INVENTORY

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INVENTORY(???????????????)
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THE FUNDAMENTAL PROBLEM OF INVENTORY MANAGEMENT
CAN BE DESCRIBED BY THE TWO QUESTIONS

• When should an order be placed?
• How much should be ordered?

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Types of Inventory

• Raw materials
• Components
• WORK-IN-PROCESS
• Finished goods

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Relevant Cost

• Holding Cost
• Order Cost
• Penalty Cost

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Holding Cost

• ( or the carrying cost or the inventory cost) is
  the sum of all costs that are proportional to the
  amount of inventory physically on hand at any
  point in time.
• Cost of providing the physical space to store the
  items.
• Taxes and insurance.
• Breakage, spoilage, deterioration, and
  obsolescence.
• Opportunity cost of alternative investment.

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Holding Cost

• An aggregated interest rate comprised of the
  four components listed above.
• Example
• cost of capital 28
• Taxes and insurance 2
• Cost of storage 6
• Breakage and spoilage 1
• Total interest charge 37

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Holding Cost

• We would assess a charge of 37cents for every
  dollar that we have invested in inventory during
  a one-year period.
• Let c the dollar value of one unit of
• inventory
• i the annual interest rate
• h holding cost in terms of dollars per
  unit per year

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Holding Cost

• The item values at 180.
• Then h ic
• 0.37 18066.60
• if we held 300of these items for five years,the
  total holding cost
• 530066.60 99,900

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Order Cost

• It depends on the amount of inventory that is
  ordered or produced.
• Two components fixed cost (K)and variable cost
  (c)
• C(x) 0 if x 0
• C(x) K cx if x gt 0

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4-3
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Penalty Cost

• The shortage cost or the stock-0ut cost
• It is the cost of not having sufficient stock on
  hand to satisfy a demand when it occurs.
• In the lost-sales, it includes the lost profit
  that would have been made from the sales.
• The symbol p is used.

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The EOQ Model

• The EOQ model (for economic order quantity) is
  the simplest and most fundamental of all
  inventory models.
• The assumptions for basic model
• ? is demand rate (units per unit time), it is
  constant and known.
• Shortage are not permitted.
• No order lead time. (this assumption will be
  relaxed.)

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The EOQ Model

• The cost include
• Setup cost at K per positive order placed.
• Proportional order cost at c per unit ordered.
• Holding cost at h per unit held per unit time.

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Basic Model

• Q the size of the order (units).
• T cycle length (year).
• In each cycle,the total fixed plus proportional
  order cost is
• C(Q) K c Q
• As Q units are consumed each cycle at a rate l ,
  then T Q/ l or (- l -Q/T)

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Basic Model

• The inventory level decreases linearly from Q to
  0 each cycle.The average inventory level Q/2
• G (Q) average annual cost

K cQ hQ ------------
----- T 2
K cQ hQ ------------
----- Q/ l 2
K l l c hQ ------
------ ----- Q
2
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Basic Model

• The three terms comprising G(Q) are annual setup
  cost, annual purchase cost, and annual holding
  cost.
• We are finding Q to minimize G(Q)
• G(Q) -K l /Q2 h/2 0
• Q sqrt (2K l /h)

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Example1

• Number 2 pencil at the campus bookstore are sold
  at a fairly steady rate of 60 per wk. The pencils
  cost the bookstore 2 cents each and sell for 15
  cents each. It costs the bookstore 12 to
  initiate an order, and holding costs are based on
  an annual interest rate of 25 percent. Determine
  the optimal number of pencils for the bookstore
  to purchase and the time between placement of
  orders. What are the yearly holding and setup
  costs for this item?

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Solution

• l 60 52 3120 units/ year
• h 0.25 0.020.005
• Substituting into the EOQ formula,
• Q sqrt (2K l /h)
• sqrt (2123120 / 0.005) 3870
• the cycle time is T Q/ l 3870/31201.24 years
• the average annual holding cost hQ/29.675
• the average annual setup cost K l /Q 9.675

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EOQ model with order lead time

• Suppose in Ex 1 that the pencils had to be
  ordered 4 months in advance.
• Let R The reorder point (the level of on-hand
  inventory at the instant an order should be
  placed.), t Lead time, l demand rate
• R t l 3120 0.3333 1040

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Example 2

• EOQ 25 uints, Demand rate 500 units/y, a lead
  time 6 wks.
• The cycle time , T 25/500 0.05 ?? 2.6
  ???????
• Lead time / Cycle time t / T 2.31 (Every
  order must be placed 2.31 cycles in advance.)
• ????? 0.31T 0.310.05 0.0155 ?? , then
• R 0.0155500 7.75 8
  units

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Case of Lead time gt Cycle time

• A. Form the ratio t / T.
• B. Consider only the fractional remainder of the
  ratio. Multiply this fractional remainder by the
  cycle time to convert back to years.
• C. Multiply the result of step (b) by the demand
  rate to obtain the reorder point.

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Finite Production Rate

• When units are produced internally, the curve
  describing inventory levels as a function of time
  is shown in next slide.
• Let Q the size of each production run.
• T Cycle length T1T2 T1
  uptime (production time) T2 downtime.
• But the maximum level of on-hand inventory during
  a cycle is not Q

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Finite Production Rate

• Number of units consumed each cycle lT Q
• the max level of on-hand inventory H
• items are produced at a rate P for a time T1 then
  
• Q PT1
• H/T1 P-l (from figure in next slide)
• H Q (1- l /P)
• The average inventory level H/2, average annual
  cost G(Q) is expressed
• G(Q) K/T hH/2 K l/Q hQ/2(1- l/P)

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Finite Production Rate

• If we define h h(1- l/P) , then
• Q sqrt (2K l /h)

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Example 3

• A local company produced a programmable EPROM for
  several industrial clients. It has experienced a
  relatively flat demand of 2500 units per year for
  the product. The EPROM is produced at a rate of
  10000 units per year. The accounting department
  has estimated that it costs 50 to initiate a
  production run, each unit costs the company 2 to
  manufacture, and the cost of holding is based on
  a 30 percent annual interest rate. Determine the
  optimal size of a production run, the length of
  each production run, and the average annual cost
  of holding and setup. What is the max level of
  the on-hand inventory of the EPROMs?

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solution

• h 0.32 0.6 per unit/ year
• h h(1- l/P) 0.6(1-2500/10000) 0.45
• Q 745
• T Q/ l 745/2500 0.298 year
• The uptime each cycle, T1 Q/ P 745/10000
  0.0745 year
• down time , T2 T-T1 0.2235 year.
• Average annual cost of holding and set up is
• K l/Q hQ/2335.41
• Max level of on-hand inventory, H Q1-l/P) 559
  units

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Quantity Discount Models

• The supplier is willing to charge less per unit
  for larger orders. The purpose of the discount is
  to encourage the customer to buy the product in
  larger batches.
• Two discount schedules all units, incremental.

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Example 4 (Case)

• The Weighty Trash Bag Company has the following
  price schedule for its large trash can liners.
  For orders of less than 500 bags, the company
  charges 30 cents per bag for orders of 500 or
  more but fewer than 1000 bags, it charges 29
  cents per bag and for orders of 1000 or more, it
  charges 28 cents per bag. In this case the
  breakpoints occur at 500 and 1000. The discount
  schedule is all-units because the discount is
  applied to all of the units in an order. gt

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Example 4 (Case )

• The order cost function C(Q) is defined as

0.30Q for 0 lt Q lt 500, 0.29Q for 500 lt Q lt
1000, 0.28Q for 1000 lt Q
C(Q)
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Discuss

• The company charges less for a larger order to
  provide an incentive for the purchaser to buy
  more.
• 499 bags would cost 149.70
• 500 bags would cost 145.00

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Optimal Policy for All-Units Discount Schedule

• Assume that the demand rate 600 u/y(l 600)
• The fixed cost of placing an order 8 (K8)
• Holding cost 20 annual interest rate (I0.2)
• Then from precious example C0 0.030, C1
  0.29, C2 0.28
• Solution
• Q0 sqrt(2K l /IC0)400 realizable
• Q1 406, Q2414

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The 3 average annual cost curves for previous
example
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Optimal Policy for All-Units Discount Schedule

• There are 3 candidates for the optimal solution
  400, 500, and 1000.
• The opt sol will be either the largest realizable
  EOQ or one of the breakpoints that exceeds it.
• The opt sol is the lot size with the lowest
  average annual cost.
• The average annual cost,
• Gj(Q) lcj lK/QIcjQ/2 for j 0,1,2.

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Optimal Policy for All-Units Discount Schedule

• As shown in figure, ,G(Q) is defined as
• Substitute Q equal 400,500, and 1000 and Cj we
  got (next page.)

G0(Q) for 0lt Q lt 500, G1(Q) for 500lt Q lt
1000, G2(Q) for 1000lt Q
G(Q)
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Optimal Policy for All-Units Discount Schedule

• G(400) G0 (400)
• 6000.36008/4000.20.3400/2 204
• G(400) G1 (500)
• 6000.296008/5000.20.29500/2 198.10
• G(1000) G2 (1000)
• 6000.286008/10000.20.281000/2 200.80
• Hence Place a standing order for 500, average
  annual cost of 198.10

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Incremental Quantity Discounts

• If the trash bags cost 30 cents each for
  quantities of 500 or lessfor quantities between
  500 and 1000, the first 500 cost 30 cents each
  and the remaining amount cost 29 cents each for
  quantities of 1000 and over the first 500 cost 30
  cents each, the next 500 cost 29 cents each, and
  the remaining amount cost 28 cents each.

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Incremental Quantity Discounts

• 0.30Q for 0 lt Q lt 500,
• 150 0.29(Q-500) for 500 lt Q lt 1000,
• 295 0.28(Q-1000) for 1000 lt Q

C(Q)
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Incremental Quantity Discounts
0.30 for 0 lt Q lt
500, 0.295/Q for 500 lt Q lt 1000, 0.2815/Q
for 1000 lt Q the average annual cost function,
G(Q), is G(Q)lC(Q)/Q Kl/QIC(Q)/QQ/2
(unit cost) C(Q)/Q
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Incremental Quantity Discounts

• The opt sol occurs at the minimum of one of the
  average annual cost curves.
• Substituting the 3 expressions for C(Q)/Q for
  G(Q), computing the 3 minima of the curves,
  determining which of these minima fall into the
  correct interval, comparing the average annual
  costs at the realizable values.

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Incremental Quantity Discounts

• G(Q)lC(Q)/Q Kl/QIC(Q)/QQ/2
• We have
• G0(Q)6000.3 8600/Q0.20.3Q/2
• which is minimized at
• Q0sqrt(2Kl/Ic0)sqrt(28600/0.20.3)400

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Incremental Quantity Discounts

• G1(Q)600(0.295/Q) 8600/Q0.2(0.295/Q)(Q/2)
• 6000.293000/Q4800/Q0.20.29Q/20.25/QQ/2
  
• 0.2960013600/Q0.20.29Q/20.25/2
• which is minimized at
• Q1sqrt(2Kl/Ic0)sqrt(213600/0.20.29)519
• G2(Q)600(0.2815/Q) 8600/Q0.2(0.2815/Q)(Q/2
  )
• 0.2860023600/Q0.20.28Q/20.215/2
• which is minimized at
• Q2sqrt(2Kl/Ic0)sqrt(223600/0.20.28)702

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Incremental Quantity Discounts

• Both are realizable. Is not realizable because lt
  1000.Substituting into the expression we got
• G0(Q0) 204.00
• G1(Q1) 204.58
• Optimum sol is ordering for 400 units at the
  highest price of 30 cents per unit.

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???????

• ???????? ????????????? ?????? ????????,
  ???????????????????????????????????????????,
  ????????????????????????, ?????????????????, ?.
  ?????????????, 2544.
• ????? ??????????, ?????????????????????????,
  ?????????? ?.?.?., 2545.
• Steven Nahmias, production and operations
  analysis , 3rd edition, IRWIN, 1997.

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• ?? slide ??????? 16
• 1. T ???????................???????????..........
  .......
• 2. Q ???????................???????????..........
  .......
• 3. ??????????????????????? 1 ??????????????
  (Cycle time) ??????????................???........
  ....
• ?? slide ??????? 18
• 1. ???????????????????????????? ???????????
  ??????????????? T
• ???????????? ????????? ?????????? ..Q 30
  ???? ???????????????? 6 ???? ???????? ???
  ?????? (lead time) 5.5 ?????...??????????????????
  ????????(Cycle time or cycle length) ??? Reorder
  point????????????????