Materials Requirements Planning

1. Materials Requirements Planning

2. MRP • Suatu perusahaan dengan beberapa produk, menerima pesanan dalam beberapa perioda. • Perusahaan perlu untuk mengatur bahan baku apa saja, berapa, dan kapan setiap bahan itu diperlukan agar biaya produksi bisa diminimalkan. • Perlu juga dipertimbangkan kuantitas pesanan agar dimungkinkan menekan biaya inventory.

3. MRP • Untuk mengetahui kebutuhan bahan baku, perlu diketahui Bill of Material ( BoM) dari setiap produk. • BOM = Resep = Formula • BOM adalah material, bagian, komponen, unit atau proses yang diperlukan untuk menghasilkan suatu produk.

4. Bill of Materials (BOM) FileA Complete Product Description • Materials • Parts • Components • Production sequence

5. Deskripsi BOM

6. Table 100 Top 023 Base 200 Frame (1) 300 Legs (4) 203 Leg Bolts (4) 220 Glue 066 Boards (3) 030 Leg Supports (4) 533 Glue 066 Ends (2) 411 Sides (2) 622 Multilevel Bill

7. Table 100 Single-level Bill Base 200 Top 023 Top 023 Base 200 Glue 066 Boards (3) 030 Legs (4) 203 Leg Bolts (4) 220 Frame (1) 300 Frame (1) 300 Leg Supports (4) 533 Glue 066 Ends (2) 411 Sides (2) 622

8. Example Problem Y X B(2) C D L M N(2) O F G(2) M(2) N P G J K(2) J K(2) Using the following product tree, construct the appropriate single level trees. How many Ks are needed to make 100 Xs and 50 Ys?

9. Pegging identify each parent item that created demand A C Pegged requirements B C D Pegged requirements Source of requirements

10. A B(4) C(2) D(2) E(1) D(3) F(2) Example of MRP Logic and Product Structure Tree Given the product structure tree for “A” and the lead time and demand information below, provide a materials requirements plan that defines the number of units of each component and when they will be needed Product Structure Tree for Assembly A Lead Times A 1 day B 2 days C 1 day D 3 days E 4 days F 1 day Total Unit Demand Day 10 50 A Day 8 20 B (Spares) Day 6 15 D (Spares)

11. LT = 1 day First, the number of units of “A” are scheduled backwards to allow for their lead time. So, in the materials requirement plan below, we have to place an order for 50 units of “A” on the 9th day to receive them on day 10.

12. LT = 2 Spares A 4x50=200 B(4) C(2) D(2) E(1) D(3) F(2) Next, we need to start scheduling the components that make up “A”. In the case of component “B” we need 4 B’s for each A. Since we need 50 A’s, that means 200 B’s. And again, we back the schedule up for the necessary 2 days of lead time.

13. 13 A Part D: Day 6 B(4) C(2) 40 + 15 spares D(2) E(1) D(3) F(2) Finally, repeating the process for all components, we have the final materials requirements plan: • The McGraw-Hill Companies, Inc., 2001

14. Additional MRP Scheduling Terminology • Gross Requirements • Scheduled receipts • Projected available balance • Net requirements • Planned order receipt • Planned order release

15. X A(2) B(1) C(2) D(5) C(3) MRP Example Requirements include 95 units (80 firm orders and 15 forecast) of X in week 10

16. X A(2) It takes 2 A’s for each X

17. X A(2) B(1) It takes 1 B for each X

18. X A(2) B(1) C(3) It takes 3 C’s for each A

19. X A(2) B(1) C(3) C(2) It takes 2 C’s for each B

20. X A(2) B(1) C(3) C(2) D(5) It takes 5 D’s for each B

21. Schedule Receipt • Kerjakan kembali bila pada perioda 6 ada schedule receipt material D sebanyak 50 unit. • Kerjakan kembali bila lot size C adalah 100.

22. Low Level Code A C Part Low Level A 0 B 1 C 1 2 D 2 B C D

23. Given the following product tree, explode, offset, and determine the gross and net requirements. All lead time are one week, and the quantities required are show in parentheses. The master production schedule call for 100 As to be available In week 5,. There are 20 Bs available. A C(1) B(2) D(2) F(1) D(1) E(1)

24. Given the following product tree, explode, offset, and determine the gross and net requirements. All lead time are one week, and the quantities required are show in parentheses. The master production schedule call for 100 As to be available in week 5, there is a schedule receipt of 100 Bs in week 1. There are 200Fs available. All order quantities are lot-for-lot. A C(1) B(2) D(2) F(1) D(1) E(1)

25. Given the following partial product tree, explode, offset, and determine the gross and net requirements for components H, I, J, and K. Lead time are 1, 2, 1, and 1 week, and the quantities required are show in parentheses. The master production schedule call for production of 50 Hs in week 3, and 80 in week5. There is a scheduled receipt of 100 Is in week 2,. There are 400 Js, and 400 Ks available. H I(2) J(2) K(3)

26. Suatu produk mempunyai Bill of Material seperti di bawah ini : Buatlah product tree dari produk A dan B. MPS menentukan agar produk A tersedia 50 unit pada perioda 4, dan produk B Tersedia 100 unit pada perioda 5. Produk A telah tersedia 10 pada awal perioda serta komponen C dijadwalkan akan diterima pada perioda 2 sebesar 100. Buatlah MRP untuk semua komponen.

27. Lot Sizing in MRP Programs • Lot-for-lot (L4L) • Economic Order Quantity (EOQ) • Period Order Quantity (POQ) • Part Period Balancing (PBB) • Least Unit Cost (LUC) • Least average Total Cost (LTC) – Silver Meal Which one to use? -> The one that is least costly!

28. Lead Time = 1 minggu Holding cost = $2/unit/minggu Setups cost =$200; Lot size = 1 Lot for Lot Biaya setup = 7*$200 = $1400 Biaya Hold = 0 ( tidak ada inv) Total biaya = $1400

29. D=(35+30+ .. + 55) = 270 Q= √((2*200*270)/20) = 74 EOQ Biaya order = 4*$200 = $800 Biaya Hold = (44+4+4+…..+61)*$2 = $790 Total biaya = $800+$790 = $1590

30. Periodic Order Quantity (POQ) EOI =EOQ/d = 74/27 = 3 periode; pemesanan setiap 3 periode Biaya order = 3*$200 = $600 Biaya Hold = (50+10+10+…..+30)*$2 = $380 Total biaya = $600+$380 = $980

31. Part Period Balancing (PPB) • Menyeimbangkan biaya setup dan biaya holding dengan menggunakan Economic Part Period (EPP). • EPP di definisikan sebagai rasio dari biaya setup terhadap biaya holding. • Teknik PPB mengkombinasikan periode kebutuhan sehingga jumlah part period mendekati nilai EPP

32. Part Period Balancing EPP = 200/2 = 100 Mencari pesanan yang mendekati EPP, dg dasar pemikiran Menyimpan 100 unit selama 1 periode sama dengan menyimpan 50 unit selama 2 periode ( biaya penyimpanan sama ) Kombinasi Lot size Part Period ( inventory) periode • 30 0 • 2,3 30+40=70 40*1p=40 • 2,3,4 30+40+0=70 40 • 2,3,4,5 30+40+0+10=80 40*1p+10*3p=70 • 2,3,4,5,6 30+40+0+10+40=120 230 Part period yg paling dekat dg EPP adalah 70, maka pesanan awal pada periode 2 sejumlah 80 unit untuk memenuhi demand pada periode 2-5.

33. Part Period Balancing Kombinasi Lot size Part Period ( inventory) periode 6 40 0 6,7 70 30 6,7,8 70 30 6,7,8,9 100 120 Part period yg paling dekat dg EPP adalah 120, maka pesanan awal pada periode 6 sejumlah 100 unit untuk memenuhi demand pada periode 6-9. Kombinasi Lot size Part Period ( inventory) periode 10 55 0

34. Part Period Balancing (PPB) Tabel MRP Biaya order = 3*$200 = $600 Biaya Hold = (50+10+10+…..+30)*$2 = $380 Total biaya = $600+$380 = $980

35. Least Unit Cost (LUC) • Metode Lot Sizing heuristik LUC menetapkan lot size yang memperhitungkan sejumlah periode demand sedemikain sehingga total biaya per unit minimum

36. Least Unit Cost Kombinasi Lot size Kumulatif Total Cost periode kumulatif Cost per unit • 30 200 200/30=6.67 • 2,3 30+40=70 200+40*2=280 280/70=4 • 2,3,4 30+40+0=70 280 280/70=4 • 2,3,4,5 30+40+0+10=80 200+70*2=340 340/80=4.25 • Replenishment pada awal periode 2, lot size = 70 untuk memenuhi kebutuhan periode 2-4. • 5 10 200 200/10=20 • 5,6 50 280 280/50=5.6 • 5,6,7 80 400 5 • 5.6.7.8 80 400 5 • 5,6,7,8,9 110 600 5.82 • Replenishment pada awal periode 5, lot size = 80 untuk memenuhi kebutuhan periode 5-8

37. Least Unit Cost Tabel MRP Biaya order = 3*$200 = $600 Biaya Hold = (40+70+30+55)*$2 = $390 Total biaya = $600+$390 = $990

38. Minimum Cost per PeriodeSilver Meal • Teknik ini mencoba mengkombinasikan beberapa periode perencanaan (trial error) untuk mendapatkan rata-rata total biaya yang minimum • Rata-rata biaya adalah jumlah order cost dan holding cost dari n periode dibagi dengan n.

39. Silver Meal Kombinasi Lot size Kumulatif Rata-rata periode kumulatif Cost Total cost • 30 200 200/1=200 • 2,3 30+40=70 200+40*2=280 280/2=140 • 2,3,4 30+40+0=70 280 280/3=93.33 • 2,3,4,5 30+40+0+10=80 200+70*2=340 340/4=85 • 2,3,4,5,6 120 660 660/5=132 • Replenishment pada awal periode 2, lot size = 80 untuk memenuhi kebutuhan periode 2-5. • 6 40 200 200/1=200 • 6,7 70 200+30*2=260 260/2=130 • 6,7,8 70 260 260/3=86.67 • 6.7.8,9 100 440 440/4=110 • Replenishment pada awal periode 6, lot size = 70 untuk memenuhi kebutuhan periode 6-9 • 30 200 200/1=200 • 10 85 310 310/2=155

40. Silver Meal Tabel MRP Biaya order = 3*$200 = $600 Biaya Hold = (50+10+10+30+55)*$2 = $310 Total biaya = $600+$310 = $910

41. Tabel Perbandingan Dipilih yang terkecil

42. Diketahui biaya pesan untuk item A adalah $20 /order, biaya simpan $2/unit/periode. Tentukan jumlah order yang harus dipesan dan biaya minimal yang harus dikeluarkan, dan tentukan dengan metode apa yang sebaiknya digunakan bila diketahui lead time 2 minggu, dan persediaan minggu lalu masih tersisa 35 unit.

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