module c6 other eoq type models. other eoq-type models quantity discount models –all units...
TRANSCRIPT
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