第四單元： Inventory Management: Cycle Inventory
Inventory Management:
Cycle Inventory
郭瑞祥教授
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－非商業性－相同方式分享」台灣3.0版授權釋出】
1
Role of Inventory in the Supply Chain
Understocking: Demand exceeds amount available
–Lost margin and future sales
Overstocking: Amount available exceeds demand
– Liquidation, Obsolescence, Holding
2
Why hold inventory?
► Economies of scale
》 Batch size and cycle time
► Stochastic
variability
》 Quantity
discountsof supply and demand
》 Short term
/ Trade
promotions
》Service
leveldiscounts
given safety
inventory
》Evaluating Service level given safety inventory
3
Role of Inventory in the Supply Chain
Improve
of Supply
Supply
ImproveMatching
Matching of
and
and Demand
Improved Forecasting
Cost
Efficiency
Reduce Material Flow Time
Availability
Responsiveness
Reduce Waiting Time
Reduce Buffer Inventory
4
Economies of Scale
Supply / Demand
Variability
Seasonal Variability
Cycle Inventory
Safety Inventory
Seasonal Inventory
Cycle Inventory
Cycle inventory
is the
average
inventory that built up in the supply
Improve
Matching
of Supply
chain because and
a stage
of the supply chain either produces or
Demand
purchases in lots that are larger than those demanded by the customer.
Improved Forecasting
Inventory
Cost
Efficiency
Reduce Material Flow Time
QReduce Waiting Time
Availability
Responsiveness
Reduce Buffer Inventory
Time t
Supply
/ Demand= lot size/2 = Q/2
Cycle
inventory
5
Economies of Scale
Variability
Seasonal Variability
Cycle Inventory
Safety Inventory
Seasonal Inventory
Little’s Law
► Average flow time = Average inventory / Average
flow rate
► For any supply chain, average flow rate equals the
demand,
Average flow time resulting from cycle inventory
= Cycle inventory / Demand
= Q / 2D
Q: Lot size
D: Demand per unit time
6
Holding Cycle Inventory for
Economies of Scale
► Fixed costs associated with lots
► Quantity discounts
► Trade Promotions
7
Economics of Scale to Exploit Fixed Costs
— Economic Order Quantity—
» D= Annual demand of the product
» S= Fixed cost incurred per order
» C= Cost per unit
» h=Holding cost per year as a fraction of product cost
» H=Holding cost per unit per year =hC
» Q=Lot size
» n=Order frequency
8
Lot Sizing for a Single Product (EOQ)
►
Annual order cost =(D/Q)S=ns
Annual holding cost = (Q/2)H =(Q/2)hC
Annual material cost = CD
TC =CD + (D/Q)S + (Q/2)hC
Cost
Total Cost
Holding Cost
Order Cost
Material Cost
Lot Size
9
Totalfor
annual
TC =CD
+ (D/Q)S + (EOQ)
(Q/2)hc
Lot Sizing
a cost,
Single
Product
Optimal lot size, Q is obtained by taking the first derivative
►
Annual orderd cost
-DS hC
(TC) =(D/Q)S
2DS
*
= 2 +
=0
=
Annual holdingdQ
cost =Q
(Q/2)H2=(Q/2)hc Q
hC
Annual material cost = CD‧
D = DhC Average flow time = Q*/2D
*
=
n
2S
Q*
TC =CD + (D/Q)S + (Q/2)hc
Cost
Total Cost
Holding Cost
Order Cost
Material Cost
Lot Size
10
Example
 Demand， D =1,000 units/month
= 12,000 units/year
 Fixed cost, S = $4,000/order
 Unit cost, C = $500
 Holding cost, h = 20% = 0.2
》Optimal order size
Q=
2X12000X4000
= 980
0.2X500
》Cycle inventory
Q/2 =490
》Numbers of orders per year
》Average flow time
11
D / Q = 12000 / 980 =12.24
Q / 2D = 490 / 12000 =0.041 (year)
=0.49(mounth)
Example - Continued
 Demand， D =1,000 units/month
► If we want to reduce the optimal lot size from 980 to 200,
= 12,000 units/year
how much the order cost per lot should be.
 Fixed cost, Sthen
= $4,000/order
 Unit cost, C = $500 h C ( Q * ) 2
0.2X500X2002
S = = 0.2
=
= $166.7
 Holding cost, h = 20%
2X12000
2D
2X12000X4000
》Optimal
= 1100),
980 what
 order
If we size
increase theQlot=size by 10% (from 980 to
0.2X500
the
total cost would be.
Annual
cost = $ 98,636
(from $ 97,980)(an increase by only 0.6%)
Q/2 =490
》Cycle
inventory
(Note: material cost is not included)
》Numbers of orders per year
》Average flow time
Microsoft。
Microsoft。
12
D / Q = 12000 / 980 =12.24
Microsoft。
Q / 2D = 490 / 12000
=0.041
(year)
Microsoft。
CoolCLIPS
=0.49(mounth)Microsoft。
Key Points from EOQ
 Total order and holding costs are relatively stable around the
economic order quantity. A firm is often better served by ordering a
convenient lot size close to the EOQ rather than the precise EOQ.
13

If demand increases by a factor of k, the optimal lot size increases
by a factor of k . The number of orders placed per year should
also increase by a factor of k . Flow time attributed to cycle
inventory should decrease by a factor of k .

To reduce the optimal lot size by a factor of k, the fixed order cost
S must be reduced by a factor of k2 .
Aggregating Multiple Products in a Single Order
► One of major fixed costs is transportation
► Ways to lower the fixed ordering and transportation costs:
► Ways
to lower receiving
or loading
》Aggregating
across the
productscosts:
from the same supplier
》Single delivery from multiple suppliers
》ASN (Advanced Shipping Notice) with EDI
》Single delivery to multiple retailers
Microsoft。
Microsoft。
Microsoft。
14
Example: Lot Sizing with Multiple Products
 Three computer models (L, M, H) are sold and the demand per year:
–DL = 12,000; DM = 1,200; DH = 120
L
M
► Common fixed (transportation) cost, S = $4,000
► Additional product specific order cost
》sL = $1,000; sM = $1,000; sH = $1,000
Microsoft。
► Holding cost, h = 0.2
► Unit cost
》CL = $500; CM = $500; CH = $500
15
Microsoft。
H
Microsoft。
Delivery Options
► No aggregation
► Complete
aggregation
》Each product
is ordered separately
► Tailored
aggregation
》All products
are delivered on each truck
》Selected subsets of products on each truck
16
Option 1: No Aggregation Result
► No aggregation
► Complete aggregation
Litepro
► Tailored aggregation
Demand per year
12000
Medpro
Heavypro
1200
120
Fixed cost / order
$5,000
$5,000
$5,000
Optimal order size
1,095
346
110
Order frequency
11.0/year
3.5/year
1.1/year
Annual holding cost
$109,544
$34,642
$10,954
Annual total cost = $155,140 (no material cost)
17
Option 2: Complete Aggregation
► No aggregation
► Complete aggregation
► Tailored
aggregation
》Combined
fixed cost per order is given by
S = S + sL + sM + sH
*
2DS
=
Q
hC
*
》Let n be the number of orders placed per year. We have
Total annual cost = Annual order cost + Annual holding cost
=
*
(S n ) + [(DL hCL / 2n ) + (DM hCM / 2n ) + (DH hCH / 2n )]
n* =
18
DL hCL +DM hCM +DH hCH
2S*
DhC
=
n
2S
*
Option 2: Complete Aggregation
► No aggregation
► Complete aggregation
► Tailored
aggregation
》Combined
fixed cost per order is given by
S = S + sL + sM + sH
*
》Let n be the number of orders placed per year. We have
Total annual cost = Annual order cost + Annual holding cost
=
*
(S n ) + [(DL hCL / 2n ) + (DM hCM / 2n ) + (DH hCH / 2n )]
n* =
19
DL hCL +DM hCM +DH hCH
2S*
DhC
=
n
2S
*
Option 2: Complete Aggregation Result
► No aggregation
► Complete aggregation
► Tailored aggregation
Litepro
Medpro
Heavypro
Demand per year
12000
1200
120
Order frequency
9.75/year
9.75/year
9.75/year
Optimal order size
1,230
123
12.3
Annual holding cost
$61,512
$6,151
$615
Annual order cost = 9.75×$7,000 = $68,250
Annual total cost = $68,250+$61,512+$6,151+$615=$136,528
20
Option 3: Tailored aggregation
► No aggregation
► Complete aggregation
► Tailored aggregation
A heuristic that yields an ordering policy whose cost is close to optimal.
► Step 1: Identify most frequently ordered product.
► Step 2: Identify frequency of other products as a multiple of the order
frequency of the most frequently ordered product.
► Step 3: Recalculate order frequency of most frequently ordered product.
► Step 4: Identify ordering frequency of all products.
21
Option 3: Tailored aggregation
n = Max { ni =
i
► No aggregation
hC L D L
= 11 . 0, n
= 3.5,
nL =
M
► Complete aggregation
2( S + s L )
n H = 1.1
hCiDi
}
2(S+si)
\ n = 11 . 0
► Tailored aggregation
►A heuristic that yields an ordering policy whose cost is close to optimal.
► Step 1: Identify most frequently ordered product.
► Step 2: Identify frequency of other products as a multiple of the order
frequency of the most frequently ordered product.
► Step 3: Recalculate order frequency of most frequently ordered product.
► Step 4: Identify ordering frequency of all products.
22
Option 3: Tailored aggregation
► Step 1: Identify most frequently ordered product.
► No aggregation
hC L D L
= 11 . 0, n
= 3.5,
nL =
M
► Complete aggregation
2( S + s L )
n H = 1.1
\ n = 11 . 0
► ►Tailored
Step 2:aggregation
Identify frequency of other products as a multiple of the order
frequency of the most frequently ordered product.
A heuristic that yields an ordering policy whose cost is close to
= optimal.
hC D
nM =
hCM DM
= 7 .7
2sM
nH = 2.4
n=
i
i
2si
 Step 2: Identify frequency of other products as a multiple of the order
11 . 0 / 7ordered
frequency
mM =ofthe
n / most
nM  =frequently
. 7  = 1product.
. 4  = 2 mH = 4 . 5  = 5
► Step 3: Recalculate order frequency of most frequently ordered product.
► Step 4: Identify ordering frequency of all products.
23
Option 3: TailoredDerivation
aggregation
of n
►► No
TC= order
cost + holding
Stepaggregation
1: Identify
most frequently
ordered cost
product.
► Complete aggregation
hC L D L
= 11 . 0, n
= 3.5,
nL =
M
2( S + s L )
► Tailored aggregation
n H = 1.1
\ n = 11 . 0

Step 2: Identify
frequency
of other
products
a multiple
order
A heuristic
that yields
an ordering
policy
whoseas
cost
is close of
to the
optimal.
frequency of the most frequently ordered product.
hCM DM
=
= 7 .7
n
nH =products
2.4
► StepM 2: Identify
frequency
of
other
as a multiple of the order
2sM
frequency of the most frequently ordered product.
mM =  n / nM  =11 . 0 / 7 . 7  = 1 . 4  = 2 mH = 4 . 5  = 5
► Step 3: Recalculate order frequency of most frequently ordered product.
► Step 3: Recalculate order frequency of most frequently ordered product.
 Step 4: Identify ordering frequency of all products.
hCi Di mi
n=
2 [ S + (si / mi )]
24
Option 3: TailoredDerivation
aggregation
of n
►► No
TC= order
cost + holding
Stepaggregation
1: Identify
most frequently
ordered cost
product.
Di
► Complete aggregation

(hC
)

n
s
hC
D
i
i
i
nS
+
)+. 0, n
= 3.5,
n L = TC=( L L i = 11
i
2n
M
i
2( S + s L )
► Tailored aggregation
n H = 1.1
\ n = 11 . 0
 hCiDiMi
n

Step 2: Identify
frequency
other
products
a multiple
order
i
 ordering
sof
A heuristic
that =
yields
an
policy
whoseas
cost
is close of
to the
optimal.
+
i
nS
+
i m
frequency of the most
frequently
ordered
product.
i
2n
mM =hCMnD/MnM  =11 . 0 / 7 . 7  = 1 . 4  = 2 mH = 4 . 5  = 5
= 7 .7
n =
nH =products
2.4
► StepM 2: Identify
frequency
of
other
as a multiple of the order
2sM
frequency of the most frequently ordered product.
mM =  n / nM  =11 . 0 / 7 . 7 
 Step 3: Recalculate order frequency of most frequently ordered product.
 Step 3: Recalculate order frequency of most frequently ordered product.
► Step 4: Identify ordering frequency of all products.
hCi Di mi
n=
2 [ S + (si / mi )]
25
Option 3: TailoredDerivation
aggregation
of n

TC= order
cost + holding
► No
Stepaggregation
1: Identify
most frequently
ordered cost
product.
Di
 Complete aggregation

(hC
)

n
s
hC
D
i
i
i
nS
+
)+. 0, n
= 3.5,
n L = TC=( L L i = 11
i
2n
M
i
2( S + s L )
 Tailored aggregation
n H = 1.1
\ n = 11 . 0
 hCiDiMi
n

Step 2: Identify
frequency
other
products
a multiple
order
i
 ordering
sof
A heuristic
that =
yields
an
policy
whoseas
cost
is close of
to the
optimal.
+
i
nS
+
i m
frequency of the most
frequently
ordered
product.
i
2n
 hCiDiMi
s

TC
i
hCM DM = 0  S+ 
- i
=0
=
=
=
n
7
.
7
n
2.4
2
m
M
H
i

2n
i
n
 Step 2: Identify
2s frequency of other products as a multiple of the order
M
frequency of the most frequently ordered product.
11 . 0 /7hC
 i mi
iD
mM =  n / nM\ =
.
7
=
n
 Step 3: Recalculate order
frequency of most frequently ordered product.
(si most
2 [ S + of
/ mi )]frequently ordered product.
► Step 3: Recalculate order frequency
 Step 4: Identify ordering frequency of all products.
hCi Di mi
n=
2 [ S + (si / mi )]
26
Option 3: TailoredDerivation
aggregation
of n
►► No
TC= order
cost + holding
cost
Stepaggregation
1: Identify
most frequently
ordered product.
Di
► Complete aggregation

(hC
)

n
s
hC
D
i
i
i
nS
+
= 3.5,
n L = TC=( L L i = 11)+. 0,i n M 2n
i
2( S + s L )
► Tailored aggregation
n H = 1.1
\ n = 11 . 0
 hCiDiMi
n
►A heuristic
Step 2: Identify
frequency
of
i products as a multiple of the order
ordering
si other
that yields
an
policy
whose cost is close to optimal.
+
nS
=
+
mi
frequency of the mosti frequently
ordered
product.
2n
 hCiDiMi
s

TC
i
hCM DM = 0  S+ 
- i
=0
=
=
=
n
7
.
7
n
2.4
2
m
M
H
i

2n
i
n
► Step 2: Identify
2s frequency of other products as a multiple of the order
M
frequency of the most frequently ordered product.
 =n11
mM =  n / nM\
.0 /
7 hC
. 7iD=i m1i . 4  = 2 mH = 4 . 5  = 5
=
 Step 3: Recalculate order frequency of most frequently ordered product.
2 [ S +of(smost
i / mi )]
► Step 3: Recalculate order frequency
frequently ordered product.
 Step 4: Identify ordering frequency of all products.
hCi Di mi
n=
=11.47
2 [ S + (si / mi )]
L
Microsoft。
27
Option 3: Tailored aggregation

aggregation
► No
Step
1: Identify most frequently ordered product.
 Complete aggregation
hC L D L
= 11 . 0, n
= 3.5,
nL =
M
2( S + s L )
 Tailored aggregation
n H = 1.1
\ n = 11 . 0
►A heuristic
Step 2: Identify
frequency
of other
products
a multiple
order
that yields
an ordering
policy
whoseas
cost
is close of
to the
optimal.
frequency of the most frequently ordered product.
hCM DM
=
= 7 .7
n
nH =products
2.4
 StepM 2: Identify
frequency
of
other
as a multiple of the order
2sM
frequency of the most frequently ordered product.
mM =  n / nM  =11 . 0 / 7 . 7  = 1 . 4  = 2 mH = 4 . 5  = 5
 Step 3: Recalculate order frequency of most frequently ordered product.
► Step 3: Recalculate order frequency of most frequently ordered product.
 Step 4: Identify ordering frequency of all products.
hCi Di mi
n=
=11.47
2 [ S + (si / mi )]
► Step 4:
28
nL=11.47/year, nM=11.47/2=5.74/year,
nH=11.47/5=2.29/year .
Option 3: Tailored Aggregation Result
► No aggregation
► Complete aggregation
► Tailored aggregation
Litepro
Medpro
Heavypro
Demand per year
12000
1200
120
Order frequency
11.47/year
5.74/year
2.29/year
Optimal order size
1,046
209
52
Annual holding cost
$52,310
$10,453
$2620
Annual order cost = nS + nLsL+ sMsM + nHsH =$65,380
Annual total cost = $130,763
Complete aggregation (Annual total cost) =$136,528
29
Option 3: Tailored aggregation
► No aggregation
► Complete aggregation
► Tailored aggregation
A heuristic that yields an ordering policy whose cost is close to optimal.
► Step 1: Identify most frequently ordered product.
A fixedfrequency
cost of (S+si)
is allocated
productofi,the
andorder
► Step 2: ─
Identify
of other
productstoaseach
a multiple
frequency of the most frequently ordered product.

hCi Di 
The most frequently order frequency = n = Max  ni =

i
+
2(S si ) 
► Step 3: Recalculate order frequency of most frequently ordered
product.

► Step 4: Identify ordering frequency of all products.
30
Option 3: Tailored aggregation
► No aggregation
► Complete aggregation
► Tailored aggregation
A heuristic that yields an ordering policy whose cost is close to optimal.
► Step 1: Identify most frequently ordered product.
► Step 2: Identify frequency of other products as a multiple of the order
frequency of the most frequently ordered product.
► Step
of most frequently ordered product.
=
= 3: Recalculate order frequency
mi = n / ni
mi = mi
ni = hCi Di
2si ordering frequency of all products.
► Step 4: Identify
31
Option 3: Tailored aggregation
► No aggregation
► Complete aggregation
► Tailored aggregation
A heuristic that yields an ordering policy whose cost is close to optimal.
► Step 1: Identify most frequently ordered product.
► Step 2: Identify frequency of other products as a multiple of the order
frequency of the most frequently ordered product.
► Step 3: Recalculate order frequency of most frequently ordered product.
hCordering
► Step =4: Identify
frequency of all products.
i Di mi
n
2 [ S + (si / mi )]
32
Option 3: Tailored aggregation
► No aggregation
► Complete aggregation
► Tailored aggregation
A heuristic that yields an ordering policy whose cost is close to optimal.
► Step 1: Identify most frequently ordered product.
► Step 2: Identify frequency of other products as a multiple of the order
frequency of the most frequently ordered product.
► Step 3: Recalculate order frequency of most frequently ordered product.
► Step 4: Identify ordering frequency of all products.
ni=n/mi
33
Impact of Product Specific Order Cost
Product specific
Product specific
order cost =$1,000 order cost =$3,000
No aggregation
$155,140
$183,564
Complete aggregation
$136,528
$186,097
Tailored aggregation
$130,763
$165,233
?
34
Lessons From Aggregation
 Aggregation allows firm to lower lot size without increasing cost
 Complete aggregation is effective if product specific fixed cost is a
small fraction of joint fixed cost
 Tailored aggregation is effective if product specific fixed cost is large
fraction of joint fixed cost
35
Why hold inventory?
► Economies of scale
》 Batch size and cycle time
► Stochastic
variability
》 Quantity
discountsof supply and demand
》 Short term discounts / Trade promotions
36
Quantity Discounts
► Lot size based
》Based on the quantity ordered in a single lot
> All units
► Volume based
> Marginal unit
》Based on total quantity purchased over a given period
 How should buyer react? How does this decision affect the supply chain
in terms of lot sizes, cycle inventory, and flow time?
 What are appropriate discounting schemes that suppliers should offer?
37
All Unit Quantity Discounts
Total Material Cost
Average Cost per Unit
C0
C1
C2
Quantity Purchased
q1
q2
q3
Order Quantity
q1
q2
q3
If an order that is at least as large as qi but smaller than qi+1 is placed,
then each unit is obtained at the cost of Ci.
38
Total Cost
Evaluate EOQ for All Unit Quantity
Discounts
Lowest
cost in the range
► Evaluate EOQ for price in range qi to qi+1 , Qi = 2DS
hCi
》 Case 1:If qi  Qi < qi+1
Total Cost
Lowest cost in the range
, evaluate cost of ordering Qi EOQi
Q
TCi = D S + i hCi + DCi
2
Qi
qi
qi+1 Order Quantity
Total Cost Lowest cost in the range
EOQi
》 Case 2:If Qi < qi, evaluate cost of ordering qi
q
TCi = D S + i hCi + DCi
2
qi
》 Case 3:If Qi  qi+1 , evaluate cost of ordering qi+1
qi+1
D
S+
TCi =
hCi + DCi+1
qi+1
2
qi qi+1
qi
qi+1
EOQi
Order Quantity
► Choose the lot size that minimizes the total cost over all price ranges.
39
Example
 Assume the all unit quantity discountsAverage Cost per Unit
Order Quantity
Unit Price
0-5,000
$ 3.00
5,000-10,000
$ 2.96
10,000 or more
$ 2.92
D = 120,000/ year
CS0 = $100/lot
h = 0.2
C
1
C2
 Based on the all unit quantity discounts, we have
Quantity
Purchased
q0=0, q1=5,000, q2=10,000
C0=$3.00, C1=$2,96, C2=$2.92
q1
q2
q3
 If i = 0, evaluate Q0 as Q0= 2DS = 6,324
hC0
Since Q0 > q1, we set the lost size at q1=5,000 and the total cost
40
Evaluate EOQ for All Unit Quantity Discounts
 Evaluate EOQ for price in range qi to qi+1 , Qi = 2DS
hCi
》 Case 1:If qi  Qi < qi+1 , evaluate cost of ordering Qi
Q
TCi = D S + i hCi + DCi
2
Qi
》 Case 2:If Qi < qi, evaluate cost of ordering qi
q
TCi = D S + i hCi + DCi
2
qi
》 Case 3:If Qi  qi+1 , evaluate cost of ordering qi+1
qi+1
D
S+
TCi =
hCi + DCi+1
qi+1
2
 Choose the lot size that minimizes the total cost over all price ranges.
41
Example
 Assume the all unit quantity discounts
Order Quantity
Unit Price
0-5,000
$ 3.00
5,000-10,000
$ 2.96
10,000 or more
$ 2.92
D = 120,000/ year
S = $100/lot
h = 0.2
 Based on the all unit quantity discounts, we have
q0=0, q1=5,000, q2=10,000
C0=$3.00, C1=$2,96, C2=$2.92
 If i = 0, evaluate Q0 as Q0= 2DS = 6,324
hC0
Since Q0 > q1, we set the lost size at q1=5,000 and the total cost
q
TC0= D S + 1 hC1+ DC1 = $359,080
2
q1
42
Example
 Assume the all unit quantity discounts
Order Quantity
Unit Price
0-5,000
$ 3.00
5,000-10,000
$ 2.96
10,000 or more
$ 2.92
D = 120,000/ year
S = $100/lot
h = 0.2
 Based on the all unit quantity discounts, we have
q0=0, q1=5,000, q2=10,000
C0=$3.00, C1=$2,96, C2=$2.92
 If i = 0, evaluate Q0 as Q0= 2DS = 6,324
hC0
Since Q0 > q1, we set the lost size at q1=5,000 and the total cost
q
TC0= D S + 1 hC1+ DC1 = $359,080
2
q1
43
All Unit Quantity Discounts
Total Material Cost
Average Cost per Unit
C0
C1
C2
Quantity Purchased
q1
q2
q3
Order Quantity
q1
q2
q3
If an order is placed that is at least as large as qi but smaller than qi+1,
then each unit is obtained at a cost of Ci.
44
Example
 Assume the all unit quantity discounts
Order Quantity
Unit Price
0-5,000
$ 3.00
5,000-10,000
$ 2.96
10,000 or more
$ 2.92
D = 120,000/ year
S = $100/lot
h = 0.2
 Based on the all unit quantity discounts, we have
q0=0, q1=5,000, q2=10,000
C0=$3.00, C1=$2,96, C2=$2.92
 If i = 0, evaluate Q0 as Q0= 2DS = 6,324
hC0
Since Q0 > q1, we set the lost size at q1=5,000 and the total cost
q
TC0= D S + 1 hC1+ DC1 = $359,080
2
q1
45
Example - Continued
 For i = 1, we obtain Q1 = 6,367
Since 5,000 < Q1 <10,000 , we set the lot size at Q1 = 6,367.
Q1
TC1= D S +
hC1+ DC1 = $358,969
2
Q1
 For i = 2, we obtain Q2 = 6,410
Since Q2 < q2 , we set the lot size at q2=10,000.
 Observe that the lowest total cost is for i = 2.
The optimal lot size = 10,000 (at the discount price of $2.92)
46
Example - Continued
 For i = 1, we obtain Q1 = 6,367
Since 5,000 < Q1 <10,000 , we set the lot size at Q1 = 6,367.
Q1
TC1= D S +
hC1+ DC1 = $358,969
2
Q1
 For i = 2, we obtain Q2 = 6,410
Since Q2 < q2 , we set the lot size at q2=10,000.
q
TC2= D S + 2 hC2+ DC2 = $354,520
2
q2
► Observe that the lowest total cost is for i = 2.
The optimal lot size = 10,000 (at the discount price of $2.92)
47
Example - Continued
 For i = 1, we obtain Q1 = 6,367
Since 5,000 < Q1 <10,000 , we set the lot size at Q1 = 6,367.
Q1
TC1= D S +
hC1+ DC1 = $358,969
2
Q1
 For i = 2, we obtain Q2 = 6,410
Since Q2 < q2 , we set the lot size at q2=10,000.
q
TC2= D S + 2 hC2+ DC2 = $354,520
2
q2
► Observe that the lowest total cost is for i = 2.
The optimal lot size = 10,000 (at the discount price of $2.92)
48
Example - Continued
 For i = 1, we obtain Q1 = 6,367
Since 5,000 < Q1 <10,000 , we set the lot size at Q1 = 6,367.
Q1
TC1= D S +
hC1+ DC1 = $358,969
2
Q1
 For i = 2, we obtain Q2 = 6,410
Since Q2 < q2 , we set the lot size at q2=10,000.
q
TC2= D S + 2 hC2+ DC2 = $354,520
2
q2
► Observe that the lowest total cost is for i = 2.
The optimal lot size = 10,000 (at the discount price of $2.92)
49
The Impact of All Unit Discounts on Supply Chain
► In the above example
》The optimal order size = 6,324 when there is no discount.
》The quantity discounts result in a higher order size = 10,000.
► If the fixed ordering cost S = $4,
》The optimal order size without discount = 1,265
》The optimal order size with all unit discounts = 10,000
► All unit quantity discounts encourage retailers to increase the size of
their lots.
► This also increases cycle inventory and average flow time.
50
Marginal Unit Quantity Discounts
Marginal Cost per Unit
Total Material Cost
C0
C1
C2
Quantity
Purchased
q1
q2
q3
Order Quantity
q1
q2
q3
If an order of size q is placed, the first q1-q0 units are priced at C0, the
next q2-q1 are priced at C1, and so on.
51
Evaluate EOQ for Marginal Unit Discounts
► Evaluate EOQ for each marginal price Ci (or lot size between qi and qi+1)
》 Let Vi be the cost of order qi units. Define V0 = 0 and
Vi=C0(q1-q0)+C1(q2-q1)+‧‧‧+Ci-1(qi-qi-1)
》 Consider an order size Q in the range qi to qi+1
Total annual cost = ( D/Q )S
(Annual order cost)
+ (Q/2)‧h‧[ Vi+(Q-qi)Ci ] / Q (Annual holding cost)
+ D‧[ Vi+(Q-qi)Ci ] / Q(Annual material cost)
Optimal lot size
Qi =
52
2D(S+Vi-qiCi)
hCi
Evaluate EOQ for Marginal Unit Discounts
► Evaluate EOQ for each marginal price Ci (or lot size between qi and qi+1)
》 Let Vi be the cost of order qi units. Define V0 = 0 and
Vi=C0(q1-q0)+C1(q2-q1)+‧‧‧+Ci-1(qi-qi-1)
》 Consider an order size Q in the range qi to qi+1
Total annual cost = ( D/Q )S
(Annual order cost)
+ (Q/2)‧h‧[ Vi+(Q-qi)Ci ] / Q (Annual holding cost)
+ D‧[ Vi+(Q-qi)Ci ] / Q(Annual material cost)
Optimal lot size
Qi =
53
2D(S+Vi-qiCi)
hCi
Example
 Assume the all unit quantity discounts
Order Quantity
Unit Price
0-5,000
$ 3.00
5,000-10,000
$ 2.96
10,000 or more
$ 2.92
D = 120,000/ year
S = $100/lot
h = 0.2
 q0=0, q1=5,000, q2=10,000
C0=$3.00, C1=$2,96, C2=$2.92
V0=0 ; V1=3(5,000-0)=$15,000
V2=3(5,000-0)+2.96(10,000-5,000)=$29,800
2D(S+V0-q0C0)
= 6,324
hC0
Since Q0 > q1, we set the lost size at q1=5,000 and the total cost
 If i = 0, evaluate Q0 as
54
Q0 =
h D
D
=
+[
V
+(q
-q
)C
]
S
TC0
1
1 1 1 2 + q [ V1+(q1-q1)C1]= $363,900
1
q1
Example - Continued
 For i = 1, evaluate Q1= 2D(S+V1-q1C1) = 11,028
hC1
Since Q1 > q2, we evaluate the cost of ordering q2=10,000
h D
D
=
+[
V
+(q
-q
)C
]
S
TC1
2
2 2 2 2 + q [ V2+(q2-q2)C2]=$361,780
2
q2
 For i = 2, evaluate Q2=
2D(S+V2-q2C2)
= 16,961
hC2
TC2= D S +[ V2+(Q2-q2)C2] h + D [ V2+(Q2-q2)C2 ]= $360,365
2 Q2
Q2

55
Optimal order size = 16,961
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作品
授權條件
作者/來源
12
本作品轉載自Microsoft Office 2007多媒體藝廊，依據Microsoft 服務合約及著
作權法第46、52、65條合理使用。
12, 14
本作品轉載自Microsoft Office 2007多媒體藝廊，依據Microsoft 服務合約及著
作權法第46、52、65條合理使用。
12, 14
本作品轉載自Microsoft Office 2007多媒體藝廊，依據Microsoft 服務合約及著
作權法第46、52、65條合理使用。
12
本作品轉載自Microsoft Office 2007多媒體藝廊，依據Microsoft 服務合約及著
作權法第46、52、65條合理使用。
12
CoolCLIPS。本作品轉載自CoolCLIPS網站
（http://dir.coolclips.com/Popular/World_of_Industry/Food/Shopping_cart_full_of_
groceries_vc012266.html），瀏覽日期2011/12/28。依據著作權法第46、52、65
條合理使用。
12
本作品轉載自Microsoft Office 2007多媒體藝廊，依據Microsoft 服務合約及著
作權法第46、52、65條合理使用。
14
本作品轉載自Microsoft Office 2007多媒體藝廊，依據Microsoft 服務合約及著
作權法第46、52、65條合理使用。
版權聲明
頁碼
57
作品
授權條件
作者/來源
15, 27
本作品轉載自Microsoft Office 2007多媒體藝廊，依據Microsoft 服務合約及著
作權法第46、52、65條合理使用。
15
本作品轉載自Microsoft Office 2007多媒體藝廊，依據Microsoft 服務合約及著
作權法第46、52、65條合理使用。
15
本作品轉載自Microsoft Office 2007多媒體藝廊，依據Microsoft 服務合約及著
作權法第46、52、65條合理使用。