Lot size/Reorder level (Q,R) Models Recap: Basic EOQ
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Transcript of Lot size/Reorder level (Q,R) Models Recap: Basic EOQ
1
© E
sma
Gel
, Pın
ar K
eski
noca
k, 2
013
Lot size/Reorder level (Q,R) Models
ISYE 3104 – Fall 2013
Recap: Basic EOQ
T 2T 3T 4T time
Inventory I(t)
d
QT
Q
Place an order when the inventory level is R. The order arrives after time periods Q was the only decision variable R could be computed easily because D was deterministic
R=d
Lead time
-d
2
Uncertain demand
Both Q and R are decision variables Cycle time is no longer constant!
Inventory
RQ
Q
s1s2
Q
s3
T1 T2 T3
time
(Q,R) Decisions
We choose R to meet the demand during lead time Service levels: Protect against uncertainties in
demand (or lead time) Balance the costs: stock-outs and inventory
Tradeoff in Q: Fixed cost versus holding cost
Objective: Minimize
fixed cost + holding cost + stockout (backorder) cost
3
Demand during lead time
Inventory
Time
Reorderlevel
What will happen if demandfollows one of these patterns?
Stockout
Excess inventoryR
Demand during lead time
Inventory
Time
Often the probability distributionof demand during lead time follows a Normal pattern
R
Expected demandduring lead time
R
P(D>R)=Probability of stockout
4
(Q,R) Model Assumptions Continuous review Demand is random and stationary. Expected demand is d per
unit time. Lead time is Costs
K: Setup cost per order h: Holding cost per unit per unit time c: Purchase price (cost) per unit p: Stockout (backorder) cost per unit
Demand during lead time is a continuous random variable Dwith pdf (density function) f(x) and cdf (distribution function) F(x) Mean= and standard deviation=
(Q,R) Model – Expected total cost per unit time
)()()()()()(0)(
0 shortage
shortage
cycleper shortage Expected )(
leadtime) during demand (expected level)(Reorder
arrivesorder an before levelinventory Average
:Recap)(
2)(
0
cost Shortagecost Fixedcost Holding
zLdxxfRxdxxfRxdxxfRn
RD
D-RRD
Rn
R-
s
d
QT
T
Rnp
T
KQshQC
R
R
R
Standard lossfunction
5
(Q,R) Model – Expected total cost per unit time
)()()()()()(0)(
0 shortage
shortage
cycleper shortage Expected )(
leadtime) during demand (expected level)(Reorder
arrivesorder an before levelinventory Average
:Recap)(
2)(
0
cost Shortagecost Fixedcost Holding
zLdxxfRxdxxfRxdxxfRn
RD
D-RRD
Rn
R-
s
d
QT
T
Rnp
T
KQshQC
R
R
R
Standard lossfunction
(Q,R) Model – Expected total cost per unit time
)()()()()()(0)(
0 shortage
shortage
cycleper shortage Expected )(
leadtime) during demand (expected level)(Reorder
arrivesorder an before levelinventory Average
:Recap)(
2)(
0
cost Shortagecost Fixedcost Holding
zLdxxfRxdxxfRxdxxfRn
RD
D-RRD
Rn
R-
s
d
QT
T
Rnp
T
KQshQC
R
R
R
Standard lossfunction
Same expression as the “expected number of stockouts” in the newsvendor model (Q replaced by R)
6
(Q,R) Model – Expected total cost per unit time
pd
QhRF
Q
RFpdh
R
Gh
RpnKdQ
Q
RpnKdh
Q
Rpdn
Q
Kdh
Q
G
Q
Rndp
Q
dKdR
Qh
QG
)(1))(1(
)]([2
)]([
20
)(
2
)(
2
cost Shortage cost Fixed cost Holding )(
222
(Q,R) Model – Expected total cost per unit time
pd
QhRF
h
RnpKdQ
Q
Rndp
Q
dKdR
Qh
QC
1)()]([2
:solution Optimal
)(
2
cost Shortage cost Fixed cost Holding )(
1 2
How do we pull Q and R from these equations? Solve iteratively!!
7
Solving for optimal Q and R
Start with a Q0 value and iterate until the Q values converge
Q0=EOQ
R1Q1
R2Q2
Q3 R3
R4
2
1
2
1
Remember: To find Q, you need n(R) = L(z)Lookup for z in the Normaltables
Example – Rainbow Colors
Rainbow Colors paint store uses a (Q,R) inventory system to control its stock levels. For a popular eggshell latex paint, historical data show that the distribution of monthly demandis approximately Normal, with mean 28 and standard deviation 8. Replenishment lead time for this paint is about 14 weeks. Each can of paint costs the store $6. Although excess demands are backordered, each unit of stockoutcosts about $10 due to bookkeeping and loss-of-goodwill. Fixed cost of replenishment is $15 per order and holding costs are based on a 30% annual interest rate. What is the optimal lot size (order quantity) and reorder level? What is the expected inventory level (safety stock) just before
an order arrives?
8
Example – Rainbow Colors Input
Monthly demand Normal mean=28 std.dev.=8 =14 weeks c=$6, p=$10, K=$15 h=ic=(0.3)(6)=$1.8/unit/year
Computed input d= ? (Expected annual demand) Expected demand during lead time
Variance of demand during lead time
?2
?
Example – Rainbow Colors Input
Monthly demand Normal mean=28 std.dev.=8 =14 weeks c=$6, p=$10, K=$15 h=ic=(0.3)(6)=$1.8/unit/year
Computed input d=(28)(12)=336 units/year (Expected annual demand) Expected demand during lead time
Variance of demand during lead time
38.1477.20652
14768demand timelead of Variance
768)8)(12( varianceAnnual 2
units90weeks)14(ar weeks/ye52
units/year336
9
Example – Rainbow Colors Input
Monthly demand Normal mean=28 std.dev.=8 =14 weeks c=$6, p=$10, K=$15 h=ic=(0.3)(6)=$1.8/unit/year
Computed input d=(28)(12)=336 units/year (Expected annual demand) Expected demand during lead time
Variance of demand during lead time
38.1477.20652
14768demand timelead of Variance
768)8)(12( varianceAnnual 2
units90weeks)14(ar weeks/ye52
units/year)12)(28(
Example – Rainbow Colors Input
Monthly demand Normal mean=28 std.dev.=8 =14 weeks c=$6, p=$10, K=$15 h=ic=(0.3)(6)=$1.8/unit/year
Computed input d=(28)(12)=336 units/year (Expected annual demand) Expected demand during lead time
Variance of demand during lead time
38.1477.20652
14768demand timelead of Variance
768)8)(12( varianceAnnual 2
units90weeks)14(ar weeks/ye52
units/year)12)(28(
As the lead time increases, so does the mean and varianceof demand during lead time
Shorter lead times Less variability of demand during lead time
10
Example – Rainbow Colors
Iteration 0: Compute EOQ
758.1
)336)(15)(2(20
h
KdQ
Example – Rainbow Colors
Iteration 1: Compute R1 (given Q0) and then compute Q1 (given R1)
11590)75.1)(38.14(
)()()()( :Recap
75.1)(96.0)336)(10(
)8.1)(75(11)(
1
01
RzRR
z
zzZPRD
PRDPRF
zzpd
hQRF
Z z Standard Normal
From standardNormal table
2
Q0=75
R1=115Q1
2
1
Safety Stock
Expected Demand during Lead time
11
Example – Rainbow Colors
Iteration 1 (continued): Compute Q1 (given R1)
808.1
)]233.0)(10(15)[336)(2(
233.0)0162.0)(38.14()()(
)]([2
1
1
Q
zLRnh
RnpKdQ
1
Q0=75
R2
2
1
2
Q1=80 R1=115
Q0 and Q1 not close, continue iterations
Example – Rainbow Colors
Iteration 2: Compute R2 (given Q1) and then compute Q2 (given R2)
11590)72.1)(38.14(
72.1)(957.0)336)(10(
)8.1)(80(11)(
2
12
RzR
zzpd
hQRF
Q0=75
Q1=80 R1=115
R2=115STOP! R values converged, optimal (Q,R)=(80,115)
12
(Q,R)=(80,115) Reorder level is larger than expected demand during
lead time. Why? Optimal order quantity is larger than EOQ. Why?
Safety stock s=R-=115-90=25 units
Example – Rainbow Colors
Weekly demand~5.64 Avg cycle time T=Q/d=80/6.64=12.38 weeks. Lead time=14 weeks. Cycle time shorter than lead time
Impact of R on the costs and Q
R inventory , therefore Q
R
Total cost
Holding cost
Fixed cost
Shortage cost
115
Q
115
As R Q
13
Optimal R as a function of Q
As the order quantity increases, the reorder level decreases
Q holding cost and setup cost , therefore R so that we can bring the holding cost , although a lower R means shortage cost
pd
QhRF 1)(
The Impact of Holding Cost on the Optimal (Q,R)
i Q R0.2 97 1160.3 81 1150.4 71 1140.5 64 1130.6 59 1120.7 55 111
As h goes up, both Q and R go down, but in this exampleQ drops at a faster rate!
14
The Impact of Stockout Cost on the Optimal (Q,R)
As p goes up, Q goes ??? and R goes ???
The Impact of Stockout Cost on the Optimal (Q,R)
p Q R2 84 1016 81 11110 81 11514 81 11718 80 11822 80 120
As p goes up, Q goes down and R goes up!
15
Summary: (Q,R) Models
Balance between holding cost, setup/fixed cost, and shortage cost To save on the shortage cost, we want large R To save on the holding cost, we want small Q and
small R To save on the fixed cost, we want large Q
Choose Q and R to strike a good balance among these three costs!!!
© E
sma
Gel
, Pın
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eski
noca
k, 2
013
Service Levels in (Q,R) Models
ISYE 3104 – Fall 2013
16
Service objectives
Type I service level () The proportion of cycles in which no stockouts occur Example: 90% Type I service level There are no
stockouts in 9 out of 10 cycles (on average)
Type II service level (fill rate, ) Fraction of demand satisfied on time
Service objectives - Example
Order cycle Demand Stock-outs1 180 02 75 03 235 454 140 05 180 06 200 107 150 08 90 09 160 010 40 0
TOTAL: 1450 55
Fraction of periodswith no stock-outs = 8/10Type I service = 80%
= 0.8
Fraction of demand satisfied on time = (1450-55)/1450=0.96 Type II service = 96%
= 0.96
In general, is it easier to achieve an x% Type I service or Type II service level?
17
Type I service level,
: Long-run average proportion of cycles with no stock-outs : Probability of having no stock-outs in a cycle : Probability of having no stock-outs during lead time : Probability that demand during lead time is less than R !!!
)()()(:Recap
)(
zzZPRD
PRDP
RDP
Set Q=EOQ Find z that satisfies (z)= Set R=z+ (safety stock + expected demand during lead time)
Type I service level,
: Long-run average proportion of cycles with no stock-outs : Probability of having no stock-outs in a cycle : Probability of having no stock-outs during lead time : Probability that demand during lead time is less than R !!!
)()()(:Recap
)(
zzZPRD
PRDP
RDP
Set Q=EOQ Find z that satisfies (z)= Set R=z+ (safety stock + expected demand during lead time)
18
Type I service level,
: Long-run average proportion of cycles with no stock-outs : Probability of having no stock-outs in a cycle : Probability of having no stock-outs during lead time : Probability that demand during lead time is less than R !!!
)()()(:Recap
)(
zzZPRD
PRDP
RDP
Set Q=EOQ Find z that satisfies (z)= Set R=z+ (safety stock + expected demand during lead time)
Why is Q=EOQ optimal in this case?
Example – Rainbow Colors
Rainbow Colors paint store uses a (Q,R) inventory system to control its stock levels. For a popular eggshell latex paint, historical data show that the distribution of monthly demandis approximately Normal, with mean 28 and standard deviation 8. Replenishment lead time for this paint is about 14 weeks. Each can of paint costs the store $6. Although excess demands are backordered, each unit of stockoutcosts about $10 due to bookkeeping and loss-of-goodwill. Fixed cost of replenishment is $15 per order and holding costs are based on a 30% annual interest rate. What is the optimal lot size (order quantity) and reorder level? What is the expected inventory level (safety stock) just before
an order arrives?
19
Example – Rainbow Colors
Rainbow Colors is not sure whether the $10 estimate for the shortage cost is accurate. Hence, they decided to use a service level approach.What are the optimal (Q,R) values if they want to
achieve no stockouts in 90% of the order cycles? satisfy 90% of the demand on time?
Example – Rainbow Colors Input
Monthly demand Normal mean=28 std.dev.=8 =14 weeks, c=$6, K=$15 h=Ic=(0.3)(6)=$1.8/unit/year = 0.9 or = 0.9
Computed input d=(28)(12)=336 units/year (Expected annual demand) Expected demand during lead time
Variance of demand during lead time
38.1477.20652
14768demand timelead of Variance
768)8)(12( varianceAnnual 2
units90weeks)14(year / weeks52
year / units)12)(28(
20
Rainbow Colors – Type I service
Find (Q,R) to have 90% Type I service level Q=EOQ=75 (z)= = 0.9 z=1.28 R= z+ R=(14.38)(1.28)+90=108 For 90% Type I service level (Q,R)=(75,108)
Remember: With unit penalty cost of $10, we found (Q,R)=(80,115). What is the Type I service level that corresponds to (Q,R)=(80,115)?
R= z+ 115=(14.38)z+90 z=1.7385 (1.7385)=0.96 96% Type I service level when (Q,R)=(80,115)
Type II service level
: Fraction of demand met on time 1- : Fraction of demand not met on time (stock-outs)
Recap:
Q
Rn
Q
Rn
d
QT
Q
Rnd
T
Rn
)(1
)(
unit timeper demand Expected
unit timeper stockouts of # Expected1
since)()(
unit timeper stockouts of # Expected
With this information, for a given (Q,R), we can compute .
4
21
Rainbow Colors
For 90% Type I service level we found (Q,R)=(75,108)
What is the Type II service level which corresponds to this policy?
99.00097.075
7276.0)(1
7276.0)0506.0)(38.14()25.1()()(
25.138.14
90108
Q
Rn
LzLRn
zR
The same policy results in 90% Type I service and 99% Type II service!!
Finding the optimal (Q,R) for a desired Type II service level
pd
QhRF
h
RnpKdQ
p
1)()]([2
:cost out -stock
have hen wesolution w Optimal:Remember
1 2
22
Finding the optimal (Q,R) for a desired Type II service level
QRn
RF
Rn
h
Kd
RF
RnQ
p
RFd
Qhp
pd
QhRF
h
RnpKdQ
p
)1()( usly with simultaneo solved be To
)(1
)(2
)(1
)(
: 1 into Substitute
)(1: 2 From
1)()]([2
:cost out -stock have hen wesolution w Optimal
2
1 2
3
4
5 Imputed shortage cost
Impact of service level on R
For a given Q As n(R)=(1- )Q i.e., R
As the service level , the reorder level as well
23
Finding the optimal (Q,R) for a desired Type II service level
QRnRF
Rn
h
Kd
RF
RnQ )1()(
)(1
)(2
)(1
)(2
3 4
Q0=EOQ
R1Q1
R2Q2
Q3 R3
R4
4
4
4
4
3
3
3
Start with a Q0 value and iterate until the Q values (or the R values) converge
Example – Rainbow Colors
Iteration 0: Compute EOQ
758.1
)336)(15)(2(20
h
KdQ
24
Example – Rainbow Colors
Iteration 1: Compute R1 (given Q0) and then compute Q1 (given R1)
Q0=75
R1=87Q1=89
4
3
89
5871.0)22.0()22.0(1
table.Normal at theLook ).(1 need we find To
8783.8690)22.0)(38.14(
22.05216.0)(
)(5.7)75)(9.01()1()(
1
11
1
01
Q
FF
R-FQ
zR
zzL
zLQRn
4
3
Example – Rainbow Colors
Iteration 2: Compute R2 (given Q1) and then compute Q2 (given R2)
25
Example – Rainbow Colors
Iteration 2: Compute R2 (given Q1) and then compute Q2 (given R2)
90
648.0)38.0()38.0(1
table.Normal at theLook ).(1 need we find To
855.8490)38.0)(38.14(
38.0619.0)(
)(9.8)89)(9.01()1()(
2
11
2
12
Q
FF
R-FQ
zR
zzL
zLQRn
4
3
Q0=75
R1=87Q1=89
4
3
R2=85Q2=904
3
Example – Rainbow Colors
Iteration 3: Compute R3 (given Q2) and then compute Q3 (given R3)
851.8590)34.0)(38.14(
34.0591.0)(
)(5.8)85)(9.01()1()(
3
23
zR
zzL
zLQRn4
Q0=75
R1=87Q1=89
4
3
R2=85Q2=904
3
R3=854STOP! R values converged, optimal (Q,R)=(90,85)