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To Accompany Ritzman & Krajewski, Foundations of Operations Management © 2003 Prentice-Hall, Inc. All rights reserved.

Chapter 10

Inventory

Management




Inventory Costs

Interest or

Opportunity Costs

Storage and Handling Costs

Taxes, Insurance, and

Shrinkage Costs Ordering and Setup Costs

Transportation Costs




Types of Inventory




Types of Inventory Cycle Inventory

Average cycle inventory =Q + 0

2





Safety Stock Inventory


2





Safety Stock Inventory Anticipation Inventory


2


http://slidepdf.com/reader/full/ch105572011c4979599169a0ceeb 7/84To Accompany Ritzman & Krajewski, Foundations of Operations Management © 2003 Prentice-Hall, Inc. All rights reserved.


Safety Stock Inventory Anticipation Inventory

Pipeline Inventory

Average cycle inventory =

Pipeline inventory = D L = dL

Q + 02



ABC Analysis



ABC Analysis

10 20 30 40 50 60 70 80 90 100

Percentage of items

P e

r c e n t a g e o f d o

l l a r v a l u e

100—

90 —

80 —

70 —

60—

50 —

40 —

30 —

20—

10 —

0 —

Figure 10.1



ABC Analysis

10 20 30 40 50 60 70 80 90 100

Percentage of items

P e


l l a r v a l u e

100—

90 —

80 —

70 —

60—

50 —

40 —

30 —

20—

10 —

0 —

Figure 10.1



ABC Analysis

10 20 30 40 50 60 70 80 90 100

Percentage of items

P e


l l a r v a l u e

100—

90 —

80 —

70 —

60—

50 —

40 —

30 —

20—

10 —

0 —

Class C

Class A

Class B

Figure 10.1



ABC Analysis

10 20 30 40 50 60 70 80 90 100

Percentage of items

P e


l l a r v a l u e

100 —

90 —

80 —

70 —

60—

50 —

40 —

30 —

20—

10 —

0 —

Class C

Class A

Class B

Figure 10.1



ABC Analysis

10 20 30 40 50 60 70 80 90 100

Percentage of items

P e


l l a r v a l u e

100 —

90 —

80 —

70 —

60—

50 —

40 —

30 —

20—

10 —

0 —

Class C

Class A

Class B

Figure 10.1



ABC Analysis

10 20 30 40 50 60 70 80 90 100

Percentage of items

P e


l l a r v a l u e

100 —

90 —

80 —

70 —

60—

50 —

40 —

30 —

20—

10 —

0 —

Class C

Class A

Class B

Figure 10.1



How Much?

When!



EconomicOrderQuantity



Economic Order Quantity

1. Demand rate is constant

2. No constraints on lot size 3. Only relevant costs are holding and

ordering/setup

4. Decisions for items are independent from other items 5. No uncertainty in lead time or supply

Assumptions







O n - h

a n d i n v e n t o r y

( u n i t s )

TimeFigure 10.2




O n - h


( u n i t s )

TimeFigure 10.2




O n - h


( u n i t s )

Time

Receiveorder

Q

Figure 10.2




O n - h


( u n i t s )

Time

Receiveorder

Q

1 cycleFigure 10.2




O n - h


( u n i t s )

Time

Receiveorder

Q

1 cycleFigure 10.2




O n - h


( u n i t s )

Time

Receiveorder

Inventory depletion(demand rate)

Q

1 cycleFigure 10.2




O n - h


( u n i t s )

Time

Receiveorder


Q

1 cycleFigure 10.2




O n - h


( u n i t s )

Time

Averagecycleinventory

Q

Q — 2

1 cycle

Receiveorder


Figure 10.2







A n n u a l c o s t ( d o l l a r s )

Lot Size (Q )Figure 10.3






Lot Size (Q )

Holding cost (HC )

Figure 10.3






Lot Size (Q )

Holding cost (HC )

Ordering cost (OC )

Figure 10.3






Lot Size (Q )

Ordering cost (OC )

Holding cost (HC )

Total cost = HC + OC

Figure 10.3









| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

A n n u a l c o s t ( d

o l l a r s )





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Current

Q

Total cost = (H ) + (S )D Q

Q 2

Holding cost = (H )Q 2

Ordering cost = (S )D Q


o l l a r s )





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Current

Q


Q 2



Bird feeder costs

D = (18 /week)(52 weeks) = 936 unitsH = 0.25 ($60/unit) = $15

S = $45 Q = EOQ

C = (H ) + (S )Q 2

D Q

EOQ =2DS

H

Example 10.1


o l l a r s )





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Current

Q


Q 2



Bird feeder costs


S = $45 Q = 75 units

C = (H ) + (S )Q 2

D Q

EOQ =2DS

H

Example 10.1


o l l a r s )





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Current

Q


Q 2



Bird feeder costs


S = $45 Q = 75 units

C = $562 + $562 = $1124

C = (H ) + (S )Q 2

D Q

EOQ =2DS

H

Example 10.1


o l l a r s )





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Current

Q


Q 2



Bird feeder costs


S = $45 Q = 75 units

C = $562 + $562 = $1124

C = (H ) + (S )Q 2

D Q

EOQ =2DS

H

Example 10.1


o l l a r s )





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Current

Q


Q 2



Bird feeder costs


S = $45 Q = 75 units

C = $562 + $562 = $1124

C = (H ) + (S )Q 2

D Q

EOQ =2DS

H

Lowestcost

Best Q (EOQ)Example 10.1


o l l a r s )





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Current

Q


Q 2

Lowestcost



o l l a r s )

Bird feeder costs


S = $45 Q = 75 units

C = $562 + $562 = $1124

C = (H ) + (S )D Q

Q 2

EOQ =2DS

H





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Current

Q


Q 2

Birdfeeder costs


S = $45 Q = 75 units

C = $562 + $562 = $1124

C = (H ) + (S )D Q

Q 2

EOQ =2DS

H

Lowestcost


D Q

Time between orders

TBOEOQ = = 75/936 = 0.080 yearEOQ

D


o l l a r s )





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Current

Q


Q 2

Birdfeeder costs


S = $45 Q = 75 units

Lowestcost

Best Q (EOQ)

Time between orders

TBOEOQ = = 75/936 = 0.080 year

TBOEOQ = (75/936)(12) = 0.96 months

TBOEOQ = (75/936)(52) = 4.17 weeks

TBOEOQ = (75/936)(365) = 29.25 days

EOQ

D

Example 10.1


o l l a r s )





| | | | | | | |50 100 150 200 250 300 350 400

Lot Size (Q )

3000 —

2000 —

1000 —

0 —

Currentcost

Lowestcost

Best Q (EOQ)

Current

Q


Q 2



Figure 10.4


o l l a r s )




How Much? When!




Continuous Review

Time

Figure 10.6

O n - h a n d i n v e

n t o r y




Continuous Review

Time

R

Orderreceived

Q

OH

Figure 10.6


n t o r y




Continuous Review

Orderreceived

Q

OH

Orderplaced

IP

TBO

L

R

Figure 10.6


n t o r y




Continuous Review

Time

Orderreceived

Orderreceived

Q Q

OH OH

Orderplaced

Orderplaced

IP IP

TBO

L

TBO

L

TBO

L

R

Orderreceived

Q

OH

Orderplaced

IP

Orderreceived

Figure 10.6


n t o r y




Uncertain Demand




Uncertain Demand

Time

R

Figure 10.7


n t o r y




Uncertain Demand

Time


n t o r y

Orderreceived

Q OH

Orderplaced

Orderplaced

Orderreceived

IP IP

R

TBO1 TBO2 TBO3

L1 L2 L3

Q

Orderplaced

Q

Orderreceived

Orderreceived

Figure 10.7




Reorder Point / Safety Stock





Averagedemand

duringlead time

Figure 10.8





Probability of stockout(1.0 – 0.85 = 0.15)

Cycle-service level = 85%

Averagedemand

duringlead time

z L

R

Figure 10.8






Averagedemand

duringlead time

z L

R


Example 10.2

Safety Stock/R






Averagedemand

duringlead time

z L

R


Example 10.2

Safety Stock/RSafety stock = zL

= 2.33(22) = 51.3= 51 boxes






Averagedemand

duringlead time

z L

R


Example 10.2

Safety Stock/RSafety stock = zL

= 2.33(22) = 51.3= 51 boxes

Reorder point = ADDLT + SS= 250 + 51= 301 boxes




Lead Time Distributions





Figure 10.9

t = 15

+75

Demand for week 1





Figure 10.9

t = 15

+

+75

Demand for week 1

75Demand for week 2

t = 15





Figure 10.9

t = 15

=

+

+75

Demand for week 1

75Demand for week 2

75

Demand for week 3

t = 15

t = 15





Figure 10.9

t = 15

=

+

+75

Demand for week 1

75Demand for week 2

75

Demand for week 3

t = 26

225Demand for

three-week lead time

t = 15

t = 15





Example 10.3

t = 15

=

+

+75

Demand for week 1

75Demand for week 2

75

Demand for week 3

t = 26

225Demand for


t = 15

t = 15

Bird feeder Lead Time Distribution





Example 10.3

t = 15

=

+

+75

Demand for week 1

75Demand for week 2

75

Demand for week 3

t = 26

225Demand for


t = 15

t = 15


t = 1 week d = 18 L = 2





Example 10.3

t = 15

=

+

+75

Demand for week 1

75Demand for week 2

75

Demand for week 3

t = 26

225Demand for


t = 15

t = 15


t = 1 week d = 18 L = 2

L = t L = 5 2 = 7.1





Example 10.3

t = 15

=

+

+75

Demand for week 1

75Demand for week 2

75

Demand for week 3

t = 26

225Demand for


t = 15

t = 15


t = 1 week d = 18 L = 2

L = t L = 5 2 = 7.1

Safety stock = zL = 1.28(7.1) = 9.1 or 9 units

Reorder point = dL + Safety stock= 2(18) + 9 = 45 units





Example 10.3

t = 15

=

+

+75

Demand for week 1

75Demand for week 2

75

Demand for week 3

t = 26

225Demand for


t = 15

t = 15


t = 1 week d = 18 L = 2

Reorder point = 2(18) + 9 = 45 units





Example 10.3

t = 15

=

+

+75

Demand for week 1

75Demand for week 2

75

Demand for week 3

t = 26

225Demand for


t = 15

t = 15


t = 1 week d = 18 L = 2

Reorder point = 2(18) + 9 = 45 units

C = ($15) + ($45) + 9($15) 75

2

936

75

C = $562.50 + $561.60 + $135 = $1259.10




Periodic Review Systems





TimeP P

T

Figure 10.10

O n - h a n d i n

v e n t o r y





Time

O n - h a n d i n

v e n t o r y

P P

T

Q 1

Orderplaced

Figure 10.10





Time

O n - h a n d i n

v e n t o r y

P P

T

Q 1

L

Orderplaced

Figure 10.10





Time

O n - h a n d i n

v e n t o r y

P P

T

Q 1

L

Orderplaced

Orderreceived

Figure 10.10





Time

O n - h a n d i n

v e n t o r y

Orderreceived

Orderreceived

Orderplaced

Orderplaced

Q 1Q 2

Q 3

L L L

P P

T Orderreceived

Figure 10.10





Time

O n - h a n d i n

v e n t o r y

IP 1

IP 3

IP 2

Orderreceived

Orderreceived

IP IP

OH OH

Orderplaced

Orderplaced

Q 1Q 2

Q 3

L L L

P P

Protection interval

T Orderreceived

IP

Figure 10.10

S




Time

O n - h a n d i n v e n t o r y

IP1

IP3

IP2

Orderreceived

Orderreceived

IP IP

OH OH

Orderplaced

Orderplaced

Q1Q2

Q3

L L L

P P

Protection interval

T Orderreceived

IP


Example 10.4

Bird feeder— Calculating P and T

P i di R i S




Time

O n - h a n d i n

v e n t o r y

IP1

IP3

IP2

Orderreceived

Orderreceived

IP IP

OH OH

Orderplaced

Orderplaced

Q1Q2

Q3

L L L

P P

Protection interval

T Orderreceived

IP


Example 10.4

EOQ = 75 units D = (18 units/week)(52 weeks) = 936 units

t = 18 units L = 2 weeks cycle/service level = 90%

Bird feeder—Calculating P and T

P i di R i S




Time

O n - h a n d i n

v e n t o r y

IP1

IP3

IP2

Orderreceived

Orderreceived

IP IP

OH OH

Orderplaced

Orderplaced

Q1Q2

Q3

L L L

P P

Protection interval

T Orderreceived

IP


Example 10.4




P = (52) = (52) = 4.2 or 4 weeksEOQ

D 75936

P+L = t P + L = 5 6 = 12 units

P i di R i S




Time


IP1

IP3

IP2

Orderreceived

Orderreceived

IP IP

OH OH

Orderplaced

Orderplaced

Q1Q2

Q3

L L L

P P

Protection interval

T Orderreceived

IP


Example 10.4

T = Average demand during the protection interval + Safety stock= d (P + L) + zP + L

= (18 units/week)(16 weeks) + 1.28(12 units) = 123 units




P = (52) = (52) = 4.2 or 4 weeksEOQ

D 75936

P+L = t P + L = 5 6 = 12 units

P i di R i S t




Time


IP1

IP3

IP2

Orderreceived

Orderreceived

IP IP

OH OH

Orderplaced

Orderplaced

Q1Q2

Q3

L L L

P P

Protection interval

T Orderreceived

IP


Example 10.4




P = 4 weeks T = 123 units

P i di R i S t




Time


IP1

IP3

IP2

Orderreceived

Orderreceived

IP IP

OH OH

Orderplaced

Orderplaced

Q1Q2

Q3

L L L

P P

Protection interval

T Orderreceived

IP


Example 10.4




C = ($15) + ($45) + 15($15) 4(18)

2

936

4(18)

C = $540 + $585 + $225 = $1350

P = 4 weeks T = 123 units




Comparison of Q and P Systems

P Systems

Q Systems

Convenient to administer Orders may be combined IP only required at review

Individual review frequencies Possible quantity discounts Lower, less-expensive safety stocks

ABC Anal sis Sol ed Problem 2




ABC Analysis — Solved Problem 2

Comparison of P and Q Systems



Comparison of P and Q Systems — Solved Problem 6

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