3 SINGLE-ECHELON SYSTEMS WITH INDEPENDENT ITEMS Different items can be controlled independently. The...

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.

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.

Inventory position

Inventory level

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

Stock level

How important is it to use the optimal order quantity? From (3.1), (3.3), and (3.4)

AdhAdhAdh

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.

Stock level

Q(1-d/p)

P = production rate (p > d).

The average inventory level is now

Q(1 - d/p)/2

. (3.6)

. (3.7))/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?

Q/p1Q/d

Stock level between machines 1 and 2 (2)

Stock level at warehouse (4)

Stock level after machine 2 (3)

Q/dTransportation

Stock level before machine 1 (1)

Holding Cost Calculation:

……

)AAA(Q

QC 3213

d)AAA(2Q

00 )(2

QQforAQ

QdvC (3.10)

(3.11)

00 )(2

QQforAQ

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)

. (3.13)

If Q´ Q0, (3.12) and (3.13) give the optimal

solution, i.e., Q* = Q´, and C* = C´.

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).

dvrvhAdC )(2 0

Incremental Discounts

Q1=1000 Q2=2500

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

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

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.

Inventory level

Q(1-x)

. (3.17)

, (3.18)

and inserting in (3.17) we get

. (3.19)

, (3.20)

.(3.21)

1* )(2

hbAd2C

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

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

Stock level after the machine

Stock level at warehouse

Q/3000 Q/1000

Q/1000

Q/1000Q/3000

Q(1-x)

100075x

Q15)x1(

Q)1510(

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)

k ff ,1

))(...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

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

rt etAdtethSV

)()()(

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

i etASV1

)1()()(

FUDGE THE DISCOUNT RATE INTO INTEREST EXPENSE

COSTAVERAGE

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

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.

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

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.

jkkdjkhAnjC

)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)

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).

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,

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

)2(1,0

)3(2,1,0

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

3.3.8 The Silver-Meal Heuristic

,2 k n, (3.24)

. (3.25)

d)1j(hA

d)1j(hA1k

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

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

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

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

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.

.,0 kjf k

, , 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

kj fff

)())((

Prob of k-1 customers with total demand i and one with the demand j-i

=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

JEJVarJEJVar

JEKJVarKEZE

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.

duexx u

)/)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

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)

)( ,1 Q

jRP ji

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.

j jip1 , 1

kDPkDpkDP

kDPkDp

kDPkDpp

1)()()(

.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

)jk)L(D(PQ

1)rRjIL(P

}rj,1max{k

}j,1rmax{k

QRj)jk)L(D(PQ

1)jIL(P

}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

1)xIL(P)x(F

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.

dvvxvxG )()()(

Figure 3.9 The loss function G(x).

-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

´xR´xQR

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)

},10max{ )!(

jkjILP

106.0)!1(

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

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

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.

3 )()0(j

jILPILPS

1k 1jk

)jIL(Pf)k,jmin(

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.

1)0(13

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

}j,10max{k

32 679.0e)!jk(

1)0IL(PSS

666.010

101)0(F1SS 32

376.010

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),

)QRu(duu1

)Ru()B(E

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

)()(IL)(1

jILPbhQ

ILEbhhEC

)1()1(

)1()1()1()()(

)1()()(

)1()1()1()(

)]1()([)(

iILPILP

jILPILP

jILPILPiILPijILjP

jILPiILPijILjP

jILPjILPjjILjP

jILPjjILjP

jILPjILPjjILjP

From (3.39),

(3.57)

Note :

)1rR1jIL(P)rRjIL(P

,)1rR0IL(P)bh(h)1r(C

)1rRjIL(P)bh()2

)1rR1jIL(P)bh()2

1Qr(h)r(C

1 )1|()()'2

11()1( rRjILPbh

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

For continuous normally distributed

demand

(3.60)

Uses (3.44), y=-x/’ , and G(∞)=0. .

.du´u

)bh()2/QR(h

dudx´xu

1)bh()2/QR(h

dx)x(F)bh()2/QR(hC

.S)bh(bS)bh(b

)1S)(bh(hR

)bh(hdR

311211

(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)

31 1 S

(3.64)

Since H´(x) = - G(x), H(x) is decreasing and

convex.

2 )x(x))x(1)(1x(2

1dv)v(G)x(H

Exercise: Use (3.42) and (3.43) to derive

Figure 3.11 The function H(x).

-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)

)bh()2/QR(hC2

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

G)cc()S(c

)Sx()cc()S(c

)Sx(cdx)x

)xS(cC

To find the optimal solution,

(3.84)

(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

Example 3.12co = 25, cu = 50, = 300, = 60,

Applying (3.85) we obtain the optimal

solution from

implying that

,6667.05025

,43.060

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

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

A)Q,R(C

).),(*(),(min)( QQRCQRCQCR

The inventory position is uniform on [R + 1, R + Q].

(3.101)

(3.102)Q = 1 R*(1) = k*- 1.

)1(min)1),1(*()1(

RgARCC

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

2)2( RgRg

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)

)]2)1(*()),1(*(min[2

)1)1(*(2

)]2)1(*()),1(*(min[2

,otherwise)Q(R)1Q(R

),1Q)Q(R(g))Q(R(gif1)Q(R)1Q(R

1)1Q)Q(R(g),Q(R(gmin

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:

A´QRH

Q)bh()2/QR(h)Q,R(C

)bh(hR

A´QRG

Q´QRH

Q)bh(2/h

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

GQ´QR

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.

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