Example
• A company purchases air filters at a rate of 800 per year
• $10 to place an order
• Unit cost is $25 per filter
• Inventory carry cost is $2/unit per year
• Shortage cost is $5
• Lead time is 2 weeks
• Assume demand during lead time follows a uniform distribution from 0 to 200
• Find (Q,R)
Solution

Partial derivative outcomes:
2  K  p n( R)
2(800)(10  5 n( R))
Q

 8000  4000 n( R)
h
2
Qh
2Q
Q
1  F ( R) 


p 5(800) 2000
Solution

From Uniform U(0,200) distribution:
1
1
U(0,200) : f ( x) 

b - a 200

200
R
R
n( R)   ( x  R) f ( x)dx 
1  x 2

 Rx
200  2

R2
 100 
R
400
x  200
xR

1
( x  R)
dx
200
2
2


1
200
R
2


 200 R 
 R 
 200  2
2


R2
n( R ) 
 R  100
400
Q1  8000  4000n(R )
Solution

Q
1  F ( R) 
2000
Iteration 1:
2K 
2(10)(800)
EOQ 

 8000F(R)
 89.44  Qo
h
2
Qo h
89
1  F(Ro ) 

 .04
p 2000
F(Ro )  .96
R o  (.96)(200  0)  192
0
200
R
Solution

Iteration 2:
R2
n( R ) 
 R  100
400
Q1  8000  4000n(R )
1  F ( R) 
(192) 2
n( R0 ) 
 192  100  .198
400
Q1  8000  4000(.198)  93.76
94
1  F ( R1 ) 
 .05
2000
 R1  (.95)( 200)  190
Q
2000
Solution

Iteration 3:
2
R2
n( R ) 
 R  100
400
Q1  8000  4000n(R )
1  F ( R) 
190
n( R1 ) 
 190  100  .2197
400
Q2  8000  4000(.2197)  94.228
94
(1  F ( R2 )) 
 .05
2000
R2  190
Q
2000
Solution


R didn’t change => CONVERGENCE
(Q*,R*) = (94,190)
I(t)
253
Slope
190
159
-800
With lead time equal to 2 weeks:
SS = R – t =190-800(2/52)=159
Example
• Demand is Normally distributed with mean of 40 per week
and a weekly variance of 8
• The ordering cost is $50
• Lead time is two weeks
• Shortages cost an estimated $5 per unit short to expedite
orders to appease customers
• The holding cost is $0.0225 per week
• Find (Q,R)
Solution


Demand is N (40,2 per
2 ) week.
Lead time is two weeks long. Thus, during the lead
time:




Mean demand is 2(40) = 80
Variance is (2*8) = 16
Demand observed in one week is independent from demand
observed in any other week:
E(demand over 2 weeks) = E (2*demand over week 1)
= 2 E(demand in a single week) = 2 μ = 80
Standard deviation over 2 weeks is σ = (2*8)0.5 = 4
Finding Q and R, iteratively
1. Compute Q = EOQ.
2. Substitute Q in to Equation (2) and compute R.
Qh
1  F ( R) 
p
3. Use R to compute average backorder level,
n(R) to use in Equation (1).

n(R)   (x  R) f (x)dx
R
4. Solve for Q using Equation (1).
2 K  p n(R) 
Q
h
5. Go to Step 2 until convergence.
Solution

Iteration 1:
2K 
2(50)( 40)
EOQ 

 421.6  Qo
h
.0225
Qo h 421.6(.0225)
1  F ( Ro ) 

 .0473
p
5(40)
F ( Ro )  .9527

From the standard normal table:
F ( Ro )  .9527
P( z  1.67)  0.9527
Ro    Z  80  1.67(4)  86.68
Solution

Iteration 2:
This is the unit normal loss expression.
Table A - 4 gives values.

n( R0 )   ( x  R) f ( x)dx
R

1
n( R0 )   ( x  R)
e
2 
R
1  x 
 

2  
2
dx
 R 
 86.68  80 
  L
  4L 
  4 L(1.67)
4
  


 4(.0197)  .0788
Solution

Iteration 2:
n( R0 )  .0788
2 K  pn( R) 
2(40)50  5(.0788) 
Q1 

 423.3
h
.0225
Q1h 423.3(.0225)
1  F(R1 ) 

 .0476
p
5(40)
F(R1 )  .9523
R1  86.68
Convergence!