LESSON 15
INVENTORY MODELS (DETERMINISTIC)
EOQ MODEL FOR PRODUCTION PLANNING
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
1
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, ….
2
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
Qj 
*
2K j j
hj '
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.
3
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.
4
Optional
Example
Problem with Separate EPQ Computation
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.
5
Optional
Example
Problem 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.
Product J - 55R :
K  set - up time  worker time  1.2  85  $102
h  Ic  0.20  20  $4/unit/year
 18,000

 0.5357
P 33,600
 
h'  h1    41  0.5357   $1.8571/unit/year
 P
6
Product J - 55R (continued from the previous slide) :
Optional
2 K
2 102 18,000
Q  EPQ 

 1,406.1404 units
h'
1.8571

*
Max Inventory, H  Q 1    1,406.141  0.5357   652.85 units
 P
*
Q* 1,406.1404
Uptime, T1 

240  10.0439 days
P
33,600
Minimum downtime (days), required
 Uptime for H - 223  setup time of H - 223
 4.2027 days (see the next slide)  0.8 hr/ 8 hr per day
 4.3027 days
H 652.8509

240

18,0001
 8.7047 days  4.3027 days (downtime demand met)
Maximum inventory lasts for (days) 
7
Product H - 223 :
Optional
K  set - up time  worker time  0.8  85  $68
h  Ic  0.20  35  $7 /unit/year
 24,000
 0.4545

P 52,800
 
h'  h1    71  0.4545  $3.8182 /unit/year
 P
2  68  24,000
2 K
 924.5848 units

Q  EPQ 
3.8182
h'

*
Max Inventory, H  Q 1    924.581  0.4545  504.32 units
 P
*
Q* 924.5848
240  4.2027 days

Uptime, T1 
52,800
P
8
Product H - 223 (continued from the previous slide) :
Optional
Minimum downtime (days), required
 Uptime for J - 55R  setup time of J - 55R
 10.0439 days  1.2 hr/ 8 hr per day  10.1939 days
Maximum inventory lasts for (days)
H 504.3190
240


24,000

 5.0432 days  10.1939 days
(downtime demand cannot be met)
Conclusion : If EPQ quantities are produced, demand for
Product H - 223 cannot be met when product J - 55R
is produced.
9
Annual production rate (units/year)
Annual Demand rate (units/year)
Setup time (hour)
K, Setup cost, $
Unit cost, $
h, Holding cost/unit/year
demand/production
h'
Separate EPQ solution
Q*
Maximum inventory
Uptime (days)
Minimum downtime (days)
Maximum inventory lasts for (days)
J-55R
33600
18000
1.2
Optional
H-223
52800
24000
0.8
20
35
10
Annual production rate (units/year)
Annual Demand rate (units/year)
Setup time (hour)
K, Setup cost, $
Unit cost, $
h, Holding cost/unit/year
demand/production
h'
Separate EPQ solution
Q*
Maximum inventory
Uptime (days)
Minimum downtime (days)
Maximum inventory lasts for (days)
J-55R
33600
18000
1.2
102
20
4
0.5357
1.8571
Optional
H-223
52800
24000
0.8
68
35
7
0.4545
3.8182
1406.1404 924.5848
652.8509 504.3190
10.0439
4.2027
4.3027 10.1939
8.7047
5.0432
11
EOQ Model for Production Planning
• 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 T
is computed. For each product j, the production
quantity Q j is the demand of the product during the
cycle time. If  j is the annual demand of product j
Q j   jT
12
EOQ Model for Production Planning
• Let T be the cycle time and Tj be the production time
of product j
Q j   jT  PjT j 
Tj
T

j
Pj
• Let sj be the setup time of product j and n be the
number of products
T  s1  T1  s2  T2    sn  Tn  Idle time
13
EOQ Model for Production Planning
T  s1  T1  s2  T2    sn  Tn  Idle time
sn Tn
s1 T1 s2 T2
       1
T T T T
T T
n s
n T
n
n 
1
j
j
j
    1   s j    1
T j 1
j 1 T
j 1 T
j 1 Pj
n
T 
s
j 1
n
1 
j 1
j
j
Pj
14
EOQ Model for Production Planning
• Two rules for T*
n
T *  Cycle1 
s
j 1
n
1 
j 1
n
j
j
Pj
T *  Cycle2 
2 K j
j 1
n
 h'
j 1
j
j
• T* is the maximum of the two. T* = max (Cycle1, Cycle2)
15
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
16
optimal production quantities.
2
n
s  s
j 1
j
j 1
 s1  s2 
j
2
n
K  K
j
j 1
n
j
j 1
2
j
P P
j 1
j
j 1
 K1  K 2 

j
1
P1

2
P2

2
n
 h'
j 1
j
j
 j   h' j  j
j 1
 h'1 1  h'2 2 
17
n
Cycle1 
s
j 1
n
1 
j 1
j
j

Pj
n
Cycle2 
2 K j
j 1
n
 h'
j 1
j

j
T *  max Cycle1, Cycle2  
18
Product J - 55R :
Q *  T * 
 
Max Inventory, H  Q*  1   
 P
Q*
Uptime, T1 

P
Downtime  T *  Uptime 
Maximum inventory lasts for (days) 
H


 downtime of J - 55R
 (uptime  setup time) of H - 223 (check)
19
Product H - 223 :
Q *  T * 
 
Max Inventory, H  Q*  1   
 P
Q*
Uptime, T1 

P
Downtime  T *  Uptime 
Maximum inventory lasts for (days) 
H


 downtime of H - 223
 (uptime  setup time) of J - 55R (check)
20
K, Setup cost, $
h, Holding cost/unit/year
demand/production
h'
h'(demand)
J-55R H-223 Total
102
68 170
4
7
0.5357 0.4545 0.9903
1.8571 3.8182
EOQ Model for Production Planning
Cycle1 in days
Cycle2 in days
Cycle time=max(cycle1, cycle2) in days
Q* = cycle time demand
Maximum inventory
Uptime (days)
Downtime (days)
Maximum inventory lasts for (days)
21
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.)
22