Transcript KRM8 Supplement D - Washburn University

Special
Inventory Models
Supplement D
© 2007 Pearson Education
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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Noninstantaneous
Replenishment
 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
occurs
 a buildup of p – d units occurs per time period,
continuing until the lot size, Q, has been
produced.
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On-hand inventory
Noninstantaneous
Replenishment
Production quantity
Q
Demand during
production interval
Imax
Maximum inventory
p–d
Time
Production
and demand
Demand
only
TBO
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Noninstantaneous
Replenishment
 Cycle inventory is no longer Q/2, as it was with the
basic EOQ method; instead, it is the maximum
cycle inventory (Imax / 2)
p–d
Q
Imax =
(p – d) = Q
p
p
(
)
 Total annual cost (C) = Annual holding cost +
annual ordering or setup cost
Q p–d
C =2
p
(
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)
D
+
(S)
Q
D = annual demand
d = daily demand
p = production rate
S = setup costs
Q = ELS
Economic Lot Size (ELS)
 Economic production lot size (ELS) is
the optimal lot size in a situation in which
replenishment is not instantaneous.
ELS =
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2DS
H
p
p–d
D = annual demand
d = daily demand
p = production rate
S = setup costs
H = annual unit holding cost
Finding the ELS
Example D.1
1.
2.
3.
4.
The manager of a chemical plant must determine
the following for a particular chemical:
Determine the economic production lot size (ELS).
Determine the total annual setup and inventory
holding costs.
Determine the TBO, or cycle length, for the ELS.
Determine the production time per lot.
• What are the advantages of reducing the setup
time by 10 percent?
Demand = 30 barrels/day
Production rate = 190 barrels/day
Annual demand = 10,500 barrels
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Setup cost = $200
Annual holding cost = $0.21/barrel
Plant operates 350 days/year
Finding the ELS for the
Example D.1 chemical
ELS =
ELS =
2DS
H
D = annual demand
d = daily demand
p = production rate
S = setup costs
H = unit holding cost
Q = ELS
p
p–d
2(10,500)($200)
$0.21
190
190 – 30
ELS = 4873.4 barrels
Demand = 30 barrels/day
Production rate = 190 barrels/day
Annual demand = 10,500 barrels
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Setup cost = $200
Annual holding cost = $0.21/barrel
Plant operates 350 days/year
Finding the
Total Annual Cost
Example D.1
D = annual demand
d = daily demand
p = production rate
S = setup costs
H = unit holding cost
Q = ELS
Q p–d
D
C =2
(H) +
(S)
p
Q
(
)
10,500
4873.4 190 – 30
C=
($0.21) +
($200)
4873.4
2
190
(
)
C = $430.91 + $430.91
Demand = 30 barrels/day
Production rate = 190 barrels/day
Annual demand = 10,500 barrels
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C = $861.82
Setup cost = $200
Annual holding cost = $0.21/barrel
Plant operates 350 days/year
Finding the TBO
Example D.1
ELS
TBOELS =
(350 days/year)
D
4873.4
TBOELS =
(350 days/year)
10,500
D = annual demand
d = daily demand
p = production rate
S = setup costs
H = unit holding cost
Q = ELS
TBOELS = 162.4, or 162 days
Demand = 30 barrels/day
Production rate = 190 barrels/day
Annual demand = 10,500 barrels
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Setup cost = $200
Annual holding cost = $0.21/barrel
Plant operates 350 days/year
Finding the
Production Time per Lot
Example D.1
D = annual demand
d = daily demand
p = production rate
S = setup costs
H = unit holding cost
Q = ELS
ELS
Production time =
p
4873.4
Production time =
190
Production time = 25.6, or 26 days
Demand = 30 barrels/day
Production rate = 190 barrels/day
Annual demand = 10,500 barrels
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Setup cost = $200
Annual holding cost = $0.21/barrel
Plant operates 350 days/year
Advantage of Reducing
Setup 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
ELS 
2DS
H
p
210,080100,000
60

 1555.38
pd
2,000
60  35
or 1555 engines
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Application D.1
continued
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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
D
Q
C=
(H) +
(S) + PD
Q
2
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D = annual demand
S = setup costs
P = per-unit price level
H = unit holding cost
Q = ELS
Quantity Discounts
EOQ 4.00
EOQ 3.50
EOQ 3.00
PD for
P = $4.00
First
price
break
0
PD for
P = $3.50
PD for
P = $3.00
Total cost (dollars)
Total cost (dollars)
C for P = $4.00
C for P = $3.50
C for P = $3.00
First
price
break
Second
price
break
100
200
Purchase quantity (Q)
Total cost curves with
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Pearson Education
materials added
300
0
Second
price
break
100
200
Purchase quantity (Q)
300
EOQs and price break quantities
Finding Q
with Quantity Discounts
 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.
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Example D.2
A supplier for St. LeRoy Hospital has introduced
quantity discounts to encourage larger order quantities
of a special catheter. The price schedule is:
Order Quantity
0 – 299
300 – 499
500 or more
Price per Unit
$60.00
$58.80
$57.00
Annual demand (D) = 936 units
Ordering cost (S) = $45
Holding cost (H) = 25% of unit price
Step 1: Start with lowest price level:
EOQ 57.00 =
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2DS
H
=
2(936)(45)
0.25(57.00)
= 77 units
Example D.2
continued
EOQ 57.00 =
EOQ 58.80 =
EOQ 60.00 =
2DS
H
=
2DS
H
=
2DS
H
=
2(936)(45)
0.25(57.00)
= 77 units
2(936)(45)
0.25(58.80)
= 76 units
2(936)(45)
0.25(60.00)
= 75 units Feasible
Not feasible
Not feasible
This quantity is feasible because it lies in the range corresponding to its price.
Order Quantity
0 – 299
300 – 499
500 or more
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Price per Unit
$60.00
$58.80
$57.00
Annual demand (D) = 936 units
Ordering cost (S) = $45
Holding cost (H) = 25% of unit price
Example D.2
continued
 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):
D
Q
C=
(H) +
(S) + PD
Q
2
936
75
C75 =
[(0.25)($60.00)] +
($45) + $60.00(936) C75 = $57,284
75
2
C300 =
936
300
[(0.25)($58.80)] +
($45) + $58.80(936) = $57,382
300
2
936
500
C500 =
[(0.25)($57.00)] +
($45) + $57.00(936) = $56,999
500
2
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.2
Solution
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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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One-Period Decisions

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.
Payoff = (Profit per unit during season) (Demand) – (Loss per unit) (Amount
disposed of after season) = pD – l(Q – D)
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Example D.3
A 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:
Q
10
20
30
40
50
Demand
10
20
30
40
50
Demand Probability
0.2
0.3
0.3
0.1
0.1
40
50
10
20
30
$100 $100 $100 $100 $100
50 200 200 200 200
0 150 300 300 300
–50 100 250 400 400
–100
50 200 350 500
Expected Payoff
100
170
195
175
140
Payoff
=ifand
30
= 40:
= $300
Expected
Q and
=D30:
0(0.2)+(150(0.3)+300(0.3+0.1+0.1)
Payoff
if payoff
Qif=Q30
=D20:
pD –pD
l(Q= –10(30)
D)=10(20)
– 5(30 – 20)= =$195
$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 perunit 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
a.
2DS
H
ELS =
p
p–d
D = 5,200 d = 100
p = 400
S = $351
H = 20% of $40
400
2(100)(52)($351)
= 780 sweaters
(400 – 100)
0.20($40)
ELS =
ELS
780
=
TBOELS =
= 0.15 year or 7.8 weeks
D
5,200
b.
c.
Q p–d
D
C =2
(H) +
(S)
p
Q
C=
(
)
780
2
5,200
400 – 100
(0.20 x $40) + 780 ($351)
400
(
)
= 2,340/year + $2,340/year = $4,680/year
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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
The highest expected payoff occurs when 400 skirts are ordered.
0.05
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0.11
Probabilities
0.34
0.34
0.11
0.05