Download - Materials Requirements Planning

Materials Requirements Materials Requirements PlanningPlanning

MRPMRP

• 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.

MRPMRP

• 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.

Bill of Materials (BOM) FileBill of Materials (BOM) FileA Complete Product DescriptionA Complete Product Description

• Materials

• Parts

• Components

• Production sequence

Part Description Quantity Number required

203 Wooden Leg 4411 Wooden End 2622 Wooden Sides 223 Table Top 1

722 Hardware Kit 1

Description : TablePart Number : 100

Simplified bill of material

Deskripsi BOM

Table100

Base 200

Top023

Legs (4)203

Boards (3)030

Leg Bolts (4)220

Frame (1)300

Sides (2)622

Ends (2)411

Leg Supports (4)

533

Glue 066

Glue 066

Multilevel Bill

Table100

Base 200

Top023

Legs (4)203

Boards (3)030Leg Bolts (4)

220Frame (1)

300

Sides (2)622

Ends (2)411

Leg Supports (4)

533

Glue 066

Glue 066

Single-level Bill

Base 200 Top

023

Frame (1)300

XY

L M N(2) O

P G

J K(2)

B(2) C D

F G(2) M(2) N

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?

Example Problem

Item Number

Week

1 2 3 4 5

C 50 125 25 50 150

A 50 25 25 50 50

B 100 100

A

DC

BC

Pegged requirements

Pegged requirements

Source of requirements

Pegging identify each parent item that created demand

Example of MRP Logic and Product Example of MRP Logic and Product Structure TreeStructure Tree

B(4)

E(1)D(2)

C(2)

F(2)D(3)

A

Product Structure Tree for Assembly A Lead TimesA 1 dayB 2 daysC 1 dayD 3 daysE 4 daysF 1 day

Total Unit DemandDay 10 50 ADay 8 20 B (Spares)Day 6 15 D (Spares)

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

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

LT = 1 day

Day: 1 2 3 4 5 6 7 8 9 10A Required 50

Order Placement 50

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.

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.

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.

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.

Day: 1 2 3 4 5 6 7 8 9 10A Required 50

Order Placement 50B Required 20 200

Order Placement 20 200

B(4)

E(1)D(2)

C(2)

F(2)D(3)

A

SparesLT = 2

4x50=200

Day: 1 2 3 4 5 6 7 8 9 10A Required 50

LT=1 Order Placement 50B Required 20 200

LT=2 Order Placement 20 200C Required 100

LT=1 Order Placement 100D Required 55 400 300

LT=3 Order Placement 55 400 300E Required 20 200

LT=4 Order Placement 20 200F Required 200

LT=1 Order Placement 200

B(4)

E(1)D(2)

C(2)

F(2)D(3)

A

40 + 15 spares

Part D: Day 6

Finally, repeating the process for all components, we have the final materials requirements plan:

Finally, repeating the process for all components, we have the final materials requirements plan:

©The McGraw-Hill Companies, Inc., 2001

13

Additional MRP Scheduling Additional MRP Scheduling TerminologyTerminology

• Gross Requirements

• Scheduled receipts

• Projected available balance

• Net requirements

• Planned order receipt

• Planned order release

MRP ExampleMRP Example

A(2) B(1)

D(5)C(2)

X

C(3)

Item On-Hand Lead Time (Weeks)X 50 2A 75 3B 25 1C 10 2D 20 2

Requirements include 95 units (80 firm orders and 15 forecast) of X in week 10

Requirements include 95 units (80 firm orders and 15 forecast) of X in week 10

A(2)

X

Day: 1 2 3 4 5 6 7 8 9 10X Gross requirements 95

LT=2 Scheduled receipts Proj. avail. balance 50 50 50 50 50 50 50 50 50 50

On- Net requirements 45hand Planned order receipt 4550 Planner order release 45A Gross requirements 90

LT=3 Scheduled receipts Proj. avail. balance 75 75 75 75 75 75 75 75

On- Net requirements 15 hand Planned order receipt 15 75 Planner order release 15 B Gross requirements 45


On- Net requirements 20 hand Planned order receipt 20 25 Planner order release 20 C Gross requirements 45 40

LT=2 Scheduled receipts Proj. avail. balance 10 10 10 10 10

On- Net requirements 35 40 hand Planned order receipt 35 40 10 Planner order release 35 40 D Gross requirements 100

LT=2 Scheduled receipts Proj. avail. balance 20 20 20 20 20 20 20

On- Net requirements 80 hand Planned order receipt 80 20 Planner order release 80

It takes 2 A’s for each X

It takes 2 A’s for each X












B(1)A(2)

X

It takes 1 B for each X

It takes 1 B for each X

A(2) B(1)

X

C(3)












It takes 3 C’s for each A

It takes 3 C’s for each A

A(2) B(1)

C(2)

X

C(3)












It takes 2 C’s for each B

It takes 2 C’s for each B

A(2) B(1)

D(5)C(2)

X

C(3)












It takes 5 D’s for each B

It takes 5 D’s for each B

Schedule ReceiptSchedule Receipt

• Kerjakan kembali bila pada perioda 6 ada schedule receipt material D sebanyak 50 unit.

• Kerjakan kembali bila lot size C adalah 100.

A

DC

BC

Low Level Code

Part Low Level

A 0 B 1 C 1 2 D 2

A

B(2)

D(1) E(1)

C(1)

D(2) F(1)

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 availableIn week 5,. There are 20 Bs available.

A

B(2)

D(1) E(1)

C(1)

D(2) F(1)

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 availablein week 5, there is a schedule receipt of 100 Bs in week 1.There are 200 Fs available. All order quantities are lot-for-lot.

H

I(2)

J(2) K(3)

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.

ProdukKompone

n Jumla

hLead Time

size

AC 2

1 L4LD 3

BE 1

1 L4LC 1

CE 1

1 L4LF 2

DE 2

1 L4LC 1

E - - 1 100F - - 1 500

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 BTersedia 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.

Lot Sizing in MRP ProgramsLot 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!

Week 0 1 2 3 4 5 6 7 8 9 10

GR 35 30 40 0 10 40 30 0 30 55

OHI 35 0 0 0 0 0 0 0 0 0 0

NR 30 40 10 40 30 30 55

POR 30 40 10 40 30 30 55

PORel 30 40 10 40 30 30 55

Lead Time = 1 mingguHolding cost = $2/unit/mingguSetups cost =$200; Lot size = 1

Lot for Lot

Biaya setup = 7*$200 = $1400Biaya Hold = 0 ( tidak ada inv)Total biaya = $1400

Week 0 1 2 3 4 5 6 7 8 9 10

GR 35 30 40 0 10 40 30 0 30 55

OHI 35 0 44 4 4 68 28 72 72 42 61

NR 30 6 2 13

POR 74 74 74 74

PORel 74 74 74 74

D=(35+30+ .. + 55) = 270Q= √((2*200*270)/20) = 74

EOQ

Biaya order = 4*$200 = $800Biaya Hold = (44+4+4+…..+61)*$2 = $790Total biaya = $800+$790 = $1590

Week 0 1 2 3 4 5 6 7 8 9 10

GR 35 30 40 0 10 40 30 0 30 55

OHI 35 0 50 10 10 0 60 30 30 0 0

NR 30 40 55

POR 80 100 55

PORel 80 100 55

EOI =EOQ/d = 74/27 = 3 periode; pemesanan setiap 3 periode

Periodic Order Quantity (POQ)


Part Period Balancing (PPB)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

Part Period Balancing

EPP = 200/2 = 100

Mencari pesanan yang mendekati EPP, dg dasar pemikiran Menyimpan 100 unit selama 1 periode sama dengan menyimpan50 unit selama 2 periode ( biaya penyimpanan sama )

Kombinasi Lot size Part Period ( inventory)periode

2 30 02,3 30+40=70 40*1p=402,3,4 30+40+0=70 402,3,4,5 30+40+0+10=80 40*1p+10*3p=702,3,4,5,6 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.

Part Period Balancing


6 40 06,7 70 306,7,8 70 306,7,8,9 100 120

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.


10 55 0

Week 0 1 2 3 4 5 6 7 8 9 10

GR 35 30 40 0 10 40 30 0 30 55

OHI 35 0 50 10 10 0 60 30 30 0 0

NR 30 40 55

POR 80 100 55

PORel 80 100 55

Tabel MRP

Part Period Balancing (PPB)


Least Unit Cost (LUC)Least Unit Cost (LUC)

• Metode Lot Sizing heuristik LUC menetapkan lot size yang memperhitungkan sejumlah periode demand sedemikain sehingga total biaya per unit minimum

Least Unit Cost

Kombinasi Lot size Kumulatif Total Cost periode kumulatif Cost per unit

2 30 200 200/30=6.672,3 30+40=70 200+40*2=280 280/70=42,3,4 30+40+0=70 280 280/70=42,3,4,5 30+40+0+10=80 200+70*2=340 340/80=4.25Replenishment pada awal periode 2, lot size = 70 untuk

memenuhi kebutuhan periode 2-4.5 10 200 200/10=205,6 50 280 280/50=5.65,6,7 80 400 55.6.7.8 80 400 55,6,7,8,9 110 600 5.82Replenishment pada awal periode 5, lot size = 80 untuk

memenuhi kebutuhan periode 5-8

Week 0 1 2 3 4 5 6 7 8 9 10

GR 35 30 40 0 10 40 30 0 30 55

OHI 35 0 40 70 30 0 0 55 0

NR 30 10 40 30

POR 70 80 85

PORel 70 80 85

Tabel MRP

Least Unit Cost

Biaya order = 3*$200 = $600Biaya Hold = (40+70+30+55)*$2 = $390Total biaya = $600+$390 = $990

Minimum Cost per PeriodeMinimum Cost per PeriodeSilver MealSilver 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.

Silver Meal

Kombinasi Lot size Kumulatif Rata-rata periode kumulatif Cost Total cost

2 30 200 200/1=2002,3 30+40=70 200+40*2=280 280/2=1402,3,4 30+40+0=70 280 280/3=93.332,3,4,5 30+40+0+10=80 200+70*2=340 340/4=852,3,4,5,6 120 660 660/5=132 Replenishment pada awal periode 2, lot size = 80 untuk

memenuhi kebutuhan periode 2-5.6 40 200 200/1=2006,7 70 200+30*2=260 260/2=130

6,7,8 70 260 260/3=86.676.7.8,9 100 440 440/4=110Replenishment pada awal periode 6, lot size = 70 untuk

memenuhi kebutuhan periode 6-99 30 200 200/1=20010 85 310 310/2=155

Week 0 1 2 3 4 5 6 7 8 9 10

GR 35 30 40 0 10 40 30 0 30 55

OHI 35 0 50 70 30 0 0 55 0

NR 30 10 40 30

POR 80 70 85

PORel 80 70 85

Tabel MRP

Silver Meal

Biaya order = 3*$200 = $600Biaya Hold = (50+10+10+30+55)*$2 = $310Total biaya = $600+$310 = $910

Metode Total Biaya

Lot for Lot 1400

EOQ 1500

POQ 980

PPB 980

LUC 990

Silver Meal 910*

Tabel Perbandingan

Dipilih yang terkecil

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 metodeapa yang sebaiknya digunakan bila diketahui lead time 2 minggu, dan persediaan minggu lalu masih tersisa 35 unit.

Periode 1 2 3 4 5 6 7 8 Tot

Permintaan 5 15 10 15 10 11 13 10 89

Download - Materials Requirements Planning

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