z

Quantity

z-scale

Averagedemand

ROP

z = (x-Average x)/(Standard Deviation of x)x = Average x +z (Standard Deviation of x)μ = Average x σ = Standard Deviation of x

Risk Service z value level0.1 0.9 1.280.05 0.95 1.650.01 0.99 2.33

x = μ +z σ

ROP and ss

Risk of astockout

Service level


Safety stock

0 z

Quantity

z-scale

Averagedemand

ROP

LTD = Lead Time Demand (Demand during the lead time)ROP = Average LTD +z (Standard Deviation of LTD)ss = z (Standard Deviation of LTD)ROP = Average LTD +ss


c) If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss


ROP

Risk ofa stockout

Service level


Expecteddemand Safety stock

x


z = 1.28z = (X- μ )/σX= μ +z σμ = 200σ = 25X= 200+ 1.28 × 25X= 200 + 32ROP = 232ss = 32


d) If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days. Compute ROP at 90% service level. Compute ss.

ROP; Variable d, Fixed LT

Previous Problem: If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss.

This Problem: If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days. Compute ROP at 90% service level. Compute ss.

If we can transform this problem into the previous problem, then we are done, because we already know how to solve the previous problem.


d) If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days. Compute ROP at 90% service level. Compute ss.

What is the average demand during the lead timeWhat is standard deviation of demand during lead time


Demand During Lead Time

: Demandd

: Average Demandd

:Standard deviation of demandd

: Lead timeLT

If demand is variable and Lead time is fixed

Average demand during lead time : ( )LT d : LTdLT d

:Standard deviation of demand during lead timedLT

: 20d : 5d :16LT

Average demand during lead time : ( )LT d

: LTdLT d

:Standard deviation of demand during lead timedLT

( ) 16 20 320LT d

: 16(5)dLT : 20dLT


The Problem originally was: If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days. Compute ROP at 90% service level. Compute ss.

We transformed it to: The average demand during the lead time is 320 and the standard deviation of demand during the lead time is 20. Compute ROP at 90% service level. Compute ss.

Which is the same as the previous problem: If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute ss.

Now it is transformed into our previous problem where total demand during lead time is variable

z = 1.28z = (X- μ )/σX= μ +z σμ = 320σ = 20X= 320+ 1.28 × 20X= 320 + 25.6X= 320 + 26ROP = 346SS = 26


e) If demand per day is 20 units and lead time is 16 days and standard deviation of lead time is 4 days. Compute ROP at 90% service level. Compute ss.

What is the average demand during the lead timeWhat is standard deviation of demand during lead time


Demand Fixed, Lead Time variable

: Lead timeLT

: Average Lead timeLT

:Standard deviation of Lead TimeLT

If lead time is variable and demand is fixed

Average demand during lead time : ( )LT d

: dLT LTd

: Demandd

Lead Time Variable, Demand fixed

e) Demand of sand is fixed and is 20 units per day. The average lead time is 16 days. Standard deviation of lead time is 4 days. Service level is 90%.

Service level ; 90% z = 1.28

LTzdLTdROP :

)4)(20(28.11620: ROP

20:d 16:LT 4:LT

4.102320: ROP

a) if average demand per day is 20 units and lead time is 10 days. assuming zero safety stock. ss=0,...

Documents