1. Chapter 13 - Planifikimi dhe kontrolli i rezervës

12. Llojet e rezervës Cycle inventory = Q /2 = 280/2 = 140 drills Pipeline inventory = D L = dL = (70 drills/week)(3 weeks) = 210 drills Example 13.1

13. Llojet e rezervës Figure 13.1

15. Classifying Items as ABC % of Inventory Items 0 20 40 60 80 100 0 50 100 % Annual $ Usage A B C Class % $ Vol % Items A 80 15 B 15 30 C 5 55

19. Analiza ABC 10 20 30 40 50 60 70 80 90 100 Percentage of items Percentage of dollar value 100 — 90 — 80 — 70 — 60 — 50 — 40 — 30 — 20 — 10 — 0 — Figure 13.2

20. Analiza ABC 10 20 30 40 50 60 70 80 90 100 Percentage of items Percentage of dollar value 100 — 90 — 80 — 70 — 60 — 50 — 40 — 30 — 20 — 10 — 0 — Figure 13.2

21. Analiza ABC 10 20 30 40 50 60 70 80 90 100 Percentage of items Percentage of dollar value 100 — 90 — 80 — 70 — 60 — 50 — 40 — 30 — 20 — 10 — 0 — Class C Class A Class B Figure 13.2

22. Analiza ABC Figure 13.2 10 20 30 40 50 60 70 80 90 100 Percentage of items Percentage of dollar value 100 — 90 — 80 — 70 — 60 — 50 — 40 — 30 — 20 — 10 — 0 — Class C Class A Class B

23. Analiza ABC Figure 13.2 10 20 30 40 50 60 70 80 90 100 Percentage of items Percentage of dollar value 100 — 90 — 80 — 70 — 60 — 50 — 40 — 30 — 20 — 10 — 0 — Class C Class A Class B

24. Analiza ABC Figure 13.2 10 20 30 40 50 60 70 80 90 100 Percentage of items Percentage of dollar value 100 — 90 — 80 — 70 — 60 — 50 — 40 — 30 — 20 — 10 — 0 — Class C Class A Class B

26. MODELI I SASISË EKONOMIKE TË POROSISË

27. Kur të bëhet porosia Sa duhet të porositet

28. Sasia ekonomike e porosis ë(EOQ) K a një aplikim mjaft të gjerë dhe mundë të përdoret per planifikim e blerjen të lëndëve të para dhe në planifikimin e blerjeve të shitësit me shumicë dhe pakicë të cilët rishesin produktet.

30. Sasia ekonomike e porosis ë kostoja e mbajtjes vjetore=(average cycle inventory)(unit holding cost) Figure 13.3 On-hand inventory (units) Koha Average cycle inventory Q Q — 2 1 cycle Receive order Inventory depletion (demand rate)

32. Economic Order Quantity Sasia ekonomike e porosis ë Figure 13.4 Annual cost (dollars) Sasia porosisë ( Q ) Kosto e porositjes ( OC ) Kosto e mbatjes ( HC ) Kosto totale = HC + OC

33. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Sasia e porosisë ( Q ) 3000 — 2000 — 1000 — 0 — Kosto totale = ( H ) + ( S ) D Q Q 2 Kosto e mbajtjes = ( H ) Q 2 Kost e porosotijes = ( S ) D Q Annual cost (dollars)

34. Economic Order Quantity Sasia ekonomike e porosis ë Annual cost (dollars) | | | | | | | | 50 100 150 200 250 300 350 400 3000 — 2000 — 1000 — 0 — Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Lot Size ( Q ) Birdfeeder costs C = ( H ) + ( S ) Q 2 D Q D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 390 units Example 13.2

35. Economic Order Quantity Sasia ekonomike e porosis ë Annual cost (dollars) | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs C = ( H ) + ( S ) Q 2 D Q D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 390 units C = $2925 + $108 = $3033 Example 13.2

36. Economic Order Quantity Sasia ekonomike e porosis ë Annual cost (dollars) | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs C = ( H ) + ( S ) Q 2 D Q D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 390 units C = $2925 + $108 = $3033 Example 13.2

37. Economic Order Quantity Sasia ekonomike e porosis ë Annual cost (dollars) | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs C = ( H ) + ( S ) Q 2 D Q D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 390 units C = $2925 + $108 = $3033 Example 13.2

38. Economic Order Quantity Sasia ekonomike e porosis ë Annual cost (dollars) | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs C = ( H ) + ( S ) Q 2 D Q D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 468 units Example 13.2

39. Economic Order Quantity Sasia ekonomike e porosis ë Annual cost (dollars) | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs C = ( H ) + ( S ) Q 2 D Q D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 468 units C = $3510 + $90 = $3600 Example 13.2

40. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Figure 13.4 Annual cost (dollars)

41. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = EOQ C = ( H ) + ( S ) Q 2 D Q EOQ = 2 DS H Example 13.3 Annual cost (dollars)

42. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 75 units C = ( H ) + ( S ) Q 2 D Q EOQ = 2 DS H Example 13.3 Annual cost (dollars)

43. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 75 units C = $562 + $562 = $1124 C = ( H ) + ( S ) Q 2 D Q EOQ = 2 DS H Example 13.3 Annual cost (dollars)

44. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 75 units C = $562 + $562 = $1124 C = ( H ) + ( S ) Q 2 D Q EOQ = 2 DS H Example 13.3 Annual cost (dollars)

45. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Bird feeder costs D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 75 units C = $562 + $562 = $1124 C = ( H ) + ( S ) Q 2 D Q EOQ = 2 DS H Lowest cost Best Q (EOQ) Example 13.3 Annual cost (dollars)

46. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Lowest cost Best Q (EOQ) Example 13.3 Annual cost (dollars) Bird feeder costs D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 75 units C = $562 + $562 = $1124 C = ( H ) + ( S ) D Q Q 2 EOQ = 2 DS H

47. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Lowest cost Best Q (EOQ) Example 13.3 Time between orders TBO EOQ = = 75/936 = 0.080 year Annual cost (dollars) Total cost = ( H ) + ( S ) D Q Q 2 Birdfeeder costs D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 75 units C = $562 + $562 = $1124 C = ( H ) + ( S ) D Q Q 2 EOQ = 2 DS H D Q EOQ D

48. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Current Q Total cost = ( H ) + ( S ) D Q Q 2 Birdfeeder costs D = (18 /week)(52 weeks) = 936 units H = 0.25 ($60/unit) = $15 S = $45 Q = 75 units Lowest cost Best Q (EOQ) Time between orders TBO EOQ = = 75/936 = 0.080 year TBO EOQ = (75/936)(12) = 0.96 months TBO EOQ = (75/936)(52) = 4.17 weeks TBO EOQ = (75/936)(365) = 29.25 days EOQ D Example 13.3 Annual cost (dollars)

49. Economic Order Quantity Sasia ekonomike e porosis ë | | | | | | | | 50 100 150 200 250 300 350 400 Lot Size ( Q ) 3000 — 2000 — 1000 — 0 — Current cost Lowest cost Best Q (EOQ) Current Q Total cost = ( H ) + ( S ) D Q Q 2 Holding cost = ( H ) Q 2 Ordering cost = ( S ) D Q Figure 13.5 Annual cost (dollars)

52. How Much? When!

53. Continuous Review Soup Soup Soup Time Figure 13.7 On-hand inventory

54. Continuous Review Soup Soup Soup Time R Order received Q OH Figure 13.7 On-hand inventory

55. Continuous Review Soup Soup Soup Order received Q OH Order placed IP TBO L R Figure 13.7 On-hand inventory

56. Continuous Review Soup Soup Soup Time Order received Order received Q Q OH OH Order placed Order placed IP IP TBO L TBO L TBO L R Order received Q OH Order placed IP Order received Figure 13.7 On-hand inventory

57. Continuous Review On-hand inventory Time Q Q OH OH Order placed Order placed IP IP TBO L TBO L TBO L R Q OH Order placed IP Example 13.4 Chicken Soup R = Average demand during lead time = (25)(4) = 100 cases IP = OH + SR – BO = 10 + 200 – 0 = 210 cases Soup Soup Soup Order received Order received Order received Order received

58. Uncertain Demand Figure 13.6 Time On-hand inventory Order received Q OH Order placed Order placed Order received IP IP R TBO 1 TBO 2 TBO 3 L 1 L 2 L 3 Q Order placed Q Order received Order received

60. Lead Time Distributions Lead Time Distribution sld =90% δ t = 4 d = 18 L = 3 dite Rezerva e sigurisë = z  L = 1.28*6,93= 8.86njesi ose 9 njesi Momentin e riporositjes = dL + rezerva e sigurisë = 3(18) + 9 = 63 njesi dL=kerkesa mesatare gjate kohes se levrimit Example 13.6  t = 15 = + + 75 Demand for week 1 75 Demand for week 2 75 Demand for week 3  t = 26 225 Demand for three-week lead time  t = 15  t = 15  L =  t L = 4 3 = 6.93

62. Reorder Point / Safety Stock Probability of stockout (1.0 – 0.90 =0.10) Example 13.5 Safety Stock/R Safety stock = z  L = 1.28*6,93 = 8.86 = 9 njesi Reorder point = ADDLT + SS = 54 + 9 = 63 njesi Probability of stockout (1.0 - 0.85 = 0.15) Cycle-service level = 90% Average demand during lead time z  L R

63. Reorder Point / Safety Stock Probability of stockout (1.0 – 0.95 =0.05) Example 13.5 Safety Stock/R Safety stock = z  L = 1.64*6,93= 11.3 ose 11njesi Reorder point = ADDLT + SS = 3(18) + 11 = 65 njesi Probability of stockout (1.0 - 0.85 = 0.15) Cycle-service level = 95% Average demand during lead time z  L R

64. Lead Time Distributions Figure 13.10  t = 15 = + + 75 Demand for week 1 75 Demand for week 2 75 Demand for week 3  t = 26 225 Demand for three-week lead time  t = 15  t = 15

65. Lead Time Distributions Bird feeder Lead Time Distribution Example 13.6  t = 15 = + + 75 Demand for week 1 75 Demand for week 2 75 Demand for week 3  t = 26 225 Demand for three-week lead time  t = 15  t = 15

66. Lead Time Distributions Bird feeder Lead Time Distribution t = 1 week d = 18 L = 2 Example 13.6  t = 15 = + + 75 Demand for week 1 75 Demand for week 2 75 Demand for week 3  t = 26 225 Demand for three-week lead time  t = 15  t = 15

67. Lead Time Distributions Bird feeder Lead Time Distribution t = 1 week d = 18 L = 2 Example 13.6  t = 15 = + + 75 Demand for week 1 75 Demand for week 2 75 Demand for week 3  t = 26 225 Demand for three-week lead time  t = 15  t = 15  L =  t L = 5 2 = 7.1

68. Lead Time Distributions Bird feeder Lead Time Distribution t = 1 week d = 18 L = 2 Safety stock = z  L = 1.28(7.1) = 9.1 or 9 units Reorder point = dL + Safety stock = 2(18) + 9 = 45 units Example 13.6  t = 15 = + + 75 Demand for week 1 75 Demand for week 2 75 Demand for week 3  t = 26 225 Demand for three-week lead time  t = 15  t = 15  L =  t L = 5 2 = 7.1

69. Lead Time Distributions Bird feeder Lead Time Distribution t = 1 week d = 18 L = 2 Reorder point = 2(18) + 9 = 45 units Example 13.6  t = 15 = + + 75 Demand for week 1 75 Demand for week 2 75 Demand for week 3  t = 26 225 Demand for three-week lead time  t = 15  t = 15

70. Lead Time Distributions Bird feeder Lead Time Distribution t = 1 week d = 18 L = 2 Reorder point = 2(18) + 9 = 45 units C = $562.50 + $561.60 + $135 = $1259.10 Example 13.6  t = 15 = + + 75 Demand for week 1 75 Demand for week 2 75 Demand for week 3  t = 26 225 Demand for three-week lead time  t = 15  t = 15 C = ($15) + ($45) + 9($15) 75 2 936 75

71. Periodic Review Systems Figure 13.11 Time On-hand inventory IP 1 IP 3 IP 2 Order received Order received IP IP OH OH Order placed Order placed Q 1 Q 2 Q 3 L L L P P Protection interval T Order received IP

72. Periodic Review Systems Example 13.7 Time On-hand inventory IP 1 IP 3 IP 2 Order received Order received IP IP OH OH Order placed Order placed Q 1 Q 2 Q 3 L L L P P Protection interval T Order received IP Order received IP TV Set - P System IP = OH + SR – BO Q t = T - IP t T = 400 BO = 5 OH = 0 SR = 0 IP = 0 + 0 – 5 = –5 sets Q = 400 – (–5) = 405 sets

73. Periodic Review Systems Example 13.7 Time On-hand inventory IP 1 IP 3 IP 2 Order received Order received IP IP OH OH Order placed Order placed Q 1 Q 2 Q 3 L L L P P Protection interval T Order received IP

74. Periodic Review Systems Example 13.8 T = Average demand during the protection interval + Safety stock = d ( P + L ) + z  P + L = (18 units/week)(16 weeks) + 1.28(12 units) = 123 units EOQ = 75 units D = (18 units/week)(52 weeks) = 936 units  t = 18 units L = 2 weeks cycle/service level = 90% Time On-hand inventory IP 1 IP 3 IP 2 Order received Order received IP IP OH OH Order placed Order placed Q 1 Q 2 Q 3 L L L P P Protection interval T Order received IP Bird feeder—Calculating P and T P = (52) = (52) = 4.2 or 4 weeks EOQ D 75 936  P+L =    P + L = 5 6 = 12 units

75. Periodic Review Systems Example 13.8 EOQ = 75 units D = (18 units/week)(52 weeks) = 936 units  t = 18 units L = 2 weeks cycle/service level = 90% P = 4 weeks T = 123 units Time On-hand inventory IP 1 IP 3 IP 2 Order received Order received IP IP OH OH Order placed Order placed Q 1 Q 2 Q 3 L L L P P Protection interval T Order received IP Bird feeder—Calculating P and T

76. Periodic Review Systems Example 13.8 EOQ = 75 units D = (18 units/week)(52 weeks) = 936 units  t = 18 units L = 2 weeks cycle/service level = 90% C = $540 + $585 + $225 = $1350 P = 4 weeks T = 123 units Time On-hand inventory IP 1 IP 3 IP 2 Order received Order received IP IP OH OH Order placed Order placed Q 1 Q 2 Q 3 L L L P P Protection interval T Order received IP Bird feeder—Calculating P and T C = ($15) + ($45) + 15($15) 4(18) 2 936 4(18)

77. Analiza ABC — Solved Problem 2