Lecture No 19

 

Held on: Wednesday, September 27, 2000

 

Notes by: J.S.Dhupia & Puneet Bhatia

 

Lot Size Reorder Point Systems: (Q,R) Problem

 

     Order Quantity Q when inventory Reaches level R

 

Assumptions:

 

  1.The system is in continuous review

  2.Demand is Random & stationary

  3.There is a positive fixed lead time t, for placing an order.

  4.The cost is as follows:

 

     Setup cost: K per order

     Holding cost: h per unit.

     Stock out cost: per unit of unsatisfied demand

 

 

 

(s,S) is Periodic Review

 

if amount is below s order up to S

 

Here we have continuos review

 

More similar to EOQ

 

In (s, S) Policy we have lead-time =0, here we also have positive lead time.

 

Stationary Demand means  ® probability fn. is fixed.

 

 

 

 

						Q
	Inv
					t
									Time

 

Final Avg. Inventory

 

For time

 

m = ED = l t

 

s 2= var D

 

s = (R-D)+ » R-D           R®Reorder Pt.

 

l is the expected demand rate per yr.

Solved Examples

 

Ex1:

Selling mustard

Cost $10 a jar

20% annual interest rate

Loss of goodwill for every jar of mustard out of stock = $25

 

Fixed cost of bookkeeping = $50

 

t = 6 months

Expected demand = ED = 100 jars

 

sD= 25

 

D~ normally distributed

 Qo =              2 K l            =                  2*50 (2*100)             =100

                          h                                        (.2)*10

 

1 – F(Ro) = Qo h  =   100L      = 0.04

                   pl         25 * 200

 

                                      D-m             Ro-m

P(D>= Ro)= 0.4 Þ  P              >=                        =0.04           From

                                           s                s                                Tables

 

Þ Ro - m  = 1.75            ÞRo= 25 (1.75) +100 =144

 

n(R) =  L     R-m   = 25L (1.75)= 25 *.0162

                      D

 

          =.405

 

Q =        2* 200 (50+25*.405)    -110

                   2

Now, 1- F(R1) - 100 * 2   = 0.44

                             25*200

Þ        R1= .143

 

 

 

 

 

R-lt = 143 –100

 

Probability that no stock out occurs in a cycle     P(D <= R) = F(R)= 0.936

 

Properties of demand                is a stock-out

                             = n(R) -.4575=0.004

                                  Q

 

99.6% of demand is satisfied as