1

Outline

• EOQ Model for Production Planning– The multi-product inventory control model with a

finite production rate – An example showing the problem with separate

EPQ computation– The procedure– An example

LESSON 15INVENTORY MODELS (DETERMINISTIC)

EOQ MODEL FOR PRODUCTION PLANNING

2

EOQ Model for Production Planning

• This model is an extension of the EPQ model• Consider the problem of producing many products in a

single facility. The facility may produce only one product at a time.

• In each production cycle there is only one setup for each product, and the products are produced in the same sequence in each production cycle. This assumption is called the rotation cycle policy.

• For example, if there are three products A, B and C, then a production sequence under the rotation cycle policy is A, B, C, A, B, C, ….

3

EOQ Model for Production Planning

• The goal is to determine the optimal production quantities of various products produced in each cycle and the optimal length of the cycle.

• Finding optimal production quantity of each product separately using the EPQ formula

may not give a good solution because a production quantity may not be large enough to meet the demand between two production runs of the product.

'2*

j

jjj h

KQ

4

EOQ Model for Production Planning

• For example, suppose that there are three products A, B and C, then a production sequence under the rotation cycle policy is A, B, C, A, B, C, ….

• The production quantity of product A obtained from the EPQ formula may not be large enough to meet the demand during the production run of products B and C.

• The next example elaborates on the problem of using EPQ formula separately for each product.

5

Example 6: Tomlinson Furniture has a single lathe for turning the wood for various furniture pieces including bedposts, rounded table legs, and other items. Two products and some relevant information appear below:

Annual Setup Time Unit Annual Piece Demand (hours) Cost Production J-55R 18,000 1.2 $20 33,600 H-223 24,000 0.8 35 52,800 Worker time for setup is valued at $85 per hour, and holding

costs are based on a 20 percent annual interest charge. Assume 8 hours per day and 240 days per year.

ExampleProblem with Separate EPQ Computation

Optional

6

ExampleProblem with Separate EPQ Computation

Find the optimal production quantities separately for each product and show that production quantity of H-223 is not large enough to meet the demand between two production runs of H-223.

/unit/year

/unit/yeartime workertime up-set

:55R-J Product

8571.1$5357.0141'

5357.0600,33000,18

4$2020.0102$852.1

Phh

P

IchK

Optional

7

met) demand (downtime days days

(days) for lastsinventory Maximum

days day per hr hr/ slide) next the (see days

223-H of time setup223-H for Uptime required (days), downtime Minimum

days Uptime,

units Inventory,Max

units EPQ

:slide) previous the from (continued 55R-J Product

3027.47047.8

2400001,188509.652

3027488020274

0439.10240600,331404.406,1

85.6525357.0114.406,11

1404.406,18571.1

000,181022'

2

*

1

*

*

H.

..

PQT

PQH

hKQ

Optional

8

days Uptime,

units Inventory,Max

units EPQ

/unit/year

/unit/yeartime workertime up-set

:223-H Product

2027.4240800,525848.924

32.5044545.0158.9241

5848.9248182.3

000,24682'

2

8182.3$4545.0171'

4545.0800,52000,24

7$3520.068$858.0

*

1

*

*

PQT

PQH

hKQ

Phh

P

IchK

Optional

9

produced. is 55R-J product whenmet be cannot 223-H Product

for demand produced, are quantities EPQ If :Conclusion

met) be cannot demand (downtime days days

(days) for lastsinventory Maximumdays day per hr hr/ days

55R-J of time setup55R-J for Uptime required (days), downtime Minimum

:slide) previous the from (continued 223-H Product

1939.100432.5

240000,243190.504

1939.1082.10439.10

H

Optional

10

J-55R H-223Annual production rate (units/year) 33600 52800Annual Demand rate (units/year) 18000 24000Setup time (hour) 1.2 0.8K, Setup cost, $Unit cost, $ 20 35h, Holding cost/unit/yeardemand/productionh'

Separate EPQ solutionQ*Maximum inventoryUptime (days)Minimum downtime (days)Maximum inventory lasts for (days)

Optional

11

J-55R H-223Annual production rate (units/year) 33600 52800Annual Demand rate (units/year) 18000 24000Setup time (hour) 1.2 0.8K, Setup cost, $ 102 68Unit cost, $ 20 35h, Holding cost/unit/year 4 7demand/production 0.5357 0.4545h' 1.8571 3.8182

Separate EPQ solutionQ* 1406.1404 924.5848Maximum inventory 652.8509 504.3190Uptime (days) 10.0439 4.2027Minimum downtime (days) 4.3027 10.1939Maximum inventory lasts for (days) 8.7047 5.0432

Optional

12

• The previous example shows that if production quantities of different products are computed separately, then demand of every product may not be met.

• Therefore, all the products must be considered at the same time.

• To solve the integrated problem, first, the cycle time is computed. For each product , the production quantity is the demand of the product during the cycle time. If is the annual demand of product

EOQ Model for Production Planning

T

TQ jj

jjQ

j j

13

• Let T be the cycle time and Tj be the production time of product j

• Let sj be the setup time of product j and n be the number of products

EOQ Model for Production Planning

j

jjjjjj PT

TTPTQ

time Idle nn TsTsTsT 2211

14

n

j j

j

n

jj

n

j j

jn

jj

n

j

jn

j

j

nn

nn

P

sT

Ps

TTT

Ts

TT

Ts

TT

Ts

TT

Ts

TsTsTsT

1

1

1111

2211

2211

1

111

1

time Idle

EOQ Model for Production Planning

15

• Two rules for T*

• T* is the maximum of the two. T* = max (Cycle1, Cycle2)

EOQ Model for Production Planning

n

j j

j

n

jj

P

sT

1

1*

1Cycle1

n

jjj

n

jj

h

KT

1

1*

'

2Cycle2

16

Example EOQ Model for Production Planning

Example 7: Tomlinson Furniture has a single lathe for turning the wood for various furniture pieces including bedposts, rounded table legs, and other items. Two products and some relevant information appear below:

Annual Setup Time Unit Annual Piece Demand (hours) Cost Production J-55R 18,000 1.2 $20 33,600 H-223 24,000 0.8 35 52,800 Worker time for setup is valued at $85 per hour, and holding

costs are based on a 20 percent annual interest charge. Assume 8 hours per day and 240 days per year. Find the optimal production quantities.

17

2211

2

11

2

2

1

12

11

21

2

11

21

2

11

''

''

hh

hh

PPPP

KKKK

ssss

jjj

n

jjj

j j

jn

j j

j

jj

n

jj

jj

n

jj

18

Cycle2Cycle1,

Cycle2

Cycle1

max

'

*T

h

K

P

s

n

jjj

n

jj

n

j j

j

n

jj

1

1

1

1

2

1

19

(check) 223-H of time) setup uptime55R-J of downtime

(days) for lastsinventory Maximum

Uptime Downtime

Uptime,

Inventory,Max

:55R-J Product

(

*

*

*

**

HTPQT

PQH

TQ

1

1

20

(check) 55R-J of time) setup uptime223-H of downtime

(days) for lastsinventory Maximum

Uptime Downtime

Uptime,

Inventory,Max

:223-H Product

(

*

*

*

**

HTPQT

PQH

TQ

1

1

21

J-55R H-223 TotalK, Setup cost, $ 102 68 170h, Holding cost/unit/year 4 7demand/production 0.5357 0.4545 0.9903h' 1.8571 3.8182h'(demand)

EOQ Model for Production PlanningCycle1 in days Cycle2 in daysCycle time=max(cycle1, cycle2) in daysQ* = cycle time demandMaximum inventoryUptime (days)Downtime (days)Maximum inventory lasts for (days)

22

READING AND EXERCISES

Lesson 15

Reading: Section 4.9 , pp. 226-229 (4th Ed.), pp. 215-220 (5th Ed.)

Exercise: 29, 30 pp. 230-231(4th Ed.), pp. 219-220 (5th Ed.)