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Module C6

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.

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Module C6

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  1. Module C6 Other EOQ Type Models

  2. 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 < 300 $10.00 300-600 $ 9.75 600-1000 $ 9.50 1000-5000 $ 9.40 5000 $ 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. Lowest point occurs at U.Qopt = U But we can see that Q* (which will have an even deeper discount) gives alower total cost, TC. L U Q* Q* for the interval > U TC Q

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

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

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

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

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

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

  12. Calculations for C = $9.00Interval (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 < 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?

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

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

  22. 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

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

  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 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

  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. 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)

  30. RESULTS

  31. OptimalValues Input Parameters Planned Shortage Worksheet Using the Template

  32. 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

  33. Reorder Point = 40 Enter Lead Time Using Template

  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

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