Post on 01-Apr-2015
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Special Inventory Models
Supplement DSupplement D
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Special Inventory Models
Three common situations require relaxation of one or more of the assumptions on which the EOQ model is based.
Noninstantaneous Replenishment occurs when production is not instantaneous and inventory is replenished gradually, rather than in lots.
Quantity Discounts occur when the unit cost of purchased materials is reduced for larger order quantities.
One-Period Decisions: Retailers and manufacturers of fashion goods often face situations in which demand is uncertain and occurs during just one period or season.
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If an item is being produced internally rather than purchased, finished units may be used or sold as soon as they are completed, without waiting until a full lot is completed.
Production rate, p, exceeds the demand rate, d. Cycle inventory accumulates faster than demand
occursa buildup of p – d units occurs per time period,
continuing until the lot size, Q, has been produced.
Noninstantaneous Replenishment
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Production quantityProduction quantity
Demand during Demand during production intervalproduction interval
Maximum inventoryMaximum inventory
Production Production and demandand demand
Demand Demand onlyonly
TBOTBO
On
-han
d i
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n-h
and
in
ven
tory QQ
TimeTime
IImaxmax
p – d
Noninstantaneous Replenishment
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Cycle inventory is no longer Q/2, as it was with the basic EOQ method; instead, it is the maximum cycle inventory (Imax / 2)
Noninstantaneous Replenishment
C = ( ) + (S)DQ
Q p – d2 p
D = annual demandd = daily demandp = production rateS = setup costsQ = ELS
Imax = (p – d) = Q( )Qp
p – dp
Total annual cost (C) = Annual holding cost + annual ordering or setup cost
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Economic production lot size (ELS) is the optimal lot size in a situation in which replenishment is not instantaneous.
Economic Lot Size (ELS)
ELS =p
p – d2DS
H
D = annual demandd = daily demandp = production rateS = setup costsH = annual unit holding cost
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Finding the ELSFinding the ELSExample D.1Example D.1
The manager of a chemical plant must determine the following for a particular chemical:
1. Determine the economic production lot size (ELS).2. Determine the total annual setup and inventory
holding costs. 3. Determine the TBO, or cycle length, for the ELS.4. Determine the production time per lot.
• What are the advantages of reducing the setup time by 10 percent?
Demand = 30 barrels/day Setup cost = $200Production rate = 190 barrels/day Annual holding cost = $0.21/barrelAnnual demand = 10,500 barrels Plant operates 350 days/year
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ELSELS = = pp
pp – – dd22DSDS
HH
Finding the ELS for the Finding the ELS for the Example D.1Example D.1 chemical chemical
Demand = 30 barrels/day Setup cost = $200Production rate = 190 barrels/day Annual holding cost = $0.21/barrelAnnual demand = 10,500 barrels Plant operates 350 days/year
ELSELS = = 190190190190 – – 3030
2(2(10,50010,500)()($200$200))$0.21$0.21
ELSELS = 4873.4 barrels = 4873.4 barrels
D = annual demandd = daily demandp = production rateS = setup costsH = unit holding costQ = ELS
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Finding the Finding the Total Annual Cost Total Annual Cost
Demand = 30 barrels/day Setup cost = $200Production rate = 190 barrels/day Annual holding cost = $0.21/barrelAnnual demand = 10,500 barrels Plant operates 350 days/year
D = annual demandd = daily demandp = production rateS = setup costsH = unit holding costQ = ELS
CC = = ( ( ))((HH) + () + (SS))DDQQ
Q Q pp – – d2 2 pp
CC = = ( ( )) ($0.21) + ( ($0.21) + ($200$200))10,50010,5004873.44873.4
4873.4 4873.4 190190 – – 30 2 2 190190
CC = $430.91 + $430.91 = $430.91 + $430.91 CC = $861.82 = $861.82
Example D.1Example D.1
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Finding the TBO Finding the TBO
Demand = 30 barrels/day Setup cost = $200Production rate = 190 barrels/day Annual holding cost = $0.21/barrelAnnual demand = 10,500 barrels Plant operates 350 days/year
D = annual demandd = daily demandp = production rateS = setup costsH = unit holding costQ = ELS
TBOTBOELSELS = (350 days/year) = (350 days/year)ELSELS
DD
TBOTBOELSELS = (350 days/year) = (350 days/year)4873.44873.410,50010,500
TBOTBOELSELS = 162.4, or 162 days = 162.4, or 162 days
Example D.1Example D.1
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Finding the Finding the Production Time per Lot Production Time per Lot
Demand = 30 barrels/day Setup cost = $200Production rate = 190 barrels/day Annual holding cost = $0.21/barrelAnnual demand = 10,500 barrels Plant operates 350 days/year
D = annual demandd = daily demandp = production rateS = setup costsH = unit holding costQ = ELS
Production time = Production time = ELSELS
pp
Production time = Production time = 4873.44873.4
190190
Production time = 25.6, or 26 daysProduction time = 25.6, or 26 days
Example D.1Example D.1
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Advantage of Reducing Advantage of Reducing Setup TimeSetup Time
OM Explorer Solver for the Economic Production Lot Size Showing the effect of a 10 Percent Reduction in setup cost.
$180 vs original $200
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Application D.1
38.1555
3560
60
000,2
000,100080,1022
dp
p
H
DSELS
or 1555 engines
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Application D.1continued
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Quantity Discounts
Quantity discounts, which are price incentives to purchase large quantities, create pressure to maintain a large inventory.
For any per-unit price level, P, the total cost is:
Total annual cost = Annual holding cost + Annual ordering or setup cost + Annual cost of materials
CC = ( = (HH) + () + (SS) + ) + PDPDQQ22
DDQQ
D = annual demandS = setup costsP = per-unit price levelH = unit holding costQ = ELS
© 2007 Pearson EducationTotal cost curves with Total cost curves with
purchased materials addedpurchased materials added
Quantity Discounts
EOQs and price break quantitiesEOQs and price break quantities
CC for for PP = $4.00 = $4.00CC for for PP = $3.50 = $3.50CC for for PP = $3.00 = $3.00
PDPD for forPP = $4.00 = $4.00 PDPD for for
PP = $3.50 = $3.50 PDPD for forPP = $3.00 = $3.00
EOQ EOQ 4.004.00
EOQ EOQ 3.503.50
EOQ EOQ 3.003.00
First First price price breakbreak
Second Second price price breakbreak
To
tal c
ost
(d
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rs)
To
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ost
(d
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rs)
To
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ost
(d
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rs)
To
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(d
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rs)
Purchase quantity (Purchase quantity (QQ))00 100100 200200 300300
Purchase quantity (Purchase quantity (QQ))00 100100 200200 300300
First price break
Second price break
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Step 1. Beginning with the lowest price, calculate the EOQ for each price level until a feasible EOQ is found. It is feasible if it lies in the range corresponding to its
price.
Step 2. If the first feasible EOQ found is for the lowest price level, this quantity is the best lot size. Otherwise, calculate the total cost for the first feasible
EOQ and for the larger price break quantity at each lower price level. The quantity with the lowest total cost is optimal.
Finding Q with Quantity Discounts
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Annual demand (D) = 936 unitsOrdering cost (S) = $45
Holding cost (H) = 25% of unit price
Order Quantity Price per Unit
0 – 299 $60.00300 – 499 $58.80500 or more $57.00
A supplier for St. LeRoy Hospital has introduced quantity discounts to encourage larger order quantities of a special catheter. The price schedule is:
Example D.2
EOQ EOQ 57.0057.00 = =22DSDS
HH2(936)(45)2(936)(45)0.25(0.25(57.0057.00))
== = 77 units= 77 units
Step 1: Start with lowest price level:
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Example D.2 continued
Annual demand (D) = 936 unitsOrdering cost (S) = $45
Holding cost (H) = 25% of unit price
Order Quantity Price per Unit
0 – 299 $60.00300 – 499 $58.80500 or more $57.00
EOQ EOQ 57.0057.00 = =22DSDS
HH2(936)(45)2(936)(45)0.25(0.25(57.0057.00))
== = 77 units= 77 units
EOQ EOQ 58.8058.80 = =22DSDS
HH2(936)(45)2(936)(45)0.25(0.25(58.8058.80))
== = 76 units= 76 units
EOQ EOQ 60.0060.00 = =22DSDS
HH2(936)(45)2(936)(45)0.25(0.25(60.0060.00))
== = 75 units= 75 units
This quantity is feasible because it lies in the range corresponding to its price.
Not feasible
Not feasible
Feasible
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Step 2: The first feasible EOQ of 75 does not correspond to the lowest price level. Hence, we must compare its total cost with the price break quantities (300 and 500 units) at the lower price levels ($58.80 and $57.00):
Example D.2 continued
CC = ( = (HH) + () + (SS) + ) + PPDDQQ22
DDQQ
CC7575 = [( = [(0.25)($60.000.25)($60.00)] + ($45) + $60.00()] + ($45) + $60.00(936936))757522
9369367575
CC7575 = $57,284 = $57,284
CC300300 = [(0.25)($58.80)] + ($45) + $58.80(936) = [(0.25)($58.80)] + ($45) + $58.80(936)300300
22936936300300
= $57,382= $57,382
CC500500 = [(0.25)($57.00)] + ($45) + $57.00(936) = [(0.25)($57.00)] + ($45) + $57.00(936)500500
22936936500500
= $56,999= $56,999
The best purchase quantity is 500 units, which qualifies for the deepest discount.
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Decision Point:If the price per unit for the range of 300 to 499 units is reduced to $58.00, the best decision is to order 300 catheters, as shown below. This shows that the decision is sensitive to the price schedule. A reduction of slightly more than 1 percent is enough to make the difference in this example.
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Application D.2
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Application D.2Solution
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One-Period Decisions
This type of situation is often called the newsboy problem. If the newspaper seller does not buy enough newspapers to resell on the street corner, sales opportunities are lost. If the seller buys too many newspapers, the overage cannot be sold because nobody wants yesterday’s newspaper.
1. List the different levels of demand that are possible, along with the estimated probability of each.
2. Develop a payoff table that shows the profit for each purchase quantity, Q, at each assumed demand level.
3. Calculate the expected payoff for each Q (or row in the payoff table) by using the expected value decision rule.
4. Choose the order quantity Q with the highest expected payoff.
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The payoff for a given quantity-demand combination depends on whether all units are sold at the regular profit margin, which results in two possible cases.
1. If demand is high enough (Q < D) then all of the cases are sold at the full profit margin, p, during the regular season.
Payoff = (Profit per unit)(Purchase quantity) = pQ
2. If the purchase quantity exceeds the eventual demand (Q > D), only D units are sold at the full profit margin, and the remaining units purchased must be disposed of at a loss, l, after the season.
One-Period Decisions
Payoff = (Profit per unit during season) (Demand) – (Loss per unit) (Amount disposed of after season) = pD – l(Q – D)
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10 $100 $100 $100 $100 $100 10020 50 200 200 200 200 17030 0 150 300 300 300 19540 –50 100 250 400 400 17550 –100 50 200 350 500 140
Q 10 20 30 40 50 Expected Payoff
Demand 10 20 30 40 50
Demand Probability 0.2 0.3 0.3 0.1 0.1
Example D.3A gift museum shop sells a Christmas ornament at a $10 profit per unit during the holiday season, but it takes a $5 loss per unit after the season is over. The following is the discrete probability distribution for the season’s demand:
Expected payoff if Q = 30: 0(0.2)+(150(0.3)+300(0.3+0.1+0.1) = $195Payoff if Q = 30 and D = 40: pD = 10(30) = $300Payoff if Q = 30 and D = 20: pD – l(Q – D)=10(20) – 5(30 – 20) = $150
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Example D.3 OM Explorer Solution
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Solved Problem 1
For Peachy Keen, Inc., the average demand for mohair sweaters is 100 per week. The production facility has the capacity to sew 400 sweaters per week. Setup cost is $351. The value of finished goods inventory is $40 per sweater. The annual per-unit inventory holding cost is 20 percent of the item’s value.
a. What is the economic production lot size (ELS)?
b. What is the average time between orders (TBO)?
c. What is the minimum total of the annual holding cost and setup cost?
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Solved Problem 1
ELS = ELS = pp
pp – – dd22DDSS
HH
ELS = ELS = 400400(400 – 100)(400 – 100)
22(100)(52)(100)(52)($351)($351)0.20($40)0.20($40)
= 780 sweaters= 780 sweaters
TBOTBOELSELS = =ELSELS
DD780780
5,2005,200= 0.15 year or 7.8 weeks= 0.15 year or 7.8 weeks==
CC = = ( ( ))((HH) + () + (SS))DDQQ
Q pQ p – – d2 p 2 p
CC = = ( ( )) ( (0.20 x $400.20 x $40) + ($351)) + ($351)5,2005,200780780
780 400 – 780 400 – 100 2 400 400
= 2,340/year + $2,340/year= 2,340/year + $2,340/year = $4,680/year= $4,680/year
D = 5,200p = 400
d = 100S = $351
H = 20% of $40
a.
c. b.
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Solved Problem 3
For Swell Productions, a concession stand will sell poodle skirts and other souvenirs of the 1950s a one-time event. Skirts are purchased for $40 each and are sold during for $75 each.
Unsold skirts can be returned for a refund of $30 each. Sales depend on the weather, attendance, and other variables.
The following table shows the probability of various sales quantities. How many skirts should be ordered?
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Solved Problem 3
Probabilities0.05 0.11 0.34 0.34 0.11 0.05
The highest expected payoff occurs when 400 skirts are ordered.