Inventory Models PDF

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Inventory Models – are used to determine how much and when to order to replenish stock.

1. Continuous Review Model -

Update stock level after every transaction. If stock level falls to or less than a specific level, the Reorder Point, order a fixed order size Q from supplier to replenish the stock.

Scenario: SMU Book opens 360 days per year and sells 10 pens per day. It replenishes the pens at a cost of $100 per pen and $10 per delivery order. It re-orders 40 pens per order. The Holding cost is 20% per year. Given Data:

Cost of a pen = P = $100 Order cost per order = S = $10 Holding cost per year = h = 20% Holding cost per pen per year = H = h x P = 0.2 x 100 = $20 per pen per year Order quantity Q = 40 Daily demand d = 10 pens/day Annual demand D = 3600 pens/yr Inventory

Q

40 30

Average inventory = Q = 20 2

20 10 0 1 2 3

4

5 6

7 8

Day

Annualcarrying cost  Average Inventory x H Q xH 2 40  x 20 2  $400 per yr 

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Annual Ordering Cost = Number of Orders placed per year x S D    S Q  3600    10  40   90 x 10  $900 per year Total Annual Inventory cost = $400 + $900 = $1,300 per yr

What is the optimal order size Q? Annual carry ing cost  Annual ordering cost D Q H  S 2 Q Q *

2DS H

2x3600x10 20  60 pens / order

i.e. Q* 

Optimal Total Annual Inventory Cost D Q    H   S 2   Q  600  600  $1200

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What is the optimal stock level to reorder? (i.e. Re-Order Level)

Case I: Constant Demand Supplier’s delivery lead time is fixed = L = 2 days Demand is constant, d = 10 pens/day

Inventory EOQ=60

ROL = 20

0

Days 6 days

L=2 days

Reorder Level = d x L = 10 x 2 = 20 pens Safety Stock = 0 Total Annual Inventory Cost D Q     H    S Q 2  600  600  $1200

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Case II: Variable demand Supplier’s delivery lead time is fixed = L = 2 days Demand is variable, d=10 pens/day

 L 2days  6 pens Probability

% of stockout

# of pens dL = 20 pens

dL+ z  L Demand during the lead time of 2 days

Reorder Level  dL  z  L 2d ays Safety Stock  z  L 2d ays z  0  50% stockout/service level z  1.3  10% stockout/90% service level z  1.64  5% stockout/95% service level e.g. To provide a 95% protection against stockout ROL

= 10 x 2 + 1.64 x 6 = 20 + 9.8 = 30 pens

SS = 9.8 = 10 pens

4

Q=60

Q=60

Q=60

ROL=30 0 2 days

2 days

2 days

Days

Total Annual Inventory Cost  Annual ordering cost  Annual carrying cost  D  Q    S    SS H  2  Q 3600 10  60 10 20       60   2   $1400 per year

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2. Periodic Review Model - Check stock level every T day s - Order enough to cover thedemand over the next T  L day s - Target stock  d(T  L)  z  T L - Order size  TS - OH

Q2=TS-OH2

Q1=TS-OH1

Q3=TS-OH3

OH3

OH2 OH1

Days

L

L T

T

Check 1

Check 2

A good T 

Check 3

Q* d

Example:

T  6 days, L  2 days, d  10 pens per day, L T  12 pens Target stock for a 5% stockout, i.e. z = 1.65 TS = d(T  L)  z  T L = 10(6+2) + 1.64x12 = 80 + 19.7 = 100 pens Order quantity Q = 100 – OH

Average Order Size  Q  dT  60 Safety Stock  z  T L  19.7 6

TotalAnnual Inventory Cost D  Q    S    SS H Q   2   3600   60    10    19.7 20  60   2   600  1000  $1600 per year Summary

Continuous Review OptimalOrder Size  Q* 

2 DS H

ROL  dL  z L ; SS  z L Total Annual Inventory Cost 

D Q  S    SS  H Q 2 

Periodic Review A Good and Recommende d Review Interval T * 

Q* d

TS  d (T  L )  z  T L ; SS  z T L Order size  TS - OH Q  dT  Q  D Total Annual Inventory Cost    S    SS  H  Q  2

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Computing  L and  T L 1. Sample daily demands  d 1, d2 ....... d30 d  d 2  ........  d 30 d 1 30  10 pens/day

 1 day 

d 1  10 2  d 2  10 2  ..... d 30  10 2 30  1

 4.2426

 2 days  2x 1 da y  6 pens  8 days  8x 1 da y  12 pens i. e.  L  L 1da y

T L  T  L 1 da y

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3. Newspaper Boy’s problem Co  overstock cost Cu  understock cost Optimalservice level to maximize profit 

Cu C o  Cu

Example: A commemorative pen is priced at $100 and cost $70. Salvage value is $20.

Cu  $100 - $70  $30 Co  $70 - $20  $50 Optimalservice level 

Cu Co  C u

30 50  30  0.375 

If demand probabilities is estimated as follows: Demand 35 36 37 38 39 40

Probability .10 .15 .25 .25 .15 .10

Cumulative Probability 0.10 0.25 0.50 0.75 0.90 1.00

Order 37 pens.

If demand is normally distributed with a mean  41 and  4

Optimal SL  0.375  z*  0.3 Order size    z   41 - 0.3x4  39.8  40

0.375 z*

0

z

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Continuous Review Model on how much and when to draw money a) How much money to draw for your pocket? Usage D = $10/day S = $10 for mother to draw money for you H = $0.02/$/day (Interest from bank) L = 2 days for mother to draw money for you

2DS 2 x10 x10   $100 H 0.02 ↑ S  draw more each time ↑ H  draw less each time ↑ D  draw more each time Q

b) How much in pocket when you ask mother to draw money?

M oneyin pocket  ROL  dL  z L  $20  z  L

Zero L

Medium L

$20

Spending during 2 days

$20

Spending during 2 days

Large  L

$20

Spending during 2 days

L  0  ROL  $20  L  0 & higher SL of not running out of money in pocket  z  ROL Larger  L  Greater variability in using money during the 2 days   ROL

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