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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.

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Chapter 10: Inventory - Part 2

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  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.

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