1 / 51

Chapter 10: Inventory - Part 2

Chapter 10: Inventory - Part 2. Types of Inventory and Demand Availability Cost vs. Service Tradeoff Pull vs. Push Reorder Point System Periodic Review System Joint Ordering Number of Stocking Points Investment Limit Just-In-Time. Optimal Inventory Control.

Thomas
Télécharger la présentation

Chapter 10: Inventory - Part 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 10: Inventory - Part 2 • Types of Inventory and Demand • Availability • Cost vs. Service Tradeoff • Pull vs. Push • Reorder Point System • Periodic Review System • Joint Ordering • Number of Stocking Points • Investment Limit • Just-In-Time

  2. Optimal Inventory Control • For perpetual (continual) demand. • Treat each stocking point independently. • Consider 1 product art 1 location. Periodic Determine: Review System How much to order: M-qi When to (re)order: T Find optimal values for: M & T.

  3. Periodic Review System Place 1st order LT3 LT1 LT2 Place 3rd order Receive 3rd order Place 2nd order Receive 1st order Receive 2nd order Order M - minus amount on hand every T time units. T = 20 days and M=90 in this example.

  4. Periodic Review • Useful when: • Inventory is reviewed on a fixed schedule (e.g., every week). • Multiple items are ordered from one supplier. • A common order interval allows transportation economies of scale. Each order must last for time T + LT -> Must protect against stockout during T + LT We will consider only constant lead time: sLT = 0 s’d = sd T + LT

  5. Inventory Variables D = demand (usually annual) d = demand rate S = order cost ($/order) LT = (constant) lead time I = carrying cost k = stockout cost (% of value/unit time) P = probability of being in C = item value ($/item) stock during lead time sd =std. deviation of demand sLT =std. deviation of lead time = 0 s’d =std. deviation of demand during lead time M = maximum level T = time between orders TC = total cost (usually annual) N = number of orders/year Q = average order quantity

  6. dT 1 S k s’d E(z) + IC z s’d + + IC TC = 2 T T Periodic Review - Optimal Ordering 2S T = Optimal order interval: Optimal number of orders/year: Maximum level : M = d(T+LT) + z s’d Optimal cost: IDC 1 N = T s’d = sd T + LT Std. deviation of demand during lead time: Average inventory: AIL =(dT)/2 + z s’d

  7. 3 Cases 1. Stockout cost k is known; P is not known. -> Calculate optimal P (similar to reorder point) We will not consider this case. 2. Stock cost k is not known; P is known. -> Can not use last term in TC. 3. Stockout cost k is known; P is known. -> Could use k to calculate optimal P.

  8. Periodic Review Example (same as Reorder Point) D = 5000 units/year  d = 96.15 units/week S = $10/order sd = 10 units/week C = $5/unit I = 20% per year LT = 2 weeks (constant)  sLT = 0

  9. = 23.0 3.29 + 2 s’d = 10 Periodic Review Example - Case 2 D = 5000 units/year  d = 96.15 units/week S = $10/order sd = 10 units/week C = $5/unit I = 20% per year LT = 2 weeks (constant)  sLT = 0 • k is not known; P =90% -> z = 1.28 2x10 = 0.0632 years = 3.29 weeks Solution: T = 0.2x5000x5 M = 96.15(3.29+2) + 1.28x23.0= 538.07 TC = 158.05 + 157.92 + 29.44 = $345.41/year

  10. Periodic Review Example - Case 3 D = 5000 units/year  d = 96.15 units/week S = $10/order sd = 10 units/week C = $5/unit I = 20% per year LT = 2 weeks (constant)  sLT = 0 • k = $2/unit; P =90% -> z = 1.28 Solution: T = 3.29 weeks as in Case 2 s’d = 23.0 as in Case 2 M = 538.07 as in Case 2 TC = 158.05 + 157.92 + 29.44 + 34.53 = $379.94/year

  11. Reorder Point Example - Case 3 Solution: • k =$2/unit; P =90% T = 3.29 weeks M = 538.07 units TC = $379.94/year • Could use k=$2/unit to find optimal P • Cost would decrease (as with Reorder Point).

  12. Service Level - Periodic Review SL = 1 - % of items out-of-stock Expected number of units out-of-stock/year = 1 - Annual demand (1/T) x s’d x E(z) = 1 - D s’d E(z) = 1 - time units must agree DT

  13. Service Levels for Cases 2 & 3 23.0(.0475) SL = 1 - = 0.9965 5000(0.0632) years units/year

  14. Comparison Reorder Point Case k P Q ROP TC($/yr) SL 1 2 .9678 322 218 347.97 .9994 2 - .90 316 210 334.33 .9979 3 2 .90 316 210 355.58 .9979 Periodic Review Case k P T(wks) M TC($/yr) SL 2 - .90 3.29 538 345.41 .9965 3 2 .90 3.29 538 379.94 .9965

  15. Comparison • Reorder Point System has: • Lower cost. • Lower average inventory. • Higher service level. • Less certainty about timing of future orders.

  16. Change T to a Convenient Value: 3 weeks T = 3.29 weeks = 0.0632 years s’d = 23.0 units M = 538.1 units TC = $379.94/year Ordering every 3.29 weeks is not very convenient! Suppose you order every 3 weeks: T = 3 weeks = 22.36 3 + 2 s’d = 10 M = 96.15(3+2) + 1.28x22.36= 509.37 TC = 173.33 + 144.23 + 28.62 + 36.82 = $383.00/year

  17. Change T to a Convenient Value: 4 weeks T = 3.29 weeks = 0.0632 years s’d = 23.0 units M = 538.1 units TC = $379.94/year Consider T = 4 weeks T = 4 weeks = 24.49 4 + 2 s’d = 10 M = 96.15(4+2) + 1.28x24.49= 607.35 TC = 130.00 + 192.30 + 31.35 + 30.25 = $383.90/year Cost for T = 3 weeks, T=4 weeks, T = 3.29 weeks are about the same!

  18. Joint Ordering • Suppose several items are ordered from the same supplier. • Each items would have an optimal order interval T. • This would require frequent orders to the same supplier. Alternative: Find a common order interval T and order all items from the same supplier together. - This is a variation of the periodic review system.

  19. Joint Ordering Variables Di = demand for item i (usually annual) di = demand rate for item i sdi =std. deviation of demand for item i Si = order cost for item i ($/order) O = common order cost ($/order) (pay once per order) Ci = value for item i($/item) Pi = probability of being in stock for item i ki = stockout cost for item i LT = (constant) lead time I = carrying cost (% of value/unit time) s’di =std. deviation of demand during lead time Mi = maximum level T = time between orders TC = total cost (usually annual)

  20. Joint Ordering - Optimal Ordering 2(O+Si) T = Optimal order interval: Maximum level : Mi = di (T+LT) + zi s’di Optimal cost: I DiCi O+Si 1 1 TC = + TI DiCi ki s’di E(zi) + I  Cizi s’d i+ 2 T T s’di = sdi T + LT Std. deviation of demand during lead time:

  21. Joint Ordering - continued s’di E(zi) SLi = 1 - Service Level: Optimal number of orders/year: Average inventory: AILi =(diT)/2 + zi s’di TDi 1 N = T

  22. Joint Ordering Example - 2 items: X & Y XY Di = Annual demand 2600 3900 di = demand rate 50/week 75/week sdi =std. deviation 10/week 15/week Si = order cost $20 $16 O = common order cost $50 Ci = value $60 $44 Pi = in-stock probability 80% 90% ki = stockout cost $30/item $10/item LT = (constant) lead time 2 weeks I = carrying cost 20%/year Consider ordering X and Y jointly.

  23. Joint Ordering Example - 2 items: X & Y XY Di = Annual demand 26003900 di = demand rate 50/week 75/week sdi =std. deviation 10/week 15/week Si = order cost $20$16 O = common order cost $50 Ci = value $60$44 Pi = in-stock probability 80% 90% ki = stockout cost $30/item $10/item LT = (constant) lead time 2 weeks I = carrying cost 20%/year 2(50 + 20 + 16) = 0.0512 years = 2.66 weeks T = (0.2)(2600x60 + 3900x44)

  24. = 21.59 s’dX = 10 2.66 + 2 Joint Ordering Example: Item X T = 0.0512 years = 2.66 weeks Optimal order interval: PX= 0.8 -> zX = 0.84 -> E(zX) = 0.1120 MX = 50(2.66+2) + 0.84(21.59) = 251.14 AILX = (50x2.66)/2 + 0.84(21.59) = 84.64 SLX = 1 - (21.59x0.1120)/(2.66x50) = 0.9818

  25. = 32.38 s’dY = 15 2.66 + 2 Joint Ordering Example: Item Y T = 0.0512 years = 2.66 weeks Optimal order interval: PY= 0.9 -> zY = 1.28 -> E(zY) = 0.0475 MY = 75(2.66+2) + 1.28(32.38) = 390.95 AILY = (75x2.66)/2 + 1.28(32.38) = 141.20 SLY = 1 - (32.38x0.0475)/(2.66x75) = 0.9923

  26. Joint Ordering Example - Total Cost T = 0.0512 years = 2.66 weeks O+Si 1 1 TC = + TI DiCi ki s’di E(zi) + I  Cizi s’d i+ 2 T T 1 50+20+16 + 0.0512(0.2)(2600x60 + 3900x44) TC = 2 0.0512 + (0.2) (60x0.84x21.59 + 44x1.28x32.38) 1 (30x21.59x0.1120 + 10x32.38x0.0475) + 0.0512 TC = 1681.20 + 1675.80 + 582.44 + 1718.79 = $5658.23/year

  27. Joint Ordering Summary For item X: Every 2.66 weeks order 251 - amount on hand. For item Y: Every 2.66 weeks order 391 - amount on hand. Total cost = $5658/year If X was ordered separately (with S = 50+20=70) TX = 3.5 weeks, MX = 295 and TCX = $3503.71/year If Y was ordered separately (with S = 50+16=66) TY = 3.22 weeks, MY = 435 and TCY = $2776.80/year Total cost for X and Y = $6280.51/year

  28. Joint Ordering - Adjust T For item X: Every 2.66 weeks order 251 - amount on hand. For item Y: Every 2.66 weeks order 391 - amount on hand. Total cost = $5658/year Could adjust T to 3 weeks: s’dY = 33.54 MY = 417.9 s’dX = 22.36 MX = 268.8 TC = 1490.67 + 1890.00 + 603.18 + 1578.39 = $5562/year

  29. Min-Max System • Approximate, but easy. • Hybrid Reorder Point/Periodic Review system. • Useful when inventory decreases in large steps (lumpy demand). Order M - amount on hand, when inventory falls below ROP. M = ROP + Q*

  30. 2DS Q = IC M ROP Min-Max System Order M - amount on hand, when inventory falls below ROP. ROP = dxLT + zs’d M = ROP + Q

  31. Min-Max System Example Q = 350; ROP = 100; LT = 2 days M = 450 Day Sales Inventory 0 112 1 8 104 2 64 40 order 410 = 450 - 40 3 10 30 4 25 415 = 30 - 25 + 410 5 35 380 … ... 32 … 121 33 23 98 order 352 = 450-98 34 3 95 35 40 407 = 95 - 40 + 352

  32. Other Issues • Stock to demand. • Estimate safety stock in terms of time. • Example: Order quantity = forecast of demand during time between orders + lead time + 1 week safety stock. • Multi-item, Multi-location inventories. • Inventories at plants, regional distribution centers, field warehouses, retail outlets, etc. • Complex, computer models (integer programming). • Pipeline inventories. • Reducing transit time can reduce inventory (regular and safety stock), but increase transportation cost.

  33. Aggregate Control of Inventories • Turnover Ratio. • Annual sales/Average inventory. • Can be for one items or all items. • Assumes sales and inventory increase proportionally. • ABC Classification • Based on (annual) sales amount. • Example on page 353 is wrong. • Give most attention to A items. (differentiation)

  34. Average Inventory and Throughput • Square Root Rule • Amount of inventory in a facility is proportional to the square root of the throughput (sales) at the facility. Average Inventory (cwt) Throughput Average inventory = k Annual Throughput (cwt)

  35. IT = k X Square Root Rule • An organization has total throughput (sales) of X units per year. • If there were n stocking locations, then each would expect throughput of X/n per year. IT = total inventory if there was a single stocking point. Ii = average inventory at one of n identical stocking points. X/n Ii = k Square root rule: IT = Ii n Total inventory in system = nIi

  36. Square Root Rule Example 1 Suppose the current system has 10 warehouses and each one has $60,000 of inventory on average. Q1: How much will the inventory investment change if there are to be 5 (not 10) warehouses? Investment with 10 warehouses = 10x60,000 = $600,000 = $189,737 IT = 60000 10 So Ii = $84,853 With 5 warehouses: 189,737 = Ii 5 Investment with 5 warehouses = 5x84853 = $424,265 Inventory change = 600,000-424,265 = $175,735 (-29%)

  37. Square Root Rule Example 2 Suppose the current system has 10 warehouses and each one has $60,000 of inventory on average. Q2: How many warehouses are needed if the average inventory in each is to be $30,000? = $189,737 IT = 60000 10 With $30,000 in each warehouse: IT = 189,737 = 30,000 n So n = (189,737/30,000)2 = 40 warehouses

  38. Square Root Rule Example 3 Suppose the current system has 10 warehouses and each one has $60,000 of inventory on average. Q2: How many warehouses are needed if the total inventory in the system is to be cut in half? Investment with 10 warehouses = 10x60,000 = $600,000 = $189,737 IT = 60000 10 With $300,000 in all warehouses: Ii = 300,000/n 300000 IT = 189,737 = (300000/n) n = n So n = (300,000/189,737)2 = 2.5 warehouses

  39. Total Investment Limit • Suppose several products are stored in the same warehouse. • A maximum total investment limit (L) for the warehouse is specified. • Order sizes (Q) may need to be decreased to reduce the total inventory investment. L = Maximum amount invested in inventory. TI = Total inventory. TI =  Ci (average inventory for item i) =  Ci (Qi/2) <= L

  40. 2DiSi Qi= ICi Total Investment Limit • Calculate order size Qi for each product. • If TI <= L, then the limit is not violated. • If TI > L, then the limit is violated, so reduce the order sizes by a fraction R: L R =  Ci(Qi/2) New order size (Q) = R x old order size

  41. Total Investment Limit Example Three products are stored in 1 warehouse. The total inventory investment can not exceed $10,000. I = 30% per year. Product Si Ci Di Qi CiQi/2 1 50 20 12000 2 50 10 25000 3 50 15 8000

  42. Total Investment Limit Example Three products are stored in 1 warehouse. The total inventory investment can not exceed $10,000. I = 30% per year. Product Si Ci Di Qi CiQi/2 1 50 20 12000 447.21 $4,472.10 2 50 10 25000 912.87 $4,564.35 3 50 15 8000 421.64 $3,162.30 $12,198.75 10,000 Investment exceeds $10,000 so R = = 0.8198 12,198.75

  43. Total Investment Limit Example Three products are stored in 1 warehouse. The total inventory investment can not exceed $10,000. I = 30% per year. Product Si Ci Di Qi new Qi 1 50 20 12000 447.21 366.6 2 50 10 25000 912.87 748.3 3 50 15 8000 421.64 345.6 R = 0.8198 Now investment equals $10,000.

  44. Just-In-Time (JIT) • Production system that originated in Japan. • Toyota is best known example. • Now used by many U.S. manufacturers and suppliers. • Producer: Produce small lots of high quality products with low inventory. • Supplier: Deliver small lots of high quality components on time.

  45. Just-In-Time Goals • Philosophy: • Synchronize the supply chain to respond to customers. • Eliminate all waste (inventory and scrap). • Goals: • Zero inventory. • Lot size of 1. • 100% quality. • Continuously strive to reduce inventory and lot size - and improve quality.

  46. JIT vs. “Traditional” Operations • Just-In-Time • Inventory = Liability • Setups = Liability • Defects/scrap = Liability • Eliminate/reduce: inventory, setups, defects, scrap. • “Traditional” Operations • Inventory = Protection against stockouts • Setups = necessary evil. • Defects/Scrap = Expected. • Optimize tradeoffs.

  47. Just-In-Time • Reducing setup cost S: • Reduces Q. • Reduces lot sizes and reduces inventory. • No inventory requires high quality components, high quality production and high quality transportation: • No defects. • No late deliveries. 2DS p Q= p-d IC JIT and “Traditional” Operations

  48. JIT Requires Frequent, Small, On-time Deliveries • Encourages suppliers to locate near customers. • Locate auto part plants near assembly plants. • Locate auto parts warehouse near assembly plants. • Requires high quality transportation carriers. • Deliveries can not be late, without inventory. • Favors truck and air. • Small shipments favor truck and air. • Example: Electronics manufacturers may fly parts from Asia several times per week.

  49. Benefits of Just-In-Time • Lower inventory. • Frees up $ for other uses. • Less space required. • Especially important when interest rates are high. • Higher quality. • Leads to higher prices, better sales, etc. • More efficient use of space. • Less scrap/defects to take up space.

  50. Concerns with Just-In-Time • Low inventory requires secure, reliable supply chains. • Reliable suppliers and transportation. • Reliable infrastructure. • Difficult to implement. • Requires major changes in business processes. • Hard to do “partial just-in-time”. • Difficult to phase in. • Works best if suppliers use just-in-time.

More Related