EOQ for Production not Order (EPQ)
We do not buy the product.
We produce it.
Total demand / year is D
Demand / day or consumption rate is u
Production rate is p / day
Example
A toy manufacturer uses 48000 parts for one of its products.
Consumption rate is uniform throughout the year.
Working days are 240 / year.
The firm can produce at a rate of 800 parts / day
Carrying cost is $1 / part / year
Setup cost for production run is $45 / setup
What is the optimal production size?
Example
Instantaneous and Incremental Replenishment
Instantaneous Replenishment
Incremental Replenishment
EPQ: Incremental Replenishment
(Production and Consumption)
p-u
p
u
u × ( number of working days ) = D = Total demand per year
EPQ: One period and One year
p-u
u
One Period
One Year
EPQ: Ordering Cost and Carrying Cost
I max
I max
I max
D
I max
S 
TC 
H
Q
2
If I can state Imax in terms of Q, then I am done.
I can easily find Economic Quantity.
I max
EPQ : Production & Consumption; rate & time
How much do we produce each time? Q
How long does it take to produce Q? d1
What is our production rate per day? p
Q  pd1
How long does it take to consume Q? d1 +d2
What is our consumption rate per day ? u
Q  u(d1  d 2 )
I max
p-u
d1
I max  ( p  u )d1 I max  ud 2
d2
u
EPQ : Optimal Q
EPQ 
2 SDp
H ( p  u)
2 SD
EPQ 
H
p
p u
2SD
EOQ 
H
p
p u
Example
A toy manufacturer uses 48000 parts for one of its products.
Consumption rate is uniform throughout the year.
Working days are 240 / year.
The firm can produce at a rate of 800 parts / day
Carrying cost is $1 / part / year
Setup cost for production run is $45 / setup
What is the optimal production size?
EPQ 
2 SD
H
EPQ  2400
p
p u
2(45)( 48000)
800
EPQ 
1
800  200
What is Run Time, What is Cycle Time
Q  pd1
2400  800d1
d1  3
Q  u(d1  d 2 )
2400  200(d1  d2 )
(d1  d2 )  12
What is the Optimal Total Cost
I Max
D
TC  S  H
Q
2
I max  ( p  u )d1
I max  (800  200)3
I max  1800
I Max
D
TC  S  H
Q
2
48000 1800
TC  45
1
2400
2
TC  900  900  1800
ROP
If setup time takes 2 days, at which level of inventory we
should start setup?
2(200) = 400
Practice
A company has a yearly demand of 120,000 boxes of its
product. The product can be produced at a rate of 2000
boxes per day. The shop operates 240 days per year.
Assume that demand is uniform throughout the year.
Setup cost is $8000 for a run, and holding cost is $10 per
box per year.
a) What is the demand rate per day
b) What is the Economic Production Quantity (EPQ)
c) What is the run time
d) What is the maximum inventory
e) What is the total cost of the system
Assignment 4.3b
a) What is the demand rate per day
Demand per year is 120,000 there are 240 days per year
Demand per day = 120000/240 = 500
u = D/240 = 500
b) What is the Economic Production Quantity (EPQ)
2 SD
EPQ 
H
p
p u
2(8000)(120000)
2000
EPQ 
10
2000  1500
=16000
Assignment 4.3b
c) What is the run time
We produce 16000 units
Our production rate is 2000 per day
It takes 16000/2000 = 8 days
d1 = EPQ/p = 8 days
d) What is the maximum inventory
We produce for 8 days. Each day we produce 2000 units and
we consume 500 units of it. Therefore we add to our
inventory at rate of 1500 per day for 8 days. That is
Imax = 8(1500) = 12000
Imax = pd1 = 8(1500) = 12000
Assignment 4.3b
e) What is the total cost of the system
I Max
D
TC  S  H
Q
2
120000
12000
TC  8000
 10
16000
2
TC  60000  60000  120000
Assignment 12.c
Problem : The Dine Corporation is both a producer and a user of brass
couplings. The firm operates 220 days a year and uses the couplings
at a steady rate of 50 per day. u = 50. Couplings can be produced at
rate of 200 per day. p=200. Annual storage cost is $1 per coupling,
H=1, and machine setup cost is $35 per run, S=35. Since there are
220 days per year and u is equal to 50, yearly demand is 50(220),
D= 11000.
a)
b)
c)
d)
e)
Determine the economic run size.
Approximately how many runs per year will there be?
Compute the maximum inventory level.
Determine the cycle time, run time and pure consumption time.
Compute the total cost.