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

2 ( ) 2(800)(10 5 ( ))8000 4000 ( )

22

1 ( )5(800) 2000

K p n R n RQ n R

hQh Q Q

F Rp

Partial derivative outcomes:

Solution From Uniform U(0,200) distribution:

RR

RR

RRxx

dxRxdxxfRxRn

abxf

x

Rx

R R

400100

2

2002

200

200

1

2200

1

200

1)()()()(

200

1

-

1)( :U(0,200)

2

2222002

200

Solution

EOQ 2K

h

2(10)(800)

2 8000 89.44 Qo

1 F(Ro) Qohp

892000

.04

F(Ro ) .96

Ro (.96)(200 0) 192

Iteration 1:

F(R)

2000

R

2000)(1

)R(40008000

100R400

R )R(

1

2

QRF

nQ

n

Solution

Iteration 2:

190)200)(95(.

05.2000

94)(1

76.93)198(.40008000

198.100192400

(192) )(

1

1

1

2

0

R

RF

Q

Rn

2000)(1

)R(40008000

100R400

R )R(

1

2

QRF

nQ

n

Solution

Iteration 3:

190

05.2000

94))(1(

228.94)2197(.40008000

2197.100190400

190)(

2

2

2

2

1

R

RF

Q

Rn

2000)(1

)R(40008000

100R400

R )R(

1

2

QRF

nQ

n

Solution R didn’t change => CONVERGENCE (Q*,R*) = (94,190)

I(t)

Slope

-

253

159

190

With lead time equal to 2 weeks:

SS = R – =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)

Demand is per 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

Solution)22,40(N

Finding Q and R, iteratively

1. Compute Q = EOQ.

2. Substitute Q in to Equation (2) and compute R.

3. Use R to compute average backorder level, n(R) to use in Equation (1).

4. Solve for Q using Equation (1).

5. Go to Step 2 until convergence.

h

p n(R)KQ

2

n(R) (x R) f (x)dxR

p

QhRF )(1

Solution

9527.)(

0473.)40(5

)0225(.6.421)(1

6.4210225.

)40)(50(2 2

o

oo

o

RF

p

hQRF

Qh

KEOQ

Iteration 1:

From the standard normal table:

( ) .9527

( 1.67) 0.9527

80 1.67(4) 86.68

o

o

F R

P z

R Z

Solution

Iteration 2:

2

0

1

20

( ) ( ) ( )

1( ) ( )

2

86.68 80 4 4 (1.67)

4

4(.0197) .0788

R

x

R

n R x R f x dx

n R x R e dx

RL L L

This is the unit normal loss expression. Table A - 4 gives values.

Solution

Iteration 2:

3.423

0225.

)0788(.550)40(2)(2

0788.)(

1

0

h

RpnKQ

Rn

1 F(R1) Q1hp

423.3(.0225)5(40)

.0476

F(R1 ) .9523

R1 86.68

Convergence !