3 SINGLEECHELON SYSTEMS WITH INDEPENDENT ITEMS

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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.

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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

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Ordering or Setup Costs

•   - setup and learning
• - administrative costs associated with the
• handling of orders
•   - transportation and material handling.

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• Shortage Costs or Service Constraints

• - 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.

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3.2 Different Ordering Systems

• 3.2.1 Inventory position 
• Inventory position stock on hand outstanding
  orders - backorders.
• Inventory level stock on hand - backorders.

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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.

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3.2.3 Different Prdering Policies

• (R, Q) policy
• Figure 3.1 (R, Q) policy with continuous review.
  Continuous demand.

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• (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) .

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• (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.

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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.

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Notation

• 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

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• Figure 3.3 Development of inventory level over
• time.

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• 
  (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)

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• 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.

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• 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.

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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.

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• P production rate (p d).
• The average inventory level is now
• Q(1 - d/p)/2
• .
  (3.6) 
• .
  (3.7)

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3.3.4 Quantity Discounts

• v price per unit for Q
• normal price,
• v price per unit for Q ? Q0, where v
• Holding cost
• h h0 rv for Q
• h h0 rv for Q ? Q0,

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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?

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• Holding Cost Calculation
• 
  (3.8)
• 
  (3.9)

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(3.10)  

(3.11)
Figure 3.6 Costs for different values of Q.
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• 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.

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• 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
• From (3.14) and (3.15), Q 69.28 and
• C 31732. Applying (3.16), C(100) 30300.
  Q Q0 100.

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• 2. If Q
• . (3.14)
• . (3.15)
• Since v v we know that Q
• . (3.16)
• The optimal solution is the minimum of (3.15) and
  (3.16).

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Incremental Discounts
d 36500 , r 0.3
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• A1A15
• A2A1 (c1-c2)Q115(1-0.8)1000215
• A3A2 (c2-c3)Q2215(0.8-0.7)2500465
• EOQi
• Procedure
• Candidate oi from each segment
• EOQi if feasible
• Oi Qi if EOQi Qi
• Qi-1 if EOQi
• Q best of all Oi

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Example

• EOQ11910 Q1 ? O1Q11000
• EOQ2 8087 ? O2Q22500
• EOQ3 12714 feasible

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TC(Qi) vidQivir /2dAi / Qi

• TC(1000)36500110000.50.313650015/1000
• 36500150547.537197.5
• TC(2500)(1000115000.8)36500/2500
  25000.50.3(1000115000.8)/2500
• 3650015/25003212033021932
  669
• TC(12714)1000115000.8
• (12714-2500)0.71536500/12
  714
• 26884.91404.728289.6
• Q12714

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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.

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• 
  . (3.17)
•  
• 
  , (3.18)
• and inserting in (3.17) we get
• 
  . (3.19)
• 
  , (3.20)
• 
  .(3.21)

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• 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.

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• The optimal solution is x 15/90 1/6, and Q
  310.

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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.

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• 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

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3.3.7 The Wagner-Whitin Algorithm

• fkminimum costs over periods 1, 2, ..., k, i.e.,
  
• when we disregard periods k 1, k 2, ...,
  T,
• fk,tminimum costs over periods 1, 2, ..., k,
  given that the last delivery is in period t (1 ?
  t ? k).
• 
  , (3.22)
• 
  . (3.23)

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• Example 3.6
• T 10, A 300, h 1 per unit and period.
• Table 3.1 Solution, fk,t , of Example 3.6.

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• Table 3.2 Optimal batch sizes in
• Example 3.6.

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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.

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DYNAMIC DETERMINISTIC MODELS
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• 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).

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• 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.

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• 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 T1,
  T2, ... PDS), THEN THE FIRST T PERIODS IS CALLED
  A FORECAST HORIZON WHILE THE FIRST J PERIODS ARE
  CALLED A DECISION HORIZON.

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• DEMAND CAN BE NETTED OUT AS SHOWN. (EXCESS OVER
  'SAFETY STOCKS') THUS I0 0

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• 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

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• THEOREM ANY OPTIMAL SOLUTION MUST SATISFY
• I) It Qt1 0 (I.E. Qt1 0 ? It
  0
• It 0
  ? Qt 1 0)
• II) IT 0
• PROOF.
• FIRST PRODUCTION WILL CERTAINLY NOT COME
  UNTIL APERIOD (t1) FOR WHICH dtl O. FOR ANY
  FOLLOWING PRODUCTION PT. IF (I) DOES NOT HOLD,
  THEN DECREASE PRECEDING PRODUCTION BY It AND
  INCREASE Qtl BY It GIVING A BETTER SOLUTION.

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• 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.

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• 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.)

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• PROOF
• I) LET t AND tk BE SUCCESSIVE P-POINTS.
• THEN Qt 0, Qtk 0 Itk-1 0 BY
  THEOREM.
• I. E., tk-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

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• FOR A COMPLETE SOLUTION, WE NEED TO KNOW
• - THAT K IS AN R-POINT
• - OPTIMAL SOLUTION FOR PERIOD K1 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
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• III) SIMPLIFY THE COST FUNCTION ?Qt ?dt ,
• AND THUS THE PORTION OF COSTS GIVEN BY
• ('SUNK' COST) IS
  CONSTANT.
• REDUCED COST C(T) A??t h ?It
• FURTHER, LET t12 AND t9 BE THE NEXT TO THE LAST
  RPOINT. THEN THE COST C(9,12).

-D10
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• 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.

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• NOTE ALSO
• CT-1(T) C(T-1) A
• Cj(T) Cj(T-1) (T-j-1)
  h dT, 1 j
• II II
  II
• C(j) C(j,T) C(j) C(j,T-1) (T-j-1)hdt

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• 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 -1
• 
  (DON'T USE FIRST TIME)
  
  

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• 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.

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• 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.

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• 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.

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• 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 dT1 TO THE T-COLUMN
• VALUES AS j DECREASES, THIS ALWAYS BENEFITS
• LARGER j'S TO BE A MINIMUM, THUS
• j(T) 11 AND j(T1) 9 IS IMPOSSIBLE SINCE
• C9(T) C11(T) ? C9 (T1) C11 (T1).
• NOTE
• C9(T1) C9(T) (T-9)hdT1
• C11(T1) C11(T) (T-11)hdT1

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• 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.

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• 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 T11. 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.

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• 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.

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• 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 (TK)-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.)

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• 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.

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• 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

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• 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.

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3.3.8 The Silver-Meal Heuristic

• ,2 ? k ?
  n, (3.24)
•  
•  
• 
  . (3.25)

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• 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
• 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.

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• 2 periods (300 30)/2 165
• 3 periods (300 30 2 ? 100)/3 176.67 165
• Starting with period 6 as the first period  
• 2 periods (300 60)/2 180
• 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

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• 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

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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

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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

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• 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

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• 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.

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• 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).

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• 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 ct1
• (I.E., IGNORING SET UP COSTS, IT IS NEVER CHEAPER
• TO PRODUCE AND HOLD.) THIS PRECLUDES THE
• SPECULATIVE MOTIVE.

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3.4 Safety stocks and reorder points

• 3.4.1 Demand processes
• Cumulative demand nondecreasing, stationary,
  independent increments
• Compound 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.

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• The size of a customer demand is
• independent of the distribution of the
• customer arrivals.
• fjprobability 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.

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• , , 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)

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

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• ? 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.

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• 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.

92

• 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.

93
3.4.2 Continuous review (R, Q) policy - inventory
level distribution

• IPinventory position.
• Order Q or nQ immediately when IP R. Thus, in
  steady state, R 1 ? IP ? R Q

94

• Proposition 3.1
• In steady state, the inventory position is
  uniformly distributed on the integers R 1, R
  2, ......,
• R Q.

95
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 8
• Positive recurrent expected time to j is finite
• Aperiodic Period 1
• Ergodic aperiodic and positive recurrent
• ? unique steady state distribution with ?j 0

96

• 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)

97

• But for a Markov chain. Given
• demand size k, pij(Dk)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.

98

• Llead time (constant)
• ILinventory 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 (R1,RQ)
• IL(tL) IP(t) RQ
• Since t is arbitrary, so is tL. So we obtain the
• steady state distribution of IL.
• ILj , IP(t)k

99

• Need to consider k j
• IL(tL) IP(t)-D(t,tL) k-(k-j) j
• Consider compound Poisson demand
• 
  (3.38)
• Translation from Rr to R0
• 
  (3.39)

100

• 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.

101

• Normally distributed demand
• IP uniformly distributed on R, RQ 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)

102

• 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.

103

• G(x)
• Figure 3.9 The loss function G(x).

104

• Using (3.43) to reformulate (3.41)
• 
  (3.44)
• From (3.41) the density is
• 
  (3.45)

105

• Example 3.9
• ?2, L5. Continuous review (R,Q) policy
• with R9 and Q5. ?L?10. Applying (3.38)
• For j 1,
• the normal approximation, ? 10 and ?101/2
• 
  see (3.44)

106

• 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.

107

• 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 tL is
  denoted the safety stock, SS,
• 
  . (3.46)
• Replace (3.44) by the equivalent expression
• 
  . (3.47)

108

• 3.4.3 Service levels
• S1probability of no stockout per order cycle,
• S2fill rate- fraction of demand that can be
  satisfied immediately from stock on hand,
• S3ready rate- fraction of time with positive
  stock on handP(IL0) or 1-F(0).

109

• 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

110

• 3.4.4 Shortage costs
• b1shortage cost per unit and time unit,
• b2shortage 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.

111

• 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 ?.

112

• 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.

113
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 f11, which implies S2S3.
  
  

114

• For continuous normally distributed demand,
• S2 S3 ,
• 
  
• 
  , (3.52)
• S2 and S3 increase with R. S2 S3 0 for R -Q,
• Since RQ 0.
• It is common to approximate (3.52) by
• 
  
• 
  , (3.53)
• It underestimates S3 . Works well for large
  values of Q.

115

• 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
• 
  

116

• 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.

117

• 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)

118

• u ? R means B 0,
• R
• R Q
• 
  (3.55)

119
3.4.8 Shortage cost per unit and time unit

• We optimize the reorder 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
• h?IL (h
  b1)(IL).

120

• 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(Q1)/2 minus
• average lead time demand ?

121
(No Transcript)
122

• From (3.39),
• Thus,
• 
  
• (3.57)
• Note

123

• Or equivalently,
• C(r 1) - C(r)h- (h b1)1- S3(r1)
• - b1 (h b1)
  S3(r1) .
• (3.58)
• where S3(r 1) is the ready rate for R r
  1. Since S3 increases with r, C(r1)-C(r)
  increases with r. Thus C(r) is convex in the
  reorder point.

124

• 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.

125

• For continuous normally distributed
• demand
• 
  
• 
  (3.60)
• Uses (3.44), y-x/? , and G(8)0.

126



(3.61) Uses S31-F(0), (3.52) and (3.44).
Chose R when Q is given. Newsboy chooses RQ.
127

• 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(IL0rR)Prob(D(L)RQ)
• ?(RQ)b1/(hb1)

128
Exercise Use (3.42) and (3.43) to derive

• 
  (3.64)
• Since H(x) - G(x), H(x) is decreasing and
• convex.

129

• Integrate G(x), we obtain

130

• H(x)
• Figure 3.11 The function H(x).

131

• Using H(x), we can express the costs in
• (3.60) as
• 
  (3.65)

132
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.

133

• For a given demand x the costs are
• 
  (3.81)
• 
  (3.82)
• and the expected cost is
• 
  (3.83)

134

• 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).

135

• Example 3.12
• co 25, cu 50, ? 300, ? 60,
• Applying (3.85) we obtain the optimal
• solution from
• implying that
• S ? 326

136

• This problem concerns a single period.
• The simple newsboy solution (3.85) is,
  however, also common in multi-period
• settings.

137
3.5 Joint optimization of order quantity and
reorder point 3.5.1 Discrete demand m
mL IL(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 at all 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 ? ? .
138
The inventory position is uniform on R 1, R
Q.
(3.101)

(3.102) Q 1 ?
R(1) k- 1.
139

• Since g(k) is convex, the second best k is k-1
  or k1.
• Case (1) g(k-1) g(k1)
• then g(k-1) g(k) g(k) g(k1)
• 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(k1)

140

• Case (1) R(2) R(1) -1
• Case (2) g(k-1) g(k1) ? R(2) R(1)
• Then g(R(1) ) g(R(1)2 ) ? R(2)
  R(1)-1
• Otherwise, R(2) R(1)
• Also,

141
General Step
(3.103)


(3.104) C(Q 1) ? C(Q) if and only if
ming(R(Q), g(R(Q) Q 1 ? C(Q).
ming(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.
142
3.5.2 An iterative technique Continuous Case By
adding the average setup cost per time unit to
the cost expression in (3.65) we have
(3.107)


(3.108) The resulting R decreases with Q.
(3.109)
143
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) Qi1 ? Qi Ri1 ? Ri
C(Ri, Qi1) - C ? h(Qi1 - Qi)
144
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.
145
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.