Module C6

Other EOQ Type Models

OTHER EOQ-TYPE MODELS

• Quantity Discount Models– All Units Discounts

– Incremental Discounts (Not Discussed Here)

• Production Lot Size Models

• Planned Shortage Model

ALL SEEK TO MINIMIZE THE TOTAL ANNUAL COST EQUATION

QUANTITY DISCOUNTS

• All-units vs. incremental discounts

ALL UNITS DISCOUNTS FOR ALLEN

Quantity Unit Cost

< 300 $10.00

300-600 $ 9.75

600-1000 $ 9.50

1000-5000 $ 9.40

5000 $ 9.00

PIECEWISE APPROACH• Consider each quantity discount for C as if it were valid

everywhere from 0 - • Then, for each value of C, calculate the corresponding value of

Q* -- it will change slightly since Ch = HC and C changes slightly.

• Now for each quantity discount for C consider the interval (from a lower limit L to an upper limit U) over which it is valid and determine the value of Q that gives the lowest cost for the interval – RULE: (See next 3 slides)– If its Q* > U-- ignore this piece (one can get a better discount)– If its Q* lies between L and U, Q*is optimal for this piece– If its Q* < L -- the lower interval limit, L, is optimal for this piece

• Now calculate the total annual cost using the best value for each interval (and the appropriate value of C), and choose the lowest

Q* for the interval > U

TC

QQ*L U

Lowest point occurs at U.Qopt = U

But we can see that Q*(which will have an evendeeper discount) gives a

lower total cost, TC.

Q* Is In the Interval (L,U)

TC

QQ*L U

Q* is the lowest point in the interval.Qopt = Q*

Q* for the interval < L

TC

QQ* L U

Lowest point occurs at L.Qopt. = L

Calculations for C = $10Interval (1,300)

327)10(14.

)6240)(12(2* Q

Since Q* > 300, the optimal solution for the model cannot come from this interval.

CDQC

Q

DCQTC opt

H

opt

Oopt

2)(

Calculations for C = $9.75Interval (300,600)

331)75.9(14.

)6240)(12(2* Q

Since Q* = 331 is in the interval (300,600), for this interval: Qopt = Q* = 331

292,61$)6240)(75.9()331(2

365.1

331

)6240)(12()331( TC

CDQC

Q

DCQTC opt

H

opt

Oopt

2)(

Calculations for C = $9.50Interval (600,1000)

336)50.9(14.

)6240)(12(2* Q

Since Q* = 336 < L = 600,for this interval: Qopt = L = 600

804,59$)6240)(50.9()600(2

33.1

600

)6240)(12()600( TC

CDQC

Q

DCQTC opt

H

opt

Oopt

2)(

Calculations for C = $9.40Interval (1000,5000)

337)40.9(14.

)6240)(12(2* Q

Since Q* = 337 < L = 1000,for this interval: Qopt = L = 1000

389,59$)6240)(40.9()1000(2

316.1

1000

)6240)(12()1000( TC

CDQC

Q

DCQTC opt

H

opt

Oopt

2)(

Calculations for C = $9.00Interval (5000,)

345)00.9(14.

)6240)(12(2* Q

Since Q* = 345 < L = 5000,for this interval: Qopt = L = 5000

325,59$)6240)(00.9()5000(2

26.1

5000

)6240)(12()5000( TC

QUANTITY DISCOUNT APPROACH FOR ALLEN

• Summarizing

Quantity Unit Cost Ch Q* Qopt TC

< 300 $10.00 $1.40 327 ---- ----

300-600 $ 9.75 $1.365 331 331 $61,292

600-1000 $ 9.50 $1.33 336 600 $59,804

1000-5000 $ 9.40 $1.316 337 1000 $59,389

5000 $ 9.00 $1.26 345 5000 $59,325

• ORDER 5000

• Note: This is over a 9 month supply -- is this OK?

Using the Template

All-Units Worksheet

Enter Values

Enter Discount Breaksand Discount Prices

Optimal Values

PRODUCTION LOT SIZE MODELS• We are producing at a rate P/yr. That is

greater than the demand rate of D/yr.

• Inventory does not “jump” to Q but builds up to a value IMAX that is reached when production is ceased

What is IMAX?

• Length of a production run = Q/P

• During a production run – Amount Produced = Q– Amount Demanded = D(Q/P)

• IMAX = Q - D(Q/P) = (1-D/P)Q

• Average inventory = IMAX/2 = ((1-D/P)/2)Q

PRODUCTION LOT SIZE -- TOTAL ANNUAL COST

• Q = The production lot size

• CO = Set-up cost rather than order cost =$/setup

• Number of Set-ups per year = D/Q

• Average Inventory = ((1-D/P)/2)Q

• Instantaneous set-up time

• TC(Q) = CO(D/Q) + Ch((1-D/P)/2)Q + CD

OPTIMAL PRODUCTION LOT SIZE, Q*

SizeLot )/1(

2

)/1(

2

,

002

)/1(

*

2

2

PDC

DCQ

PDC

DCQ

Solving

PDC

Q

DC

dQ

dTC

h

O

h

O

hO

TC(Q) = CO(D/Q) + Ch((1-D/P)/2)Q + CD

EXAMPLE-- Farah Cosmetics

• Production Capacity 1000 tubes/hr.

• Daily Demand 1680 tubes

• Production cost $0.50/tube (C = 0.50)

• Set-up cost $150 per set-up (CO = 150)

• Holding Cost rate: 40% (Ch = .4(.50) = .20)

• D = 1680(365) = 613,200

• P = 1000(24)(365) = 8,760,000

OPTIMAL PRODUCTION LOT SIZE

449,31)

000,760,8200,613

1(20.

)200,613)(150(2

)/1(

2*

PDC

DCQ

h

O

TOTAL ANNUAL COST

• TOTAL ANNUAL COST = TC(Q) = TV(Q) + CD

TV(Q) = CO(D/Q) + Ch((1-D/P)/2)Q =

(150)(613,200/31,449) + .2((1-(613,200/8,760,000))(31,449)/2)

= $5,850

TC(Q) = TV(Q) + CD =

5,850 + .50(613,200) = $312,450

OTHER QUANTITES

• Length of a Production run = Q*/P =

31,449/8,760,000 = .00359yrs. = .00359(365)

= 1.31 days• Length of a Production cycle = Q*/D =

31,449/613,200 = .0512866yrs. = .00512866(365)

= 18.72 days• Number of Production runs/yr. = D/Q* = 19.5

• IMAX = (1-(613,200/8,760,000))(31,449) = 29,248

Using the Template

Production Lot SizeWorksheet

Enter Parameters

Optimal Values

PLANNED SHORTAGE MODEL

• Assumes no customers will be lost because of stockouts

• Instantaneous reordering

• Stockout costs:– Cb -- fixed administrative cost/stockout

– Cs -- annualized cost per unit short • Acts like a holding cost in reverse

• Reorder when there are S backorders

PROPORTION OF TIME IN/OUT OF STOCK

• T1 = time of a cycle with inventory

• T2 = time of a cycle out of stock

• T = T1 + T2 = time of a cycle

• IMAX = Q-S = total demand while in stock.

• Proportion of time in stock = T1/T. Multiply by D/D.

– T1D/TD = (Demand while in stock)/(Demand for cycle) = (Q-S)/Q

• Proportion of time out of stock=T2/T. Multiply by D/D.

– T2D/TD = (Demand while out of stock)/(Demand for cycle) = S/Q

Average InventoryAverage Number of Backorders

• Average Inventory = (Average Inv. When In Stock)(Proportion of time in stock)

=(IMAX/2)((Q-S)/Q) = ((Q-S)/2)((Q-S)/Q) =

(Q-S)2/2Q

• Average Backorders = (Average B/O When Out of Stock)(Proportion of time out of stock)

= (S/2)(S/Q) =

S2/2Q

TOTAL ANNUAL COST EQUATION

• TC(Q,S) = CO(Avg. Cycles Per Year) + CH(Average Inv.) + Cs (Average Backorders) + Cb (Number B/Os Per Cycle) (Avg. Cycles Per Year) +CD

TC(Q,S) =

CO(D/Q) + Ch((Q-S)2/2Q) + Cs(S2/2Q) + CbS(D/Q) + CD

• Take partial derivatives with respect to Q and S and set = 0. Solve the two equations for the two unknowns Q and S.

OPTIMAL ORDER QUANTITY, Q*OPTIMAL # BACKORDERS, S*

sh

bh

sh

b

s

sh

h

O

CC

DCQCS

CC

DC

C

CC

C

DCQ

**

)(2*

2

EXAMPLESCANLON PLUMBING

• Saunas cost $2400 each (C = 2400)

• Order cost = $1250 (CO = 1250)

• Holding Cost = $525/sauna/yr. (Ch = 525)

• Backorder Good will Cost $20/wk (CS = 1040)

• Backorder Admin. Cost = 10/order (Cb = 10)

• Demand = 15/wk (D = 780)

RESULTS

backorders20 are e when ther74order Re

201040525

)10)(780()74)(525(*

74)1040)(525(

)10*780(

1040

1040525

525

)780)(1250(2*

2

S

Q

Using the Template

Planned ShortageWorksheet

InputParameters

OptimalValues

What If Lead Time Were 4 Weeks?

• Demand over 4 weeks = 4(15) = 60– 4 weeks = .07692 years (for template)

• Want order to arrive when there are 20 backorders.

• Thus order should be placed when there are 60 - 20 = 40 saunas left in inventory

Using Template

Reorder Point = 40Enter Lead Time

Module C6 Review• All-Units Quantity Discount Model

– Q* modified for each piece– Best point for interval is Q* if Q* is in the interval– If Q* < L, L is best point for interval

• Production Lot Size Model– P > D, else optimal solution is to run machine continuously

– Same as EOQ except IMAX Q/2

• Planned Shortage– Time-dependent and time-independent shortage costs– 2 unknowns -- Q*, S*

• Use of template