3 single-echelon systems with independent items different items can be controlled independently. the...
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3 SINGLE-ECHELON SYSTEMS WITH INDEPENDENT ITEMS
Different items can be controlled independently.
The items are stocked at a single location, i.e., not in a multi-stage inventory system.
3.1 Costs
Holding costs
- Opportunity cost for capital tied up in
inventory
- Material handling costs
- Costs for storage
- Costs for damage and obsolescence
- Insurance costs
- Taxes
Ordering or Setup Costs
- setup and learning
- administrative costs associated with the
handling of orders
- transportation and material handling.
- extra costs for administration - price discounts for late deliveries - material handling and transportation.
If the sale is lost, the contribution of the sale is also lost. In any case it usually means a loss of goodwill.
- component missing - rescheduling, etc.
Because shortage costs are so difficult to estimate, it is very common to replace them by a suitable service constraint.
Shortage Costs or Service ConstraintsShortage Costs or Service Constraints
3.2 Different Ordering Systems
3.2.1 Inventory position
Inventory position = stock on hand + outstanding orders - backorders.
Inventory level = stock on hand - backorders.
3.2.2 Continuous or Periodic Review
As soon as the inventory position is sufficiently low, an order is triggered. We denote this continuous review.
L = lead-time.
T = review period, i. e., the time interval
between reviews.
3.2.3 Different Prdering Policies (R, Q) policy
Figure 3.1 (R, Q) policy with continuous review. Continuous demand.
R
R+Q
LL
Inventory position
Inventory level
(R, Q) Policy
When the inventory position declines to or below the reorder point R, a batch quantity of size Q is ordered. (If the inventory position is sufficiently low it may be necessary to order more than one batch to get above R) .
(s, S) Policy
Figure 3.2 (s, S) policy, periodic review.
When the inventory position declines to or below s, we order up to the maximum level S.
LL
Inventory position
Inventory level
Time
3.3.1 Classical Economic Order Quantity Model
Harris (1913), Wilson (1934), Erlenkotter (1989)
- Demand is constant and continuous. - Ordering and holding costs are constant over time. - The batch quantity does not need to be an integer. - The whole batch quantity is delivered at the same time. - No shortages are allowed.
H = holding cost per unit and time unit
A = ordering or setup cost
D = demand per time unit
Q = batch quantity
C = costs per time unit
Notation:
Figure 3.3 Development of inventory level over
time.
Time
Stock level
Q
Q/d
(3.1)
(3.2)
(3.3)
(3.4)
How important is it to use the optimal order quantity? From (3.1), (3.3), and (3.4)
(3.5)
AQ
dh
QC
2
02 2
AQ
dh
dQ
dC
h
AdQ
2*
Q
Q
Q
Q
h
Ad
QAd
hQ
C
C *
** 2
12
2
1
22
AdhAdhAdh
C 222
*
If Q/Q*= 3/2 (or 2/3), then C/C* = 1.08
from (3.5) The cost increase is only 8 percent.Costs are even less sensitive to errors in
the cost parameters. For example, if ordering cost A is 50 percent higher than correct ordering cost, from (3.3), batch quantity is Q/Q*= (3/2)1/2 =1.225 , relative cost increase 2 percent.
Example 3.1 A = $200, d = 300,
unit cost $100, holding cost is 20
percent of the value. Then,
h = 0.20 100 =$20
Applying (3.3), Q* = (2Ad/h)1/2 =
(2 200 300/20)1/2 = 77.5 units.
In practice Q often has to be an integer.
3.3.2 Finite Production Rate If there is a finite production rate, the
whole batch is not delivered at the same time
Figure 3.4 Development of the inventory level over
time with finite production rate.
Time
Stock level
Q/d
Q/p
Q(1-d/p)
Q
P = production rate (p > d).
The average inventory level is now
Q(1 - d/p)/2
. (3.6)
. (3.7))/1(
2*
pdh
AdQ
AQ
dh
pdQC
2
)/1(
3.3.4 Quantity Discounts v = price per unit for Q < Q0, i.e., the normal price,
v´= price per unit for Q Q0, where v´ < v.
Holding cost
h = h0 + rv for Q < Q0,
h´= h0 + rv´ for Q Q0,
3.3.3 More General Models Example 3.2
d: constant customer demand
Two machine production rates: p1 > p2 > d,
Two machine set up costs: A1 and A2
The fixed cost of the transportation of a batch of goods from machine 2 to the warehouse: A3
The holding cost before machine 1 is h1 per unit and time unit. The holding cost is h2 and after machine 2 it is h3.
Optimal Batch size?
Time
Q/p1Q/d
Q
Stock level between machines 1 and 2 (2)
Time
Time
Time
Q/p1
Q/p2
Stock level at warehouse (4)
Stock level after machine 2 (3)
Q/p2
Q/dTransportation
1
M2M1
4
3
2
Stock level before machine 1 (1)
Holding Cost Calculation:
……
(3.8)
(3.9)
hp
dh
dQ
pQQ
1
1
/2
)/(
)11
(2
)(2/2
)1(
1221
2121
2
pp
Qd
pp
ppQdh
dQ
pQ
pp
Q
)AAA(Q
dh)1
p
d(h)
p
d
p
d(h
p
d
2
QC 3213
22
121
1
32
212
11
321*
h)1p
d(h)
p
d
p
d(h
p
d
d)AAA(2Q
00 )(2
QQforAQ
drvh
QdvC (3.10)
(3.11)
00 )(2
QQforAQ
dvrh
QvdC
Q
C
Q0Q´´Q´
Figure 3.6 Costs for different values of Q.
Two steps 1. First we consider (3.11) without the
constraint Q Q0. We obtain
(3.12)
and
. (3.13)
If Q´ Q0, (3.12) and (3.13) give the optimal
solution, i.e., Q* = Q´, and C* = C´.
vrh
AdQ
0
2
vdvrhAdC )(2 0
Example 3.3 v = $100, v´ = $95 for
Q Q0 = 100. h0 = $5 per unit and year,
and r = 0.2, i.e., h = $25 and h´= $24
per unit and year. d = 300 per year, A = $200. From (3.12) and (3.13), Q´= 70.71 and
C´= 30197. Since Q´< Q0, go to step 2.
From (3.14) and (3.15), Q´´ = 69.28 and
C´´ = 31732. Applying (3.16), C(100) = 30300. Q* = Q0 = 100.
2. If Q´ < Q0 we need to determine
. (3.14)
. (3.15) Since v > v´ we know that Q´´ < Q´< Q0
. (3.16) The optimal solution is the minimum of (3.15)
and (3.16).
rvh
AdQ
0
2
dvrvhAdC )(2 0
AQ
dvrh
QvdQC
00
00 )(
2)(
Incremental Discounts
A3
A2
A1
Q1=1000 Q2=2500
v1=1
v2=0.8
v3=0.7
d =36500 , r =0.3
A1=A=15
A2=A1 +(c1-c2)Q1=15+(1-0.8)*1000=215
A3=A2 +(c2-c3)Q2=215+(0.8-0.7)*2500=465
EOQi=
Procedure Candidate oi from each segment
EOQi if feasible
Oi = Qi if EOQi > Qi
Qi-1 if EOQi < Qi-1
Q* =best of all Oi
rv
dA
i
i2
Example EOQ1=1910 > Q1 O1=Q1=1000
EOQ2= =8087 O2=Q2=2500
EOQ3= =12714 feasible
8.0*3.0
215*36500*2
7.0*3.0
465*36500*2
TC(Qi) =vid+Qivir /2+dAi / Qi
TC(1000)=36500*1+1000*0.5*0.3*1+36500*15/1000
=36500+150+547.5=37197.5
TC(2500)=(1000*1+1500*0.8)*36500/2500 +2500*0.5*0.3*(1000*1+1500*0.8)/2500
+36500*15/2500=32120+330+219=32669
TC(12714)=[1000*1+1500*0.8
+(12714-2500)*0.7+15]*36500/12714
=26884.9+1404.7=28289.6
Q*=12714
3.3.5 Backorders Allowed b1= shortage penalty cost per unit and time unit.
x = fraction of demand that is backordered.
Figure 3.7 Development of inventory level over time
with backorders.
Time
Inventory level
Q(1-x)
Q/d
-Qx
. (3.17)
, (3.18)
and inserting in (3.17) we get
. (3.19)
, (3.20)
.(3.21)
AQ
db
Qxh
xQC
1
22
22
)1(
1
*
bh
hx
AQ
d
bh
hbQxC
1
1*
2)(
1
1* )(2
hb
bhAdQ
)bh(
hbAd2C
1
1*
Example 3.4
demand = 1000 units per year,
production rate = 3000 units per year,
holding cost before the machine = $10 per unit
and year,
holding cost after the machine =$15 per unit and
year,
shortage cost = $75 per unit and year,
ordering cost = $1000 per batch.
The optimal solution is x* = 15/90 = 1/6, and Q* = 310.
Stock level before the machine
Time
Stock level after the machine
Time
Stock level at warehouse
Time
Q/3000 Q/1000
Q/1000
Q/1000Q/3000
Q
Q(1-x)
-Qx
Q
1000Q
100075x
2
Q15)x1(
2
Q)1510(
6
QC 22
3.3.6 Time-varying Demand
T = number of periods
Di = demand in period i, i = 1,2,...,T, (Assume
that d1 > 0, since otherwise we can just
disregard period 1)
A = ordering cost,
h = holding cost per unit and time unit.
A replenishment must always cover the demand in an integer number of consecutive periods.
The holding costs for a period demand should
never exceed the ordering cost.
Case: when backorders are not allowed
3.3.7 The Wagner-Whitin Algorithm
fk=minimum costs over periods 1, 2, ..., k, i.e.,
when we disregard periods k + 1, k + 2, ..., T,
fk,t=minimum costs over periods 1, 2, ..., k, given that
the last delivery is in period t (1 t k).
, (3.22)
. (3.23)
tkkt
k ff ,1
min
))(...2( 211, kttttk dtkddhAff
Example 3.6
T =10, A = $300, h = $1 per unit and period.
Table 3.1 Solution, fk,t , of Example 3.6.
Period tdt
150
260
390
470
530
6100
760
840
980
1020
k=t 300 600 660 840 1030 1090 1370 1450 1530 1770
k=t+1 360 690 730 870 1130 1150 1410 1530 1550
k=t+2 540 830 790 1070 1250 1230 1570 1570
k=t+3 750 920 1090 1250 1370 1470 1630
k=t+4 870 1330 1410 1550
k=t+5 1530
Table 3.2 Optimal batch sizes in
Example 3.6.
Period t 1 2 3 4 5 6 7 8 9 10
Solution 1 110 190 200 100
Solution 2 110 190 300
Rolling horizon
Whether it is possible to replace an infinite horizon by a sufficiently long finite horizon such that we still get the optimal solution in the first period.
DYNAMIC DETERMINISTIC MODELS
- NONSTATIONARY DEMAND (POS OR NEG)- ADDITIONAL MOTIVES FOR HOLDING INVENTORY SMOOTHING,SPECULATIVE- DISCRETE TIME MODELS FOR COMPUTATIONAL REASONS- FINITE HORIZON - INITIAL DECISION SHOULD NOT BE VERY SENSITIVE TO THE HORIZON LENGTH T
IF T 'LARGE'.- ROLLING PROCEDURE - USE THESE INITIAL DECISIONS, REPEAT THE PROCESS USING UPDATED FORECASTS IF AVAILABLE.
FORWARD ALGORITHM
AN EFFICIENT PROCEDURE FOR SOLVING LONGER AND LONGER PROBLEMS.
STOPPING RULE: PROCEDURE TO STOP.DECISION HORIZON IS A STOPPING RULE.(WILL ELABORATE ON IT LATER.)
FIG. 2. NONSTATIONARY DEMAND
HOLDING COSTS ARE FLOWS h(t)SETUP COSTS ARE LUMP SUM AMOUNTS A(ti)LET S BE A POLICY AS SHOWN IN FIG. 2
TOTAL DISCOUNTED COST
irt
ii
rt etAdtethSV
0
1
)()()(
Q
tt1 t2 t3 …
GENERAL PROBLEM MIN V(S) (VERY COMPLICATED) S
ALSO V(S) = FOR ALL S, THEN WHAT?
1ST SIMPLIFICATION
- DISCRETE TIME- PRODUCTION AT THE BEGINNING OF THE PERIOD.- HOLDING COST BASED ON AVERAGE INVENTORY IN EACH PERIOD- FINITE TIME
T
i
ir
i etASV1
)1()()(
FUDGE THE DISCOUNT RATE INTO INTEREST EXPENSE
T
i
iASV1
)()(
T
i
iAT
SV
COSTAVERAGE
1
)(1
)(
2ND SIMPLIFICATION
- DYNAMIC PROGRAMMING INSTEAD OF THE BRUTE FORCE METHOD OF EXAMINING EVERY S.
- STRUCTURE OF THE PROBLEM; IT IS POSSIBLE TO SHOW IN THE DYNAMIC LOT SIZE MODEL THAT IT IS
OPTIMAL TO PRODUCE ONLY WHEN INVENTORY IS ZERO. (WAGNER-WHITIN ALGORITHM)
WHY FINITE T ?
1) IT IS CONVENIENT MAKES COMPUTATIONS FEASIBLE DOES NOT REQUIRE ALL FUTURE FORECAST REAL WORLD PROBLEMS ARE FINITE BUT T IS
USUALLY UNKNOWN: ALSO THE SALVAGE VALUE IS UNKNOWN.
2) IT IS REASONABLE ROLLING HORIZON PROCEDURE SOLVE WITH T=40 AT t=0. USE Q1, Q2 , Q3 , Q4 DECISIONS. AT t=4, SOLVE A 40-PD PROBLEM FOR [5, 44].
USE Q5 THRU Q8 AS 'OPTIMAL' DECISIONS: COMPUTATIONAL TIME CONSIDERATIONS.
. - AT THE VERY LEAST, WE MUST BELIEVE THAT THE LARGER. THE VALUE OF T, THE LESS WILL BE THE INFLUENCE OF FINAL PERIODS ON THE INITIAL DECISIONS.
- SMALL VALUES OF T ARE COMPUTATIONALLY CONVENIENT; LARGE T ARE SAFE IN GETTING GOOD ANSWERS.
- A COMPROMISE: FORWARD ALGORITHM SOLVE PROBLEMS FOR T =1,2,.. .,40. IF SOLVING THESE FORTY PROBLEMS
REQUIRE LITTLE OR NO MORE WORK THAN SOLVING THE 40-PD PROBLEM, THEN SUCH A PROCEDURE IS CALLED A FORWARD ALGORITHM.
STOPPING RULE
A PROCEDURE WHICH CHECKS WHETHER OR
NOT THE INITIALDECISION IS A 'GOOD ENOUGH‘
APPROXIMATION EVERY TIME T IS INCREASED BY ONE
PERIOD.
A) APPARENT DECISION HORIZON: Q1 HAS CHANGED LITTLE OR NONE FOR THE LAST FEW VALUES OF T.
B) DECISION HORIZON: Q1 CAN BE GUARANTEED
NEVER TO CHANGE IF T WERE FURTHER
INCREMENTED (INDEPENDENTOF DEMAND IN PDS AFTER T).
STOPPING RULE (CONTINUED)
C) NEAR-COST DECISION HORIZON: Q1 CAN BE
GUARANTEED TO BE WITHIN 5% OF THE
OPTIMAL COST.
D) NEAR-POLICY DECISION HORIZON: Q1 CAN BE
GUARANTEED TO WITHIN 5% OF Q1*.
FORECAST HORIZON, DECISION HORIZON
SUPPOSE AFTER SOLVING A FORWARD ALGORITHM OUT TO T, WE CAN GUARANTEE THAT DECISIONS FOR THE FIRST J PERIODS ARE CORRECT FOR ANY (T +K) - PROBLEM, K ≥ 1 (INDEPENDENT OF DEMAND IN T+1, T+2, ... PDS), THEN THE FIRST T PERIODS IS CALLED A FORECAST HORIZON WHILE THE FIRST J PERIODS ARE CALLED A DECISION HORIZON.
DEMAND CAN BE NETTED OUT AS SHOWN. (EXCESS OVER 'SAFETY STOCKS') THUS I0 = 0
NET DEMAND dl d2 d3 . . .
Qt PRODUCTION AT THE BEGINNING OF t TH
PERIOD.
A + cQt Qt > 0
COST = 0 OTHERWISE
ENDING INVENTORY It = It-1 + Qt - dt, t > 0
It = AVERAGE INVENTORY
= It + ½dt = It-1 + Qt - ½dt
][
1tt
T
tt IhcQAVCOST
THEOREM: ANY OPTIMAL SOLUTION MUST SATISFY
I) It Qt+1 = 0 (I.E. Qt+1 > 0 It = 0
It > 0 Qt+ 1 = 0)
II) IT = 0
PROOF.
FIRST PRODUCTION WILL CERTAINLY NOT COME UNTIL APERIOD (t+1) FOR WHICH dt+l > O. FOR ANY FOLLOWING PRODUCTION PT. IF (I) DOES NOT HOLD, THEN DECREASE PRECEDING PRODUCTION BY It AND INCREASE Qt+l BY It GIVING A BETTER SOLUTION.
EXERCISE: COMPLETE THE PROOF IF THE
PREVIOUS PRODUCTION IS LESS THAN It.
ALSO SHOW IT MUST BE ZERO.
IF It = 0, THEN PERIOD t IS CALLED A REGENERATION POINT.
IF Qt > 0, THEN PERIOD t IS CALLED A PRODUCTION POINT.
REGENERATION - PRODUCTION ALTERNATION PROPERTY
I) BETWEEN ANY TWO SUCCESSIVE P-POINTS, THERE IS
AT LEAST ONE R-POINT.
II) BETWEEN ANY TWO SUCCESSIVE R-POINTS, THERE IS
AT MOST ONE P-POINT.
(CONVENTION: IF PERIOD t IS BOTH R AND P POINT, THEN P IS SAID TO BE BEFORE R.)
PROOF: I) LET t AND t+k BE SUCCESSIVE P-POINTS.
THEN Qt > 0, Qt+k > 0 => It+k-1 = 0 BY THEOREM. I. E., t+k-1 IS AN R POINT. ALSO NOTE THAT MORE R-POINTS POSSIBLE IF THERE ARE PERIODS WITH NO DEMAND. II) , WHICH R P P R CONTRADICTS WITH (I). NOTE: NO P-POINT CAN ONLY
OCCUR IF TWO SUCCESSIVE R-POINTS ARE SEPARATED BY ZERO DEMAND.
EXERCISE: CONSTRUCT SITUATIONS
P R R P R R
FOR A COMPLETE SOLUTION, WE NEED TO KNOW
- THAT K IS AN R-POINT
- OPTIMAL SOLUTION FOR PERIOD K+1 TO T.
BOTH OF THESE ARE EASILY TAKEN CARE OF.
NOTES I) 0 AND T ARE R-POINTS
II) EASY TO OBTAIN AN OPTIMAL SOLUTION
BETWEEN TWO SUCCESSIVE R-POINTS.
R R
III) SIMPLIFY THE COST FUNCTION Qt = dt ,
AND THUS THE PORTION OF COSTS GIVEN BY
('SUNK' COST) = IS CONSTANT.
REDUCED COST C(T) = At + h It
FURTHER, LET t=12 AND t=9 BE THE NEXT TO THE LAST R~POINT. THEN THE COST C(9,12).
tt dhQc2
1
-D10
GENERAL FORNULA
WHERE j AND n ARE TWO SUCCESSIVE R-POINTS.
E.G. C(8,12) = A + h(d10 + 2d11+ 3d12)
C(8,13) = C(8,12) + h (4d13)
C(T) =
Cj(T) = OPTIMAL T PERIOD COST IF j IS PERCURSOR
R-POINT OF T
= C(j) + C(j,T)
OPT. COST FROM COST FROM j TO T
t =0 TO t =j WHERE j PRECEDES
(IMMEDIATELY) T.
n
jkkdjkhAnjC
1
)1(),(
)()1...(1,0
TCMIN jTj
NOTE ALSO
CT-1(T) = C(T-1) + A
Cj(T) = Cj(T-1) + (T-j-1) h dT, 1 ≤ j < T – 1
II II II
C(j) + C(j,T) = C(j) + C(j,T-1) + (T-j-1)hdt
ALGORITHM
LET j*(T) BE IMMEDIATELY PRECEDING R-POINT
FOR THE OPTIMAL SOLUTION TO THE T-PERIOD
PROBLEM. LET f*(T) BE THE FIRST
1) SET T = 0, C(0) = 0, j*(0) = -1, f*(0) = 0
(UNDEFINED)
2) SAVE C(T), j*(T), f*(T), Cj(T), 0 ≤ j ≤ T -1
(UNDEFINED FIRST TIME)
3) INCREASE T BY 1
4) CT-1(T) = C(T-1) + A
Cj(T) = Cj (T-1) + (T-j-1)hdT , 0 ≤ j < T -1
(DON'T USE FIRST TIME)
5)
6) SAVE THE LARGEST VALUE OF j* WHICH MINIMIZES AS j*(T)
7) IF j*(T) = 0, THEN f*(T) = T
IF j*(T) > 0, THEN f*(T) = f*(j*(T))
8) GO TO STEP 2.
)]([)()1...(1,0
TCMINTC jTj
REMARKS
I) T = 0 IS JUST INITIALIZATION. WE DO NOT
BOTHER TO WRITE IT DOWN.
II) BOTTOM ELEMENT IN EACH COLUMN IS ALWAYS
OBTAINED BY ADDING A TO THE CIRCLED ITEM IN THE
COLUMN TO THE LEFT.
III) REMAINING ITEM, WORKING UP THE COLUMN, ARE
OBTAINED BY ADDING hdT TO THE ITEM DIRECTLY TO
THE LEFT, NEXT ITEM ADDING 2hdT TO THE ITEM
DIRECTLY TO ITS LEFT, THEN 3hdT , 4hdT AND SO ON.
IV) f* GIVES US OUR FIRST DECISION: PRODUCE ENOUGH TO RUN OUT BY THE END OF f*.
V) WE WILL ALSO SHOW HOW TO AVOID COMPUTING NUMBERS IN THE SHADED AREA.
VI) NOTE THAT ROW WHICH IS CIRCLED ALWAYS INCREASES AS T INCREASES. REGENERATION MONOTONICITY: j*(T) WITH T.
REGENERATION MONOTONICITY THEOREM:
FOR THE DYNAMIC LOT SIZE MODEL WITH
LINEAR COSTS
T1 > T2 j*(T1) ≥ j*(T2)
PROOF: RECALL THAT IN FORMING THE T + 1 COLUMN
WE ADD LARGER MULTIPLES OF dT+1 TO THE T-COLUMN
VALUES AS j DECREASES, THIS ALWAYS BENEFITS
LARGER j'S TO BE A MINIMUM, THUS
j*(T) = 11 AND j*(T+1) = 9 IS IMPOSSIBLE SINCE
C9(T) > C11(T) C9 (T+1) > C11 (T+1).
NOTE:
C9(T+1) = C9(T) + (T-9)hdT+1
C11(T+1) = C11(T) + (T-11)hdT+1
THUS, WE DON'T HAVE TO COMPLETE THE WHOLE
TABLE. WE START AT THE BOTTOM AND STOP
OPPOSITE TO THE CIRCLED ITEM IN THE PREVIOUS
COLUMN. NOTE THAT TOTAL ENTRIES IN THE TABLE
ARE 1/2(T)(T) = 1/2T2. IT NOW REDUCES TO kT WHERE k
IS AVERAGE NUMBER OF ENTRIES IN A COLUMN.
DECISION HORIZON COROLLARY:
LOOK AT THE TABLE FOR T = 7 COLUMN j*(7) = 5.
BY COROLLARY, AT LEAST ONE OPTIMAL SOLUTION
FOR T ≥ 8-PROBLEMS WILL HAVE AN R-POINT IN [5,6].
I. E., EITHER THE 5 OR 6-PERIOD PROBLEMS WILL
ALWAYS BE A PART OF ANY LONGER PROBLEM. SINCE
FOR THESE PROBLEMS, f*(5) = 5, f*(6) = 6. THUS, AFTER
7-PERIODS, IT WILL BE OPTIMAL TO PRODUCE 13 OR 17
IN THE FIRST PERIOD. LOOK AT T=11. j*(11) = 6 [6,7,8,9,10] IS AN R-SET FOR T ≥ 12. THUS WE STILL KNOW
f* = 5 OR f* = 6. NOW LOOK AT T = 15, j*(15) = 12.
THUS [12,13,14] IS AN R-SET FOR T ≥ 16,
BUT ALL THREE PROBLEMS (I.E. 12, 13, 14 PERIOD)
HAVE f* = 5 PRODUCING EXACTLY 13 IN PERIOD 1. THUS
WE HAVE A FORECAST HORIZON OF 15 AND DECISION
HORIZON OF 5. WE NOW HAVE FOR A T-PERIOD
PROBLEM, {j*(T), j*(T)+1, ..., (T-1)} IS AN R-SET. IF j*(T) 0,
THEN
A) f*(j*(T)) = f*(j*(T)+1) ... = f*(T-1) T IS A FORECAST
HORIZON AND f* IS A DECISION HORIZON.
B) IN GENERAL, THE MAXIMUM AND MINIMUM VALUES OF
f* GIVES BOUNDS ON THE INITIAL DECISION FOR ANY
LONGER PROBLEM.
COROLLARY: IF j*(T) IS THE NEXT TO THE LAST
R-POINT IN A T -PROBLEM, THEN { j* (T), j*(T) + 1...,
(T -1) } IS GUARANTEED TO CONTAIN AN R-POINT
OF EVERY (T+K)-PROBLEM FOR K ≥ 1 (FOR AT LEAST
ONE OPTIMAL SOLUTION).
(DEFINITION: A GROUP OF PERIODS GUARANTEED TO
CONTAIN AN R-POINT FOR ANY LONGER PROBLEM WILL
BE TERMED AN R-SET.)
PROOF:
IF j*(20) = 15 THEN {15, 16, 17, 18, 19} IS .AN R-SET.
CONSIDER T + K = 28. BY THEOREM, j*(28) [15,27].
IF j*(28) [15,19]. WE ARE DONE. IF NOT, SUPPOSE
j*(28) = 24, THUS OPTIMAL SOLUTION OF 24-PROBLEM
IS A PART OF THE OPTIMAL SOLUTION TO THE
28-PROBLEM. BY THEOREM, j*(24) [15,23]. LET
j*(24) = 21. THEN j*(21) [15,20]. IF j*(21) [15,19].
WE ARE DONE. IF NOT, THEN j*(21) = 20. BUT WE KNOW
THAT j*(20) = 15. THIS COMPLETES THE PROOF.
TV SPEAKER PROBLEM SOLUTION
d1 = 3, d2 = 2, d3 = 3, d4 = 2, A = 2, h =0.2
T = 0, C(0) = 0, j*(0) = -1, f*(0) = 0
T = 1 C(1) = = C0(1) = C(0) + A = 2
j*(1) = 0, f*(1) = 1
T = 2, C(2) = = 2.4, j*(2) = 0, f*(2) = 2
C0(2) = C0(1) + (2-0-1)hd2 = 2 + 1(.2)2 = 2.4*
C1(2) = C(1) + A = 2 + 2 = 4
T = 3, C(3) = = 3.6, j*(3) = 0, f*(3) = 3
C0(3) = C0 (2) + (3-0-1)hd3 = 2.4 + 2(.2)3 = 3.6*
C1(3) = C1(2) + 1(.2)3 = 4 + .6 = 4.6
C2 (3) = C(2) + A = 2.4 + 2 =4.4
)1(0
jj
CMIN
)2(1,0
jj
CMIN
)3(2,1,0
jj
CMIN
T = 4, C(4) =
C0(4) = C0 (3) + (4-0-1)hd2 = 3.6 + 3(.2)2 = 4.8*
C1(4) = C1 (3) + 2(.2)2 = 4.6 + .8 = 5.4
C2(4) = C2 (3) + 1(.2)2 = 4.4 + .4 = 4.8*
C3(4) = C(3) + A = 3.6 + 2 = 5.6.
NOTE A TIE IN j* SO WE CHOOSE j*(4) = 0 f*(4) = 4
IF WE CHOOSE j* = 2, THEN j*(4) = 2,
f*(4) = f*(2) = 2.
)4(3,2,1,0
jj
CMIN
3.3.8 The Silver-Meal Heuristic
,2 k n, (3.24)
. (3.25)
1k
d)1j(hA
k
d)1j(hA1k
2jj
k
2jj
n
djhA
n
djhAn
jj
n
jj
2
1
2)1(
1
)1(
Example 3.7 A = $300 and h = $1 per unit and period.
Table 3.3 Demands in Example 3.7.
If the delivery in period 1 covers only the demand in period 1, the ONLY cost for this period is A = 300. Then,
2 periods (300 + 60)/2 = 180 < 300, 3 periods (300 + 60 + 2 90)/3 = 180 180, 4 periods (300 + 60 + 2 90 + 3 70)/4 = 187.5 >
180 The same procedure is now applied with period
4 as the first period.
Period tdt
150
260
390
470
530
6100
760
840
980
1020
2 periods (300 + 30)/2 = 165 < 300, 3 periods (300 + 30 + 2 100)/3 = 176.67 > 165
Starting with period 6 as the first period 2 periods (300 + 60)/2 = 180 < 300, 3 periods (300 + 60 + 2 40)/3 = 146.67 180, 4 periods (300 + 60 + 2 40 + 3 80)/4 = 170 >
146.67
Starting with period 9 as the first period 2 periods (300 + 20)/2 = 160 < 300.
Table 3.4 Solution with the Silver- Meal
Heuristic.
In most situations the cost increase is ONLY about 1-2 %
The relative error can be arbitrarily large
Period t 1 2 3 4 5 6 7 8 9 10
Quantity 200 100 200 100
3.3.9 A heuristic that balancesholding and ordering costs
- Classical economic order quantity formula Optimal solution
- Holding costs = ordering costs First delivery quantity covers n periods, where n
is determined by
, (3.26)
The relative error is bounded by 100 percent
1n
2jj
n
2jj d)1j(hAd)1j(h
Example 3.8 A = $300 and h = $1 per unit and period.
Demand table
Starting with period 2
2 periods 60 300,
3 periods 60 + 2 90 = 240 300,
4 periods 60 + 2 90 + 3 70 = 450 > 300
Period tdt
150
260
390
470
530
6100
760
840
980
1020
period 4 as the first period.
2 periods 30 300,
3 periods 30 + 2 100 = 230 300,
4 periods 30 + 2 100 + 3 60 = 410 > 300
period 7 as the first period
2 periods 40 300,
3 periods 40 + 2 80 = 200 300,
4 periods 40 + 2 80 + 3 20 = 260 30.
Table 3.5 Solution with the heuristic.
Period t 1 2 3 4 5 6 7 8 9 10
Quantity 200 200 200
AVERAGE DEMAND: AVERAGE DEMAND FOR 10
PERIODS IS 60. IF STATIONARY, EOQ MODEL WOULD
IMPLY T* = . ROUNDING THIS TO 3
YIELD THE PLAN (1-3) (4-6) (7-9)(10).
10/2 dhA
REMARKS
I) HEURISTICS SEEM TO WORK WELL;
SUGGESTS THAT CONCLUSIONS OF STATIONARY
DEMAND MODELS MAY BE FAIRLY WELL APPLIED FOR
THE NONSTATIONARY CASE.
II) IF PRODUCTION COSTS ARE NON-STATIONARY THEN
ALL THE RESULTS GO THRU PROVIDED
ct + ht > ct+1 (I.E., IGNORING SET UP COSTS, IT IS NEVER CHEAPER
TO PRODUCE AND HOLD.) THIS PRECLUDES THE
SPECULATIVE MOTIVE.
3.4 Safety stocks and reorder points
3.4.1 Demand processesCumulative demand: nondecreasing,
stationary, independent incrementsCompound Poisson Process: probability for k
customers in a time interval of length t is
(3.27)
Both the average and the variance are equal to t.
......2,1,0,!
)()( ke
k
tkP t
k
The size of a customer demand is
independent of the distribution of the
customer arrivals.
fj=probability of demand size j (j = 1, 2, ... ),
∑ fj =1.
This assumes no demands of size zero without loss of generality (division by 1-f0).
=probability that k customers give the total demand j;
D(t)=stochastic demand in the time interval of length t.
kjf
.,0 kjf k
j
, , and the j-fold convolution
of fj:
k =2, 3, 4,....; j ≥ k. (3.28)
Note that i goes from k-1 to j-1,since for each customer minimum demand is 1. Using (3.27)
. (3.29)
100 f jj ff 1
,1
1
1ij
j
ki
ki
kj fff
0 !
)())((
k
kj
tk
fek
tjtDP
Prob of k-1 customers with total demand i and one with the demand j-i
1
1
j
k
i ff
=average demand per unit of time,
=standard deviation of the demand per unit of time
K =the stochastic number of customers during one time unit
J =the stochastic demand size of a single customer
Z = the stochastic demand during the time unit considered.
Recall that E(K) = Var(K) =
.(3.30) To determine
(3.31)Using ,
(3.32)
Items with relatively low demand, use this Poisson demand model in practice.
1)()()()}({})({)(
jjjfJEJEKEJEKEKZEEZE
}.))(()({}))(()({)}({)( 22222 JEKJKVarEKZEKZVarEKZEEZE
222 ))(()()( KEKVarKE
jj
fjJE
JEJVarJEJVar
JEKJVarKEZE
1
22
2222
222222
)(
))(()())()(()(
}))(()({)(
Items with higher demand, use a continuous distribution. If the time period is long enough, use normal
distribution. Standardized normal distribution has the density
(3.33) and the distribution function,
(3.34) For values of >0 and , the density is
,the distribution function is .
Note: from (3.33). Note: Normal distribution can give negative demand
with a small probability.
2
2
2
1)(
x
ex
duexx u
2
2
2
1)(
)/)x(()/1( )/)(( x)()(' xxx
3.4.2 Continuous review (R, Q) policy - inventory level distribution IP=inventory position.
Order Q or nQ immediately when IP ≤ R. Thus, in steady state, R + 1 IP R + Q
Proposition 3.1
In steady state, the inventory position is uniformly distributed on the integers R + 1, R + 2, ......,
R + Q.
Markov Chain:
Irreducible: all states communicate;
j transient: finite number of transitions to j;
j recurrent: infinite number of transitions to j;
Null recurrent: expected time to j is ∞ ;
Positive recurrent: expected time to j is finite;
Aperiodic: Period= 1;
Ergodic = aperiodic and positive recurrent;
unique steady state distribution with j > 0
Proof
Let pi,j = the probability for a jump from R + i to R + j. Markov chain: irreducible and ergodic. Thus, it
has a unique steady state distribution. Sufficient to show that the uniform distribution is a steady state distribution.
Uniform
j = 1, 2, ..., Q
(3.35)
,11
)( ,1 Q
pQ
jRP ji
Q
i
1,
1
ji
Q
ip
,Q
i)P(R1
But for a Markov chain. Given
demand size k, pij(D=k)=0 or 1, and for a
given j it is one for exactly one i.
So, Then,
. (3.36)With normally distributed demand, the
continuous inventory position is uniformly distributed on the interval [R, R + Q], if we can ignore the possibilities of negative demand.
Q
j jip1 , 1
k
Q
iij
k
k
Q
iij
kij
Q
i
Q
iji
kDPkDpkDP
kDPkDp
kDPkDpp
1)()()(
)()(
)()(
1
1
11,
.1)(1 kDpQi ij
L=lead time (constant) IL=inventory level D(t, t + ) = D( ) =stochastic demand in the
interval ( t, t + ]. Consider that the system has reached a steady
state by time t .
IL(t + L) = IP(t) - D(t, t + L). (3.37) IP(t) uniform distribution on (R+1,R+Q)
IL(t+L) ≤ IP(t) ≤ R+Q Since t is arbitrary, so is t+L. So we obtain the
steady state distribution of IL.
IL=j , IP(t)=k
Need to consider k ≥ j IL(t+L) = IP(t)-D(t,t+L) =k-(k-j) =jConsider compound Poisson demand
(3.38)
Translation from R=r to R=0
(3.39)
).0RrjIL(P))rj(k)L(D(PQ
1
)jk)L(D(PQ
1)rRjIL(P
Q
}rj,1max{k
Qr
}j,1rmax{k
QRj)jk)L(D(PQ
1)jIL(P
QR
}j,1Rmax{k
In the special case of an S policy,
R = S - 1 and Q = 1 and (3.38) can be
simplified to
j S (3.40)
Note: The steady state is S with Prob 1.
)jS)L(D(P)jIL(P
Normally distributed demand IP uniformly distributed on [R, R+Q]; density 1/Q ´ = L = the average of the lead-time demand, ´ = L1/2 = the standard deviation of the lead-
time demand, f(x) = the density of the inventory level in steady
state, F(x)= the distribution function of the inventory
level in steady state: IP(t)=u, D(L) ≥ u-x
(3.41) du
´xu1
Q
1)xIL(P)x(F
QR
R
Loss function G(x):
G´(x) , (3.43)
which means that G´(x) is decreasing and that G(x) is convex. See Figure 3.9.
G(x) given in Table in Appendix 2.
x
dvvxvxG )()()(
1)()(
xdvvx
G(x)
Figure 3.9 The loss function G(x).
0
1
2
3
4
-3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5
Using (3.43) to reformulate (3.41)
(3.44) From (3.41) the density is
(3.45)
´xQRG
´xRG
Qdu
´xuG
Q
1)x(F
QR
R
´xR´xQR
Q
1du
´xu1
Q
1)x(f
QR
R
Example 3.9 =2, L=5. Continuous review (R,Q) policy
with R=9 and Q=5. ’=L=10. Applying (3.38)
For j =1,
the normal approximation, ’= 10 and ’=101/2
see (3.44)
1014
},10max{ )!(
10
5
1)(
e
jkjILP
jk
jk
14
10
101
106.0)!1(
10
5
1)1(
k
k
ek
ILP
;)10
4()
10
1(
5
10)(
xG
xGxF
It is reasonable to compare P(IL = j ) by F( j + 0.5) - F( j - 0.5)
Figure 3.10 Probability distributions in Example 3.9.
In case of continuous demand and
continuous review, order is triggered at t
when IP(t)=R. Thus, the average stock on
hand just before the order arrives at t+L is denoted the safety stock, SS,
. (3.46)
Replace (3.44) by the equivalent expression
. (3.47)
.LRRSS
xQSS
GxSS
GQ
)x(F
3.4.3 Service levels
S1=probability of no stockout per order cycle,
S2=“fill rate”- fraction of demand that can be satisfied immediately from stock on hand,
S3=“ready rate”- fraction of time with positive stock on hand=P(IL>0) or 1-F(0).
From a practical point of view it is usually
most important that the service level is
clearly defined and interpreted in the same
way throughout the company. A common solution is to group the items in some way and specify service levels for each group.
The choice of service level should be based on customer expectations
3.4.4 Shortage costs
b1=shortage cost per unit and time unit,
b2=shortage cost per unit.
Shortage cost of type b1: spare parts
Shortage cost of type b2: overtime production The optimal reorder point will increase with the
service level or the shortage cost used.
3.4.5 Determining the safety stock
for given S1
In the continuous demand case,
(3.48) k denotes the safety factor, and
(3.49) Finally we get the reorder point as R = SS + ´.
)())(( 1 kSSR
SRLDP
kSS
Table 3.6 illustrates how the safety factor grows with service level S1.
Table 3.6 Safety factors for different values of service level S1.
Note: the safety factor is increasing rapidly for large service levels.
Service level S1 0.75 0.80 0.85 0.90 0.95 0.99
Safety factor k 0.67 0.84 1.04 1.28 1.64 2.33
3.4.6 Fill rate and ready rate constraints
The ready rate:
. (3.50) The fill rate: Average filled/average demand
, (3.51)
For Poisson Demand f1=1, which implies S2=S3.
1
3 )()0(j
jILPILPS
1kk
1k 1jk
2
fk
)jIL(Pf)k,jmin(
S
For continuous normally distributed demand,
S2 = S3 ,
, (3.52)
S2 and S3 increase with R. S2 = S3 =0 for R ≤ -Q,
Since R+Q ≤ 0.
It is common to approximate (3.52) by
, (3.53)
It underestimates S3 . Works well for large values of Q.
´
13
RG
QS
´´
1)0(13
QRG
RG
QFS
Example 3.10
Consider pure Poisson demand with = 2,
L = 5, continuous review (R, Q) policy with
R = 9 and Q = 5. We want to determine the fill
rate S3 , which in this case are equal because we
have pure Poisson demand. Applying (3.38) and
(3.50) we obtain
14
1j
14
}j,10max{k
10jk
32 679.0e)!jk(
10
5
1)0IL(PSS
666.010
4G
10
1G
5
101)0(F1SS 32
376.010
1RS1
3.4.7 Fill rate - a different approach
Consider a batch Q that is ordered when the
inventory position is R.
The considered batch will be consumed by the
demand for Q units following after these first R
units. When the batch arrives in stock, a part of
this demand may already occurred, i.e., there are
backorders waiting for the batch.
Let B= backordered quantity that will be covered by the batch.
Note that we are only considering the
backorders that are covered by the batch.
Therefore, we must have 0 ≤ B ≤ Q. If the
backorders exceed Q when the batch arrives in
stock, the quantity exceeding Q is covered by
future batches.
.(3.54)
Q/)B(E1SS 32
u R means B = 0,
R < u R + Q means B = u - R,
R + Q < u means B = Q.
(3.55),
QRG
RG
duu1
)QRu(duu1
)Ru(
duu1
Qduu1
)Ru()B(E
QRR
QR
QR
R
3.4.8 Shortage cost per unit and time unitWe optimize the re order point by balancing
backorder costs against holding costs(x)+ = max(x, 0),(x)- = max(-x, 0).
Using x+- x- = x,
the total cost rate =h IL+ + b1IL¯
= hIL + (h + b1)(IL)¯.
For the compound Poisson demand, we can
use (3.38) to obtain the average costs per
time unit,
(3.56)
Where E(IL)=average inventory R+(Q+1)/2 minus
average lead time demand ’
)()()2
1(
)()(IL)(1
1
1
jILPbhQ
Rh
ILEbhhEC
j
)(
)()1(
)1()1(
)1()1()1()()(
)1()()(
)1()1()1()(
)1()(
)]1()([)(
1
2
1
1 11
1 12
1 11
1 1
1 1
jILP
iILPILP
jILPILP
jILPILPiILPijILjP
jILPiILPijILjP
jILPjILPjjILjP
jILPjjILjP
jILPjILPjjILjP
From (3.39),
Thus,
(3.57)
Note :
)1rR1jIL(P)rRjIL(P
,)1rR0IL(P)bh(h)1r(C
)1rRjIL(P)bh()2
1Qr(h
)1rR1jIL(P)bh()2
1Qr(h)r(C
1
0
j1
1
j1
1
1 )1|()()'2
11()1( rRjILPbh
QrhrC
Or equivalently,
C(r + 1) - C(r)=h- (h + b1)[1- S3(r+1)]
= - b1 + (h + b1) S3(r+1) .
(3.58)where S3(r + 1) = is the ready rate for R = r +1.
Since S3 increases with r, C(r+1)-C(r) increases with r. Thus C(r) is convex in the reorder point.
To find the optimal R, start with R = - Q
and increase R by one unit at a time until
the costs are increasing. If R* is the optimal reorder point
. (3.59) In case of pure Poisson demand this is also
true for the fill rate.
)1R(Sbh
b)R(S 3
1
13
For continuous normally distributed
demand
(3.60)
Uses (3.44), y=-x/’ , and G(∞)=0. .
.du´u
GQ
)bh()2/QR(h
dudx´xu
GQ
1)bh()2/QR(h
dx)x(F)bh()2/QR(hC
QR
R1
0 QR
R1
0
1
.S)bh(bS)bh(b
)1S)(bh(hR
GQR
GQ
)bh(hdR
dC
311211
211
(3.61)
Uses S3=1-F(0), (3.52) and (3.44). Chose R when Q is given. Newsboy chooses R+Q.
Since dC/dR is increasing with R, C is a convex function of R. The optimal R is obtained for dC/dR = 0,
(3.62)
(3.63)
S3(R*)=P(IL>0|r=R*)=Prob(D(L)≤R*+Q)
(R*+Q)=b1/(h+b1)
1
13 bh
bS
3
31 1 S
hSb
(3.64)
Since H´(x) = - G(x), H(x) is decreasing and
convex.
x
2 )x(x))x(1)(1x(2
1dv)v(G)x(H
Exercise: Use (3.42) and (3.43) to derive
H(x)
Figure 3.11 The function H(x).
0
1
2
3
4
-3.5 -2.5 -1.5 -0.5 0.5 1.5 2.5 3.5
Using H(x), we can express the costs in
(3.60) as
(3.65)
´QR
H´R
HQ
)bh()2/QR(hC2
1
3.4.12 The newsboy model Single period with stochastic demand Penalty costs associated with ordering both too
much and too little
x = stochastic period demand,
S = ordered amount,Co = overage cost, i.e., the cost per unit for a
remaining inventory at the end of the period, cu = underage cost, i.e., the cost per unit for
unsatisfied period demand.
For a given demand x the costs are
(3.81)
(3.82)and the expected cost is
(3.83)
,Sxc)xS()x(C o
,Sxc)Sx()x(C u
.S
G)cc()S(c
dx)x
(1
)Sx()cc()S(c
dx)x
(1
)Sx(cdx)x
(1
)xS(cC
uo0
Suo0
Su
S
o
To find the optimal solution,
(3.84)
or,
(3.85) It is easy to see that C is convex, so (3.85)
provides the optimal solution.Compare with (3.62).
,0)1)S
()(cc(cdS
dCuoo
.cc
c)
S(
uo
u
Example 3.12co = 25, cu = 50, = 300, = 60,
Applying (3.85) we obtain the optimal
solution from
implying that
S 326
,6667.05025
50)
60
300(
S
,43.060
300
S
This problem concerns a single period.
The simple newsboy solution (3.85) is, however, also common in multi-period
settings.
.kj),jk)L(D(P)jIL(P
.)jIL(P)bh()k(h)IL(E)bh()IL(Eh)k(g1
j11
3.5 Joint optimization of order quantity and reorder point 3.5.1 Discrete demand´ = LIL(t + L) = IP(t) - D(L). (3.98)(S - 1, S) policy with S = k, i.e., R = k - 1 and Q = 1. Then IP(t)=k atall time.
(3.99)Let g(k) be the average holding and shortage costs per time unit.
(3.100)From section 3.4.8, g(k) is convex, g(k) |k| .
.)k(gQ
1
Q
A)Q,R(C
QR
1Rk
).),(*(),(min)( QQRCQRCQCR
The inventory position is uniform on [R + 1, R + Q].
(3.101)
(3.102)Q = 1 R*(1) = k*- 1.
*)(
)(min
)1(min)1),1(*()1(
kgA
kgA
RgARCC
k
R
Since g(k) is convex, the second best k is k*-1 or k*+1.
Case (1): g(k*-1) ≤ g(k*+1)
then g(k*-1)+ g(k*) ≤ g(k*)+ g(k*+1)
if R*(2) = R*(1)-1= k*-2
g(R*(2) +1)+ g(R *(2) +2) = g(k*-1)+ g(k*)
if R*(2) = R*(1)= k*-1
g(R*(2) +1)+ g(R *(2) +2) = g(k*)+ g(k*+1)
)]2()1([min2
1
2)2( RgRg
AC
R
Case (1): R*(2) = R*(1) -1
Case (2): g(k*-1) > g(k*+1) R*(2) = R*(1)
Then g(R*(1) )≤ g(R*(1)+2 ) R*(2) = R*(1)-1
Otherwise, R*(2) = R*(1)
Also,
)]2)1(*()),1(*(min[2
1
2
)1(
)1)1(*(2
1
)]2)1(*()),1(*(min[2
1
2)2(
RgRgC
Rg
RgRgA
C
,otherwise)Q(R)1Q(R
),1Q)Q(R(g))Q(R(gif1)Q(R)1Q(R
**
****
.1Q
1)1Q)Q(R(g),Q(R(gmin
1Q
Q)Q(C)1Q(C **
(3.103)
(3.104)C(Q + 1) C(Q) if and only if min{g(R*(Q), g(R*(Q) + Q + 1} C(Q). min{g(R*(Q), g(R*(Q) + Q + 1} is increasing with Q. Let Q* be the smallest Q such that C(Q + 1) C(Q). C(Q) C(Q*) for any Q Q*. Q* and R*(Q*) provides the optimal solution.
General Step:
.Q
A´QRH
´RH
Q)bh()2/QR(h)Q,R(C
2
1
.´R
G´QR
GQ
)bh(hR
C1
.Q
A´QRG
Q´QRH
´RH
Q)bh(2/h
Q
C22
2
1
3.5.2 An iterative technique: Continuous CaseBy adding the average setup cost per time unitto the cost expression in (3.65) we have
(3.107)
(3.108)The resulting R decreases with Q.
(3.109)
h/A2Q0
0/ RC
.´QR
GQ´QR
H´R
Hh
)bh(2
h
A2Q
iiiiii211i
Start by determining the batch quantity
(3.110)
Determine reorder point R0 from (3.108) and
(3.111) Determine the reorder point R1 corresponding to Q1 from (3.108)
Qi+1 Qi Ri+1 Ri
C(Ri, Qi+1) - C* h(Qi+1 - Qi)
Example 3.14 Let A = 100, h = 2, and b1 = 20.
The demand per time unit is normally distributed with = 50 and = 20. The lead-time L = 4. We obtain ´= L= 200 and ´= L1/2 = 40.
Table 3.7 Results from the iterations for the data in Example 3.14.
Iteration i0 1 2 3 4 5
Order quantity Qi
70.71 87.91 93.08 94.59 95.03 95.15
Reorder point Ri
224.76 219.60 218.16 217.75 217.63 217.60
Costs Ci 232.01 226.63 226.24 226.21 226.20 226.20
3.6 Optimality of ordering policies
The (R, Q) type or of the (s, S) type.
Without ordering costs and given a fixed batch quantity, an (R, Q) policy is optimal.
(s, S) policies are not necessarily optimal for problems with service constraints.