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Prof. Yuan-Shyi Peter Chiu Feb. 2011

Material Management Class Note # 1-A MRP – Capacity Constraints. Prof. Yuan-Shyi Peter Chiu Feb. 2011. § M1: Push & Pull Production Control System MRP : Materials Requirements Planning (MRP) ~ PUSH JIT : Just-in-time (JIT) ~ PULL Definition (by Karmarkar, 1989)

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Prof. Yuan-Shyi Peter Chiu Feb. 2011

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

  2. § 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.

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

  4. § 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 • The master production schedule (MPS) • The materials requirements planning (MRP) • 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.

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

  6. § M2: MRP ~ Push Production Control System ( p.4 ) • The data sources for determining the MPS include • Firm customer orders • Forecasts of future demand by item • Safety stock requirements • Seasonal plans • 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

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

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

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

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

  11. § M4: The Explosion Calculus (page 3) =>Steps • Predicted Demand (Final Items) • Net demand (or MPS) • Forecasts • Schedule of Receipts • Initial Inventory • 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. • Push all the way down

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

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

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

  15. MRP Calculations for the valves( 3 valves for each valve casing • assembly)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 2 3 4 5 6 7 8 9 10 11 12 13 Gross Requirements 126 126 96 36 78 336 135 42 228 114 Scheduled Receipts 96 On-hand inventory 18660 30 Net Requirements 0 0 66 36 78 336 135 42 228 114 Time-Phased Net Requirements 66 36 78 336 135 42 228 114 Planned Order Release (lot for lot) 66 36 78 336 135 42 228 114

  16. §. M4.1: Class Work # CW.1 • 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 Show the MRP Calculations for the slide assemblies ! ◆1g-s-62 Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes

  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.

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

  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?

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

  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

  22. § 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 y1 = r1+r2+…+rj-1 and begin again starting at period j

  23. § M5: Alternative Lot-sizing schemes • The silver-meal heuristicWill Not Alwaysresult 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

  24. Starting in Period 3 : • r = (18, 30, 42, 5, 20) 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)

  25. § 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.

  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 ∵ C(3) >C(2) ∴ STOP ; Set 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

  27. r = (18, 30, 42, 5, 20) 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

  28. § 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.

  29. ● PPB example : r = (18, 30, 42, 5, 20) h = $2 K = $80 Starting in Period 1 Order Horizon Total Holding cost 1 2 3 0 60 (2*30) 228 (2*30+2*2*42) K=80 ∵ K is closer to period 2 ∴

  30. r = (18, 30, 42, 5, 20) h = $2 K = $80 Starting in Period 3 : Order Horizon Total Holding cost 1 2 3 0 10 (2*5) 90 (2*5+2*2*20) K=80 ∵ K is closer to period 3 ∴ ∴ Solution = (48, 0, 67, 0, 0) cost = 2(80)+2(30)+2(5)+2(2)(20) = $310 #

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

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

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

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

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

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

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

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

  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)∴

  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 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Net Requirements 42 42 32 12 26 112 45 14 76 38 Time-Phased Net Requirements 42 42 32 12 26 112 45 14 76 38 Planned Order Release (S-M) 128 0 0 0 197 0 0 0 114 0 Planned deliveries 128 0 0 0 197 0 0 0 114 0 Ending inventory 86 44 12 0 171 59 14 0 38 0

  41. ▓ Compute the total costs • S-M : Total cost = 132(3)+(0.6)(86+44+12+171+59+14+38) = $650.4 • Lot-For-Lot : $132*10 = $1320 g-s-14 • E.O.Q : 4(132)+(0.6)(653) = $919.80 g-t-20 • for optimal schedule by Wagner-Whitin algorithm it is • y4=154 , y9=171 , y12=114 ; Total costs= ＄610.20 ▓ push down to lower level…

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

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

  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?)

  45. ◇ § 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’)’]

  46. ◇ § 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

  47. ◇ § 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)

  48. ◇ § M8:Lot sizing with Capacity Constraints(page 5) ◆ 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 5th period 242 192 142 58 108 158

  49. ◇ § M8:Lot sizing with Capacity Constraints(page 6) • ∵ 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 6th period to the 5th • Result : → γ’ = (100,109,200,105,158,0,120,0,0)

  50. ∵ Furthermore, it is okay to shift 158 from the 5th period to the 4th period • 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 ∵ 158 units shifts from the 5th period to the 4th increase holding cost by $2*158=$316 < $K “ okay ’’ → final γ’ = (100,109,200,263,0,0,120,0,0)

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