2. 1 Prof. Yuan-Shyi Peter Chiu Feb. 2012 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