2. 1 Prof. Yuan-Shyi Peter Chiu Feb. 2011 Material Management Class Note # 1-A MRP – Capacity Constraints

3. 2 § M1: Push & Pull § M1: Push & Pull Production Control System  MRP: Materials Requirements Planning (MRP) ~ PUSH  JIT: Just-in-time (JIT) ~ PULL  Definition (by Karmarkar, 1989) A pull system initiates production as a reaction to present demand, while A push system initiates production in anticipation of future demand Thus, MRP incorporates forecasts of future demand while JIT does not.

4. 3 § M2: MRP ~ Push § M2: MRP ~ Push Production Control System We determine lot sizes based on forecasts of future demands and possibly on cost considerations  A top-down planning system in that all production quantity decisions are derived from demand forecasts.  Lot-sizing decisions are found for every level of the production system. Item are produced based on this plan and pushed to the next level.

5. 4 § M2: MRP ~ Push § M2: MRP ~ Push Production Control System ( p.2 )  A production plan is a complete spec. of  The amounts of final product produced  The exact timing of the production lot sizes  The final schedule of completion  The production plan may be broken down into several component parts 1) The master production schedule (MPS) 2) The materials requirements planning (MRP) 3) The detailed Job Shop schedule  MPS - a spec. of the exact amounts and timing of production of each of the end items in a production system.

6. 5 § M2: MRP ~ Push § M2: MRP ~ Push Production Control System ( p.3 ) P.405 Fig.8-1

7. 6 § M2: MRP ~ Push § M2: MRP ~ Push Production Control System ( p.4 )  The data sources for determining the MPS include 1) Firm customer orders 2) Forecasts of future demand by item 3) Safety stock requirements 4) Seasonal plans 5) Internal orders  Three phases in controlling of the production system Phase 1: gathering & coordinating info to develop MPS Phase 2: development of MRP Phase 3: development of detailed shop floor and resource requirements from MRP

8. 7 § M2: MRP ~ Push § M2: MRP ~ Push Production Control System ( p.5 )  How MRP Calculus works: 1. Parent-Child relationships 2. Lead times into Time-Phased requirements 3. Lot-sizing methods result in specific schedules

9. 8 § M3: JIT ~ Pull § M3: JIT ~ Pull Production Control System Basics : 1. WIP is minimum. 2. A Pull system ~ production at each stage is initiated only when requested. 3. JIT extends beyond the plant boundaries. 4. The benefits of JIT extend beyond savings of inventory-related costs. 5. Serious commitment from Top mgmt to workers. Lean Production ≈ JIT

10. 9 § M4: The Explosion Calculus § M4: The Explosion Calculus (BOM Explosion) Gross Requirements of one level Push down Lower levels

11. 10 § M4: The Explosion Calculus § M4: The Explosion Calculus (page 2) Eg. 7-1 Fig.7-5 p.353 Trumpet ( End Item ) Bell assembly (1) Lead time = 2 weeks Valve casing assembly (1) Lead time = 4 weeks Valves (3) Lead time = 3 weeks Slide assemblies (3) Lead time = 2 weeks b-t-14 b-t-15 b-t-13

12. 11 § M4: The Explosion Calculus § M4: The Explosion Calculus (page 3) =>Steps 1. Predicted Demand (Final Items) 2. Net demand (or MPS) Forecasts Schedule of Receipts Initial Inventory 3. Push Down to the next level (MRP) Lot-for-lot production rule (lot-sizing algorithm) – no inventory carried over. Time-phased requirements May have scheduled receipts for different parts. 4. Push all the way down

13. 12 1 Trumpet 1 Bell Assembly 1 Valve casing Assembly 3Slide Assemblies 3 Valves 7 weeks to produce a Trumpet ? To plan 7 weeks ahead The Predicted Demands: Expected schedule of receipts Week Demand Week Scheduled receipts 8 9 10 11 12 13 14 15 16 17 77 42 38 21 26 112 45 14 76 38 8 9 10 11 12 0 6 9 Eg. 7-1

14. 13 Beginning inventory = 23, at the end of week 7 Accordingly the net predicted demands become MRP calculations for the Bell assembly (one bell assembly for each Trumpet) & Lead time = 2 weeks go-see-10 go-see-10 Master Production Schedule (MPS) for the end product (i.e. Trumpet) Week Net Predicted Demands 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 Week Gross Requirements Time-Phased Net Requirements Planned Order Release (lot for lot) Net Requirements 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38

15. 14 MRP Calculations for the valve casing assembly (1 valves casing assembly for each Trumpet) & Lead time = 4 weeks go-see-10 go-see-10 Week Gross Requirements Net Requirements Time-Phased Net Requirements Planned Order Release (lot for lot) 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 b-t-20 b-t-38

16. 15 MRP Calculations for the valves ( 3 valves for each valve casing assembly) go-see-10 go-see-10 Lead Time = 3 weeks On-hand inventory of 186 valves at the end of week 3 Receipt from an outside supplier of 96 valves at the start of week 5 MRP Calculations for the valves Week Gross Requirements Net Requirements Time-Phased Net Requirements Planned Order Release (lot for lot) Scheduled Receipts On-hand inventory 2 3 4 5 6 7 8 9 10 11 12 13 126 126 96 36 78 336 135 42 228 114 96 186 60 30 0 0 66 36 78 336 135 42 228 114 66 36 78 336 135 42 228 114

17. 16 Show the MRP Calculations for the slide assemblies ! §. M4.1: Class Work # CW.1 Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes What is the MRP Calculations for the slide assemblies ? ( 3 slide assemblies for each valve casing ) Lead Time = 2 weeks Assume On-hand inventory of 270 slide assemblies at the end of week 3 & Scheduled receipts of 78 & 63 at the beginning of week 5 & 7 ◆ 1 g-s-62 g-s-62

18. 17 To Think about … Lot-for-Lot may not be feasible ?! e.g. 336 Slide assemblies required at week 9 may exceeds plant’s capacity of let’s say 200 per week. Lot-for-Lot may not be the best way in production !? Why do we have to produce certain items (parts) every week? why not in batch ? To minimize the production costs.

19. 18 §. M4.2: Class Problems Discussion Chapter 7 : ( # 4, 5, 6 ) Chapter 7 : ( # 4, 5, 6 ) p.356-7 ( # 9 (b,c,d) ) ( # 9 (b,c,d) ) p.357 Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes

20. 19 § M5: Alternative Lot-sizing schemes  Log-for-log : in general, not optimal  If we have a known set of time-varying demands and costs of setup & holding, what production quantities will minimize the total holding & setup costs over the planning horizon?

21. 20 (1) MRP Calculation for the valve casing assembly when applying E.O.Q. lot sizing Technique instead of lot-for-lot ( g-s-14) g-s-14 Week Net Requirements Time-Phased Net Requirements Planned order release (EOQ) Planned deliveries Ending inventory 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 139 0 0 0 139 0 139 0 0 139 97 55 23 11 124 12 106 92 16 117 (1) EOQ Lot sizing (page 2)

22. 21 Ending Inventory Beginning Inventory Planning Deliveries Net Requirements = +- Total ordering ( times ) = 4 ; cost = $132 * 4 = $528 Total ending inventory = = 653 ; cost = ($0.6) (653) = $391.80 Total Costs = Setup costs + holding costs = 4*132+$0.6*653 = $919.80 vs. lot-for-lot 10*132 = $1320 (setup costs) g-b-41

23. 22 § M5: Alternative Lot-sizing schemes § M5: Alternative Lot-sizing schemes (page 3) (2) The Silver-Meal Heuristic (S-M)  Forward method ~ avg. cost per period (to span)  Stop when avg. costs increases.  i.e. Once c(j) > c(j-1) stop Them let y 1 = r 1 +r 2 +…+r j-1 and begin again starting at period j

24. 23 § M5: Alternative Lot-sizing schemes  The silver-meal heuristic Will Not Always result in an optimal solution (see eg.7.3; p.360)  Computing Technology enables heuristic solution ● S-M example 1 : Suppose demands for the casings are r = (18, 30, 42, 5, 20) Holding cost = $2 per case per week Production setup cost = $80 Starting in Period 1 : C(1) = $80 C(2) = [$80+$2(30)] /2 = $70 C(3) = [$80+$2(30)+$2(2)(42)] /3 =308/3 = $102.7 ∵ C(3) >C(2) ∴ STOP ; Set

25. 24 Starting in Period 3 : C(1) = 80 C(2) = [80+2(5)] /2 = 45 C(3) = [80+2(5)+$2(2)(20)] /3 = 170/3 = 56.7 ∵ C(3) >C(2) ∴ STOP ; Set ∴ Solution = (48, 0, 47, 0, 20) cost = $310 ● S-M example 2 : (counterexample) Let r = (10, 40, 30), k=50 & h=1 Silver-Meal heuristic gives the solution y=(50,0,30) but the optimal solution is (10,70,0) Conclusion of Silver-Meal heuristic It will not always result in an optimal solution The higher the variance (in demand), the better the improvement the heuristic gives (versus EOQ) r = (18, 30, 42, 5, 20)

26. 25 § M5: Alternative Lot-sizing schemes § M5: Alternative Lot-sizing schemes (page 4) (3) Least Unit Cost (LUC)  Similar to the S-M except it divided by total demanded quantities.  Once c(j) > c(j-1) stop and so on.

27. 26 ● LUC example: r = (18, 30, 42, 5, 20) h = $2 K = $80 Solution : in period 1 C(1) = $80 /18 = $4.44 C(2) = [80+2(30)] /(18+30) = 140/48 = $2.92 C(3) = [80+2(30)+2(2)(42)] /(18+30+42) = 308/90 = $3.42 Starting in period 3 C(1) = $80 /42 = 1.90 C(2) = [80+2(5)] /(42+5) = 90 /47 = 1.91 ∵ C(3) >C(2) ∴ STOP ; Set

28. 27 Starting in period 5 C(1) = $80 /5 = 16 C(2) = [80+2(20)] /(5+20) = 120 /25 = 4.8 ∴ Set ∴ Solution = ( 48, 0, 42, 25, 0) cost = 3(80)+2(30)+2(20) = $340 r = (18, 30, 42, 5, 20)

29. 28 § M5: Alternative Lot-sizing schemes § M5: Alternative Lot-sizing schemes (page 5) (4) Part Period Balancing (PPB)  More popular in practice  Set the order horizon equal to “# of periods” ~ closely matches total holding cost closely with the setup cost over that period.  Closer rule Eg. 80 vs. (0, 10, 90) then choose 90 Last three : S-M, LUC, and PPB are heuristic methods ~ means reasonable but not necessarily give the optimal solution.

30. 29 ● PPB example : r = (18, 30, 42, 5, 20) h = $2 K = $80 Starting in Period 1 Order Horizon Total Holding cost 123123 0 60 (2*30) 228 (2*30+2*2*42) K=80 ∵ K is closer to period 2 ∴

31. 30 Starting in Period 3 : Order Horizon Total Holding cost 123123 0 10 (2*5) 90 (2*5+2*2*20) ∵ K is closer to period 3 ∴ ∴ Solution = (48, 0, 67, 0, 0) cost = 2(80)+2(30)+2(5)+2(2)(20) = $310 # K=80 r = (18, 30, 42, 5, 20) h = $2 K = $80

32. 31 §. M5.1: Class Problems Discussion Chapter 7 : ( # 14, 17 ) Chapter 7 : ( # 14, 17 ) p.363 Preparation Time : 25 ~ 40 minutes Discussion : 15 minutes

33. 32 § M6: Wagner – Whitin Algorithm ~ § M6: Wagner – Whitin Algorithm ~ guarantees an optimal solution to the production planning problem with time-varying demands. Eq.

34. 33 § M6: Wagner – Whitin Algorithm (page 2) Eg. A four periods planning ◆ 2 g-t-63 g-t-63

35. 34 § M6: Wagner – Whitin Algorithm (page 3)  Enumerating vs. dynamic programming ◆ Dynamic Programming

36. 35 § M6: Wagner – Whitin Algorithm (page 4) See ‘ PM00c6-2 ‘ for Example

37. 36 § M6.1: Dynamic Programming Eq 7.2 r =(18,30,42,5,20) h=$2 k=$80

38. 37 §. M6.2: Class Problems Discussion Preparation Time : 10 ~ 15 minutes Discussion : 10 minutes 300 200 K=$20 C=$0.1 h=$0.02 #1: Inventory model when demand rate λ is not constant ( Chapter 7: # 18 ) #2: ( Chapter 7: # 18 (a),(b) ) p.363

39. 38 § M7: Incorporating Lot-sizing Algorithms into the Explosion calculus ▓ From Time-phased net requirements applies algorithm p.364 Example 7.6 from the time-phased net requirements for the valve casing assembly : Week Time-Phased Net Requirements 4 5 6 7 8 9 10 11 12 13 42 42 32 12 26 112 45 14 76 38 Setup cost = $132 ; h= $0.60 per assembly per week Silver-Meal heuristic : g-s-14

40. 39 Starting in week 4 : C(1) = $132 C(2) = [132+(0.6)(42)] /2 = 157.2/2 = $78.6 C(3) = [132+(0.6)(42)+(0.6)(2)(32)] /3= 195.6/3 =$65.2 C(4) = [195.6+(0.6)(3)(12)] /4 = 217.2/4 = $54.3 C(5) = [217.2+(0.6)(4)(26)] /5 = 279.6/5 = $55.9 (STOP) ∴ Starting in week 8 : C(1) = $132 C(2) = [132+(0.6)(112)] /2 = 199.2/2 = $99.6 C(3) = [199.2+(0.6)(2)(45)] /3= 253.2/3 =$84.4 C(4) = [253.2+(0.6)(3)(14)] /4 = 278.4/4 = $69.6 C(5) = [278.4+(0.6)(4)(76)] /5 = 460.8/5 = $92.2 (STOP) ∵ C(5) >C(4) ∴

41. 40 Starting in week 12 : C(1) = $132 C(2) = [132+(0.6)(38)] /2 = $77.4 ∴ ∴ y = (128, 0, 0, 0, 197, 0, 0, 0, 114, 0) MRP Calculation using Silver-Meal lot-sizing algorithm : Week Net Requirements Time-Phased Net Requirements Planned Order Release (S-M) Planned deliveries Ending inventory 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 128 0 0 0 197 0 0 0 114 0 86 44 12 0 171 59 14 0 38 0

42. 41 S-M : Total cost = 132(3)+(0.6)(86+44+12+171+59+14+38) = $650.4 Lot-For-Lot : $132*10 = $1320 E.O.Q : 4(132)+(0.6)(653) = $919.80 ▓ Compute the total costs for optimal schedule by Wagner-Whitin algorithm it is y 4 = 154, y 9 = 171, y 12 = 114 ; Total costs= ＄ 610.20 ▓ push down to lower level… g-t-20 g-s-14

43. 42 §. M7.1: Class Work # CW.2 Applies Part Period Balancing in MRP Calculation for the valve casing assembly. Applies Wagner-Whitin algorithm in MRP for the valve casing assembly. Applies Least Unit Cost in MRP Calculation for the valve casing assembly. Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes ◆ 3 g-t-64  

44. 43 §. M 7.2: Class Problems Discussion Preparation Time : 15 ~ 20 minutes Discussion : 10 minutes Chapter 7 : ( # 24, 25 ) Chapter 7 : ( # 24, 25 ) p.365-6 ( # 49 ) ( # 49 ) p.393

45. 44 § M8: Lot sizing with Capacity Constraints ▓ Requirements vs production capacities. ’’realistic’’~more complex. ▓ True optimal is difficult, time-consuming and probably not practical. ▓ Even finding a feasible solution may not be obvious. ▓ Feasibility condition must be satisfied ◇ e.g. Demand r = ( 52, 87, 23, 56 ) Total demands = 218 Capacity C= ( 60, 60, 60, 60 ) Total capacity = 240 though total capacity > total demands ; but it is still infeasible (why?)

46. 45 § M8: Lot sizing with Capacity Constraints § M8: Lot sizing with Capacity Constraints (page 2) ▓ Lot-shifting technique to find initial solution ▓ Eg. #7.7 (p.376) γ=(20,40,100,35, 80,75,25) C =(60,60, 60,60, 60,60,60) ◆ First tests for Feasibility condition → satisfied ◆ Lot-shifting C = (60,60, 60,60,60,60,60) γ = (20,40,100,35,80,75,25)  demand (C-γ) = (40,20,-40,…) (C-γ)’ =(20, 0, 0,…) (production plan) γ’= (40,60,60,35,80,75,25) [γ’=C- (C- γ)’] (C-γ’)’ = (20,0,0,25,-20,…) (C-γ’)’ = (20,0,0,5,0,…) γ’ = (40,60,60,55,60,75,25) [γ’=C- (C- γ’)’]◇

47. 46 § M8: Lot sizing with Capacity Constraints § M8: Lot sizing with Capacity Constraints (page 3) (C-γ’)’ = (20,0,0,5,0,-15,…) (C-γ’)’ = (10,0,0,0,0,0,…) γ’ = (50,60,60,60,60,60,25) [γ’=C- (C- γ’)’] (C-γ’)’ = (10,0,0,0,0,0,35) (production plan) γ’= (50,60,60,60,60,60,25) ∴ lot-shifting technique solution (backtracking) gives a feasible solution. ▓ Reasonable improvement rules for capacity constraints ◆ Backward lot-elimination rule ◇

48. 47 § M8: Lot sizing with Capacity Constraints § M8: Lot sizing with Capacity Constraints (page 4) ◆ Eg. 7.8 Assume k=$450, h=$2 C = (120,200,200,400,300,50,120, 50,30) γ= (100, 79,230,105, 3,10, 99,126,40) from lot-shifting γ’= ？ γ’ = (100,109,200,105,28,50,120,50,30) [ How ? ] costs = (9*$450)+2*(216)=$4482 ◆ Improvement Find Excess capacity first. C = (120,200,200,400,300,50,120, 50,30) γ’ = (100,109,200, 105, 28, 50,120, 50,30) (C - γ’) = ( 20, 91, 0, 295,272, 0, 0, 0, 0) ◇

49. 48 ◆ Is there enough excess capacity in prior periods to consider shifting this lot? excess capacity: (C –γ’) = (20,91,0,295,272,0,0,0,0) γ’ = (100,109,200,105,28,50,120,50,30) ∵ 30 units shifts from the 9th period to the 5 th period 242 192 142 58 108 158 § M8: Lot sizing with Capacity Constraints § M8: Lot sizing with Capacity Constraints (page 5) ◇

50. 49 ∵ 50 units shifts from the 8th period to the 5th ∵ 120 units shifts from the 7th period to the 5th [not Okay] ∵ okay to shift 50 from the 6 th period to the 5th Result : → γ ’ = (100,109,200,105,158,0,120,0,0) § M8: Lot sizing with Capacity Constraints § M8: Lot sizing with Capacity Constraints (page 6) ◇

51. 50 263 0 → γ ’ = (100,109,200,105,158,0,120,0,0) (C-γ ’ ) = (20,91,0,295,142,50,0,50,30) 137 300 Excess capacity ∵ Furthermore, it is okay to shift 158 from the 5 th period to the 4 th period ∵ 158 units shifts from the 5 th period to the 4 th increase holding cost by $2*158=$316 < $K “ okay ’’ → final γ’ = (100,109,200,263,0,0,120,0,0)

52. 51 ◆ after improvement; total cost = [ 5*$450+ $2*(694) ] = $3638 vs { $4482 (before improvement) where γ’ = (100,109,200,105,28,50,120,50,30) } ◆ improvement save 20% of costs § M8: Lot sizing with Capacity Constraints § M8: Lot sizing with Capacity Constraints (page 7) ◇

53. 52 §. M 8.1: Class Problems Discussion Chapter 7 : # CW.3 ; # 28 (a) (b) Chapter 7 : # CW.3 ; # 28 (a) (b) p.369 # CW.5 ; #CW.4 # CW.5 ; #CW.4 Preparation Time : 25 ~ 30 minutes Discussion : 15 minutes

54. 53 # CW.5 Consider problem #28 (a), suppose the setup cost for the construction of the base assembly is $200, and the holding cost is $0.30 per assembly per week, and the time-phased net requirements and production capacity for the base assembly in a table lamp over the next 6 weeks are: Week Time-Phased Net Requirements r = 1 2 3 4 5 6 335 200 140 440 300 200 Production c= Capacity 600 600 600 400 200 200 (a)Determine the feasible planned order release (b)Determine the optimal production plan

55. 54 § M 9: Shortcoming of MRP ■ Uncertainty ◆ forecasts for future sales ◆ lead time from one level to another ■ Safety stock to protect against the uncertainty of demand ◆ not recommended for all levels ◆ recommended for end products only, they will be transmitted down thru the explosion calculus. ■ Two implication in MRP  all of the lot-sizing decisions could be incorrect.  former decisions that are currently being implemented in the production process may be incorrect.

56. 55 § M 9: Shortcoming of MRP § M 9: Shortcoming of MRP ( page 2 ) ■ Applies the coefficient variation σ/μ ◆ obtain σ, find → ratio = ∴ σ=μx ratio ◆ obtain safety stock σx z (e.g. z = 1.28 → 90 ％ ) ◆ obtain (μ+σ*z ) as planned production schedule.

57. 56 Example 7.9 (p.381) [ Using a Type 1 service lever of 90 %] Consider example 7.1 (p.362) Demands for Trumpets If analyst finds that the ratio σ/μ (coefficient of variation) is 0.3 Harmon co. decided to produce enough Trumpets to meet all weekly demand with probability 0.90 0.90 for Normally Distributed demand has a Z = 1.28 Week Predicted Demand ( μ ) Standard Deviation ( σ= μ*0.3 ) Mean demand Plus safety stock ( μ+ z σ ) 8 9 10 11 12 13 14 15 16 17 77 42 38 21 26 112 45 14 76 38 23.1 12.6 11.4 6.3 7.8 33.6 13.5 4.2 22.8 11.4 107 58 53 29 36 155 62 19 105 53 [ i.e. μ+(1.28) σ ]

58. 57 ■ Capacity Planning ◆ Feasible solution at one level may result in an ‘’ infeasible ’’ requirements schedule at a lower level. ◆ CRP – Capacity requirements planning by using MRP planned order releases. ~ If CRP results in an ‘’ infeasible ’’ case then to correct it by ◇ schedule overtime, outsourcing ◇ revise the MPS ~ Trial & Error between CRP and MRP until fitted. § M 9: Shortcoming of MRP § M 9: Shortcoming of MRP (page 3)

59. 58 ▓ Rolling Horizons and System Nervousness ◆ MRP is not always treated as a static system. ~ may need to rerun each period for 1st period decision ▓ Other considerations ◆ Lead times is not always dependent on lot sizes ~ sometimes lead time increases when lot size increases ◆ MRP Ⅱ： Manufacturing Resource Planning ◇ MRP converts an MPS into planned order releases. ◇ MRP Ⅱ： Incorporate Financial, Accounting, & Marketing functions into the production planning process § M 9: Shortcoming of MRP § M 9: Shortcoming of MRP (page 3)

60. 59 Ultimately, all divisions of the company would work together to find a production schedule consistent with the overall business plan and long-term financial strategy of the firm. ◇ MRP Ⅱ：～ incorporation of CRP ◆ Imperfect production Process ◆ Data Integrity § M 9: Shortcoming of MRP § M 9: Shortcoming of MRP (page 4)

61. 60 §. M 9.1: Class Problems Discussion Chapter 7 : ( # 33 ) Chapter 7 : ( # 33 ) p.376 Preparation Time : 15 ~ 20 minutes Discussion : 10 minutes

62. 61 § M 10: J I T ◆ Kanban ◆ SMED (Single minute exchange of dies) ‧ IED (inside exchange of dies ) ‧ OID (out side exchange of dies ) ◆ Advantages vs. Disadvanges (See Table 6-1) § M 11: MRP & JIT 36 distinct factors to compare JIT, MRP, & ROP (reorder point) [Krajewski et al 1987]

63. 62 The End

64. 63 Solution: MRP Calculations for the Slide assemblies ( 3 ) Lead Time = 2 weeks On-hand inventory of 270 valves at the end of week 3 Receipt from an outside supplier of 78 & 63 at the start of week 5 & 7 MRP Calculations for the valves Week Gross Requirements Net Requirements Time-Phased Net Requirements Planned Order Release (lot for lot) Scheduled Receipts On-hand inventory 2 3 4 5 6 7 8 9 10 11 12 13 126 126 96 36 78 336 135 42 228 114 78 270 144 96 27 0 0 0 0 51 336 135 42 228 114 63 51 336 135 42 228 114 ◆1◆1 # CW.1 g-b-16

65. 64 ◆2◆2 g-b-33

66. 65 ◆3◆3 Solution: Applies Part Period Balancing in MRP Calculation for the valve casing assembly. Week Net Requirements Time-Phased Net Requirements Planned Order Release (PPB) 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 ? MRP Calculation using Part Period Balancing lot-sizing algorithm : Starting in Period 4: Order Horizon Total Holding cost 1234512345 0 $25.2 (0.6)*(42) $63.6 $25.2+2(0.6)(32) $85.2 $63.6+3(0.6)(12) $147.6 $85.2+4(0.6)(26) K=132 ∵ K is closer to period 5 ∴ # CW.2

67. 66 ◆3◆3 MRP Calculation using Part Period Balancing lot-sizing algorithm : Week Net Requirements Time-Phased Net Requirements P.O.R. (PPB) Planned deliveries Ending inventory 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 154 0 0 0 0 247 0 0 0 38 112 70 38 26 0 135 90 76 0 0 Starting in Period 9: Order Horizon Total Holding cost 12341234 0 $27 (0.6)*(45) $43.8 $27+2(0.6)(14) $180.6 $43.8+3(0.6)(76) K=132 ∵ K is closer to period 4 ∴ 154 0 0 0 0 247 0 0 0 38 # CW.2

68. 67 S-M : $650.4 Lot-For-Lot : $132*10 = $1320 E.O.Q : $919.80 ▓ Compute the total costs for optimal schedule by Wagner-Whitin algorithm it is y 4 = 154, y 9 = 171, y 12 = 114 ; Total costs= ＄ 610.20 ◆3◆3 PPB : Total cost = $132(3)+(0.6)(547) = $724.2 # CW.2 g-s-42

69. 68 # CW.3 Consider the example presented previously for the scheduling of the valve casing assembly. Suppose that the production capacity in any week is 50 valve casings. Determine the feasible planned order release for the valve casings. Recall that the time phased net requirements for the valve casings as followed: Week Net Requirements Time-Phased Net Requirements r = 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50

70. 69 Week Net Requirements Time-Phased Net Requirements r = 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 Production c= Capacity # CW.3 excess (c-r) = Capacity 8 8 18 38 24 (62) 5 36 (26) 12 (c-r)’ = [2] Lot-shifting technique (back-shift demand from r j > c j ): 50 50 50 50 50 8 8 18 38 24 (62) 5 36 (26) 12 8 8 18 0 0 0 5 10 0 12 (c-r)’ = final r ’ = 42 42 32 50 50 50 45 40 50 38 [1] First test for: It is okay!

71. 70 # CW.4 ( continues on #CW.3) Determine the optimal production plan. Suppose that with overtime work on 2 nd shifts, the company could increase the weekly production capacity to 120 valve casings, however, extra cost per week = $105. Where K=$100, is the regular setup cost. The holding cost per valve casing per week is estimated to be $0.65. Determine the optimal production plan. Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Production c= Capacity (O-T) 120 120 120 120 120

72. 71 # CW.4 ( continues on #CW.3) Determine the optimal production plan. Suppose that with overtime work on 2 nd shifts, the company could increase the weekly production capacity to 120 valve casings, however, extra cost per week = $105. Where K=$100, is the regular setup cost. The holding cost per valve casing per week is estimated to be $0.65. Determine the optimal production plan. Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 final r ’ = 42 42 32 50 50 50 45 40 50 38 (using regular shift) Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0 0 0 38 62 0 0 26 0 0 Σ= 126 Ending Inventories = [1] First, the cost for using regular shift is $100(10) + $0.65 (126) [ lot for lot ] = $1,081.9 [ lot for lot ]

73. 72 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity (O-T) excess (c-r) = Capacity 120 120 120 120 120 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 # CW.4 [1] First, the cost for using regular shift is $100(10) + $0.65 (126) [ lot for lot ] = $1,081.9 [ lot for lot ] 78 78 88 108 94 8 75 106 44 82 42 42 32 12 26 112 45 14 76 38 r = 78 78 88 108 94 8 75 106 44 excess (c-r)’= Capacity 35 77 0 31 43 0 6 42 42 32 12 26 112 45 14 76 38 85 43 120 89 77 120 59 0 114 0 0 0 0 0 85 0 120 0 0 120 0 0 114 0 final r ’ =

74. 73 # CW.4 Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 85 0 120 0 0 120 0 0 114 0 final r ’ = [2] T he cost for using Overtime shift is $205(4) + $0.65(372) = $1061.8 43 1 89 77 51 59 14 0 38 0 Ending Inventories = Less than the cost for using regular shift $1,081.9,  Saved $ 20.10

75. 74 # CW.4 Ending Inventories = [3] To think about the following solution: Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Suppose r ’= 74 32 0 38 62 0 0 26 0 0 Σ= 232 116 0 0 50 50 50 45 40 50 38 [ One OT, 7 regular ] T he cost for using only one Overtime shift on week 4 is $205(1) + $100(7) + $0.65(232) = $1055.8 Less than the cost for using regular shift $1,081.9,  Saved $ 26.1 Less than the cost for using all Overtime shift $1061.8  Saved $ 6.0 WHY ?

76. 75 # CW.4 Ending Inventories = [4] A Better Solution : Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Suppose r ’= 74 32 0 27 1 9 14 0 38 0 Σ= 195 116 0 0 39 0 120 50 0 114 0 T he cost for using the above solution is $205 (3) + $100 (2) + $0.65(195) = $ 941.75 Less than the cost for using regular shift $1,081.9,  Saved $ 140.05 Wow ! 42 113 0 0 0 120 50 0 114 0 $944.35

77. 76 §. M4.2: Class Problems Discussion Chapter 7 : ( # 4, 5, 6 ) Chapter 7 : ( # 4, 5, 6 ) p.356-7 ( # 9 (b,c,d) ) ( # 9 (b,c,d) ) p.357 Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes p.18

78. 77 §. M5.1: Class Problems Discussion Chapter 7 : ( # 14, 17 ) Chapter 7 : ( # 14, 17 ) p.363 Preparation Time : 25 ~ 40 minutes Discussion : 15 minutes p.31

79. 78 §. M6.2: Class Problems Discussion Preparation Time : 10 ~ 15 minutes Discussion : 10 minutes 300 200 K=$20 C=$0.1 h=$0.02 #1: Inventory model when demand rate λ is not constant ( Chapter 7: # 18 ) #2: ( Chapter 7: # 18 (a),(b) ) p.363 p.37

80. 79 §. M7.1: Class Work # CW.2 Applies Part Period Balancing in MRP Calculation for the valve casing assembly. Applies Wagner-Whitin algorithm in MRP for the valve casing assembly. Applies Least Unit Cost in MRP Calculation for the valve casing assembly. Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes ◆ 3 g-t-64   p.42

81. 80 §. M 7.2: Class Problems Discussion Preparation Time : 15 ~ 20 minutes Discussion : 10 minutes Chapter 7 : ( # 24, 25 ) Chapter 7 : ( # 24, 25 ) p.365-6 ( # 49 ) ( # 49 ) p.393 p.43

82. 81 §. M 8.1: Class Problems Discussion Chapter 7 : # CW.3 ; #CW.5 ; #CW.4 Preparation Time : 25 ~ 30 minutes Discussion : 15 minutes

83. 82 # CW.5 Consider problem #28 (a), suppose the setup cost for the construction of the base assembly is $200, and the holding cost is $0.30 per assembly per week, and the time-phased net requirements and production capacity for the base assembly in a table lamp over the next 6 weeks are: Week Time-Phased Net Requirements r = 1 2 3 4 5 6 335 200 140 440 300 200 Production c= Capacity 600 600 600 400 200 200 (a)Determine the feasible planned order release (b)Determine the optimal production plan p.53

84. 83 # CW.5 Solution Week Time-Phased Net Requirements r = 1 2 3 4 5 6 335 200 140 440 300 200 Production c= Capacity 600 600 600 400 200 200 (b) Determine the optimal production plan p.53 (a)Determine the feasible planned order release 265 400 460 -40 -100 0 ( c-r) = 265 400 320 0 0 0 Adj.( c-r) = 335 200 280 400 200 200 r’ = 335 200 280 400 200 200 r’ = 265 400 320 0 0 0 ( c-r’) =

85. 84 Production c= Capacity 600 600 600 400 200 200 (b) Determine the optimal production plan 335 200 280 400 200 200 r’ = 265 400 320 0 0 0 ( c-r’) = # CW.5 Solution (b) 535 0 480 400 0 200 r’’ = 65 600 120 0 200 0 Increase holding cost = $0.3*(200) + $0.3*2*(200)=$180 Saving setup cost = 2*K = 2*$200= $400 Overall savings = $220 535 0 480 400 0 200 Final production plan r’’ =

86. 85 §. M 9.1: Class Problems Discussion Chapter 7 : ( # 33 ) Chapter 7 : ( # 33 ) p.376 Preparation Time : 15 ~ 20 minutes Discussion : 10 minutes p.60

87. 86 # CW.3 Consider the example presented previously for the scheduling of the valve casing assembly. Suppose that the production capacity in any week is 50 valve casings. Determine the feasible planned order release for the valve casings. Recall that the time phased net requirements for the valve casings as followed: Week Net Requirements Time-Phased Net Requirements r = 4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 p.68

88. 87 # CW.4 ( continues on #CW.3) Determine the optimal production plan. Suppose that with overtime work on 2 nd shifts, the company could increase the weekly production capacity to 120 valve casings, however, extra cost per week = $105. Where K=$100, is the regular setup cost. The holding cost per valve casing per week is estimated to be $0.65. Determine the optimal production plan. Time-Phased Net Requirements r = 42 42 32 12 26 112 45 14 76 38 Production c= Capacity 50 50 50 50 50 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Production c= Capacity (O-T) 120 120 120 120 120 p.70