1. Module C6
Other EOQ Type Models

2. ALL SEEK TO MINIMIZE THE TOTAL ANNUAL COST EQUATION
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

3. QUANTITY DISCOUNTS All-units vs. incremental discounts
ALL UNITS DISCOUNTS FOR ALLEN
Quantity Unit Cost
< $10.00
$ 9.75
$ 9.50
$ 9.40
 $ 9.00

4. 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

5. Q* for the interval > U
Lowest point occurs at U. Qopt = U TC L U
But we can see that Q*
(which will have an even
deeper discount) gives a lower total cost, TC. Q* Q

6. Q* Is In the Interval (L,U)
Q* is the lowest point in the interval. Qopt = Q* TC L U Q* Q

7. Q* for the interval < L
Lowest point occurs at L. Qopt. = L TC L U Q* Q

8. Calculations for C = $10 Interval (1,300)
Since Q* > 300, the optimal solution for the model cannot come from this interval.

9. Calculations for C = $9.75 Interval (300,600)
Since Q* = 331 is in the interval (300,600), for this interval: Qopt = Q* = 331

10. Calculations for C = $9.50 Interval (600,1000)
Since Q* = 336 < L = 600, for this interval: Qopt = L = 600

11. Calculations for C = $9.40 Interval (1000,5000)
Since Q* = 337 < L = 1000, for this interval: Qopt = L = 1000

12. Calculations for C = $9.00 Interval (5000,)
Since Q* = 345 < L = 5000, for this interval: Qopt = L = 5000

13. QUANTITY DISCOUNT APPROACH FOR ALLEN
Summarizing
Quantity Unit Cost Ch Q* Qopt TC
< $ $
$ $ $61,292
$ $ $59,804
$ $ $59,389
 $ $ $59,325
ORDER 5000
Note: This is over a 9 month supply -- is this OK?

14. Using the Template Enter Values Optimal Values Enter Discount Breaks
and Discount Prices
All-Units Worksheet

15. 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

16. 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

17. 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

18. OPTIMAL PRODUCTION LOT SIZE, Q*
TC(Q) = CO(D/Q) + Ch((1-D/P)/2)Q + CD

19. 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

20. OPTIMAL PRODUCTION LOT SIZE

21. 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, (613,200) = $312,450

22. OTHER QUANTITES Length of a Production run = Q*/P =
31,449/8,760,000 = yrs. = (365)
= 1.31 days
Length of a Production cycle = Q*/D =
31,449/613,200 = yrs. = (365)
= days
Number of Production runs/yr. = D/Q* = 19.5
IMAX = (1-(613,200/8,760,000))(31,449) = 29,248

23. Using the Template Optimal Values Enter Parameters Production Lot Size
Worksheet

24. 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

25. 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

26. Average Inventory Average Number of Backorders
(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

27. 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.

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

29. EXAMPLE SCANLON PLUMBING
Saunas cost $2400 each (C = 2400)
Order cost = $ (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)

30. RESULTS

31. Using the Template Optimal Values Input Parameters Planned Shortage
Worksheet

32. What If Lead Time Were 4 Weeks?
Demand over 4 weeks = 4(15) = 60
4 weeks = years (for template)
Want order to arrive when there are 20 backorders.
Thus order should be placed when there are = 40 saunas left in inventory

33. Using Template
Reorder Point = 40
Enter Lead Time

34. 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