Introduction to Materials Management Chapter 4 – Material Requirements Planning

Links To Other MPC Functions

Independent versus Dependent Demand Independent Demand Not related to demand for other assemblies or products, instead from outside sources Generally forecasted demand Dependent Demand Generally related to production of an end product (as defined on the MPS) Can be calculated instead of forecasted

Dependent Demand Approach – Materials Requirement Planning (MRP) Major Objectives of MRP Determine Requirements – Calculated to meet product requirements defined in the MPS What to order How much to order When to order When to schedule delivery Keep Priorities Current

Major Inputs to MRP Master Production Schedule quantities and times Inventory records of all items to be planned Planning factors such as lead times, order quantities, and safety stock Current status of each item Bills of material for MPS items

Sample Bill of Materials

Bill of Material Points The BOM shows all parts to make one of the item Each part has one, and only one, part number A part is defined by form, fit, and function – any change requires a new part number

Sample Product Tree for the BOM

Breaking Down the BOM into More Detail – the Multilevel Bill

Indented BOM – Use Indentation to Show Parent-Component Relationships

Planning Bills of Material Artificial grouping of components for Planning Purposes Used to simplify Forecasting Planning Master Scheduling Represent an average, not buildable product

Sample Planning BOM

Where-Used and Pegging Where-used reports - shows the parents for a component (contrast with a bill of materials that show the components for a parent) Pegging report – shows the parents for a component, but only those parents where there is an existing requirement

Major Uses for Bills of Material Defines the product Provides method for design change control Planning – What is needed and when Order entry – order configuration and pricing Production – Parts needed to assemble a product Costing – material cost of goods sold

Packing box and installation kit of wire, bolts, and screws BOM Example Product structure for “Awesome” (A) A Level B(2) Std. 12” Speaker kit C(3) Std. 12” Speaker kit w/ amp-booster 1 E(2) F(2) Packing box and installation kit of wire, bolts, and screws Std. 12” Speaker booster assembly 2 D(2) 12” Speaker G(1) Amp-booster 3 Determine the number of units of each item required to satisfy demand for a new order of 50 Awesome speaker kits.

If there are 100 Fs in stock, how many Ds do you need? Part B: 2 x number of As = (2)(50) = 100 Part C: 3 x number of As = (3)(50) = 150 Part D: 2 x number of Bs + 2 x number of Fs = (2)(100) + (2)(300) = 800 Part E: 2 x number of Bs + 2 x number of Cs = (2)(100) + (2)(150) = 500 Part F: 2 x number of Cs = (2)(150) = 300 Part G: 1 x number of Fs = (1)(300) = 300 If there are 100 Fs in stock, how many Ds do you need? Part D: 2 x number of Bs + 2 x number of Fs =2(100) + 2(300 – 100) =600.

Bills of Material Low-Level Coding Item is coded at the lowest level at which it occurs BOMs are processed one level at a time Item Low level code D 3 G E 2 F B 1 C A

Try This! A company assembles desks using bought-in parts of a top and four legs. These have lead times of one and two weeks respectively, and assembly takes a week. Design the Bill Of Material for this parts.

Desk LT =1 Leg (4) LT =2 Top (1) Level 0 Level 1

Figure below shows the BOMs for two end items, A and B Determine low level coding for the BOM. A B C(2) D(1) D(2) E(2) C(1)

A B C(2) D(1) D(2) E(2) C(1) Item Low level code A 1 E 2 C 3 D

Parent Code Part Number Level Code Description Quantity Req 377 #377 Lawn Sprinkler M 1 Water motor assembly F Frame assembly H #699 hose recept. Assembly A 2 ½” diameter. 1/32” aluminium tube 10 B ½” X 1/16” metal screws 3 C Water motor 40 D ½” X ½” #115 plastic cup ½” X 1/16” metal screws

1. Transform the above BOM for Lawn Sprinkler to :- i) Indented BOM format. ii) Multilevel Bill format. 2. Classify low level coding Multilevel Bill format for Lawn Sprinkler. 3. Determine the number of units of each item required to satisfy demand for a new order of 100 units Lawn Sprinkler.

1. i) Indented BOM Format Part Number Descriptions Quantity Requirement 377 #377 Lawn Sprinkler M Water motor assembly 1 A ½” diameter. 1/32” aluminium tube 10 B ½” X 1/16” metal screws 3 C Water motor F Frame assembly 40 D ½” X ½” #115 plastic cup ½” X 1/16” metal screws H #699 hose recept. Assembly

1 2 2. ii) Multilevel BOM Format Low Level Code 377 M(1) A(10) F(1) 1 2 2. ii) Multilevel BOM Format 377 M(1) A(10) F(1) C(1) B(3) A(40) D(3) H (1)

3. Part M: 1 x number of 377s = (1)(100) = 100 Part F: 1 x number of 377s = (1)(100) = 100 Part H: 1 x number of 377s = (1)(100) = 100 Part A: 10 x number of Ms + 40 x number of Fs = (10)(100) + (40)(100) =5,000 Part B: 3 x number of Ms + 3 x number of Fs = (3)(100) + (3)(100) = 600 Part C: 1 x number of Ms = (1)(100) = 100 Part D: 3 x number of Fs = (3)(100) = 300

Material Requirement Planning MRP

MRP Planning Sheet Lot Size Technique 1. Lot For Lot (LTL) = The order for each period is that exact quantity in period. How much to order. 2. Wagner and Whitin = a dynamic programming procedure to determine the optimum varying order size. = Method that yields optimal results. 3. Silver and Meal = a heuristic algorithm for order size determination. = Provide good cost performance and are very efficient to use.

Lead time The time required to purchase, produce, or assemble an item For purchased items – the time between the recognition of a need and the availability of the item for production For production – the sum of the order, wait, move, setup, store, and run times span of time for a process On hand Number of stocks in inventory.

Safety Stock Allocated item A back-up supply of products which are held for use in an emergency. Allocated item Number of units in inventory that have been assigned to specific future production but not yet used or issued from the stockroom. Allocated item increase requirement and may be included in an MRP planning sheet.

Low Level Code Item Identification Gross Requirement Item is coded at the lowest level at which it occurs Item Identification ID for item Gross Requirement Number of units to be produced X materials required for each unit. Scheduled Receipts Stock on order

Planned order Receipts Project On Hand Current stock Net requirement = (Gross Req + Allocations) – (Project on Hand + Schedule Receipts) = Gross Req – [Project on hand - Safety Stock – Inventory allocated to other uses] Planned order Receipts A plan of the quantity of each material to be ordered in each time period.

Planned order Releases A plan of the quantity of each material to be received in each time period.

A company assembles desks using bought-in parts of a top and four legs A company assembles desks using bought-in parts of a top and four legs. These have lead times of one and two weeks respectively, and assembly takes a week. The company receives orders for 10 tables to be delivered in a week 5 of a production period and 20 tables in week 7. It has stock of only 2 complete desks, 11 tops and 20 legs. Lot size = 1. When should it order parts? Desk LT =1 Leg (4) LT =2 Top (1) Level 0 Level 1

Level 0 - Desks Week PD 1 2 3 4 5 6 7 10 20 8 Gross Requirements Project on Hand Net Requirement 8 Sch Recp Start Assembly

Level 1-Tops 8 20 11 17 Week PD 1 2 3 4 5 6 7 Gross Requirements Project on Hand 11 Net 17 Schedule Receipts Place Order

Level 1 - Legs Week PD 1 2 3 4 5 6 7 32 80 20 12 Gross Req Project On Hand 20 Net Req 12 Schedule Receipts Place Order

Result. Weeks 2 : order 12 legs Week 4 : order 80 legs and assemble 8 desks Week 5 : order 17 tops Week 6 : assemble 20 desk

Redo Example above in the lot size equal to 100 unit. Level 0 - Desks Week PD 1 2 3 4 5 6 7 Gross Requirements 10 20 Project on Hand 92 28 Net Requirement 8 72 Sch Recp 100 Start Assembly Level 1-Tops Week PD 1 2 3 4 5 6 7 Gross Requirements 100 Project on Hand 11 89 Net Sch Recp Place Order Level 1 - Legs Week PD 1 2 3 4 5 6 7 Gross Req 200 Project On Hand 20 180 80 Net Req Schedule Receipts 100 Place Order

Result. Week 4 : assemble 100 desks, order 100 legs Week 5 : order 100 tops Week 6 : assemble 100 desk

Sample Multiproduct MRP Explosion

Example Problem Lead time for this component is 2 weeks and order quantity is 200. Complete the table. What action should be taken?

Example For the MRP table below, indicate the project on hand, the planned order receipts and the planned order releases. The lead time is two periods, and the lot size is the same as the net requirement (lot size = 1) Week PD 1 2 3 4 5 6 7 8 Gross Requirement 10 18 14 Schedule Receipts 20 Project On Hand Net Requirement Planned order Receipts Planned order Releases

Solution: Week Gross Requirement Schedule Receipts Project On Hand PD 1 2 3 4 5 6 7 8 Gross Requirement 10 18 14 Schedule Receipts 20 Project On Hand 15 25 Net Requirement Planned order Receipts Planned order Releases

Example For the MRP table below, indicate the project on hand, the planned order receipts and the planned order releases. The lead time is two periods, and the lot size is the same as the net requirement (lot size = 1) Redo Example above in the lot size equal to 15 units. Week PD 1 2 3 4 5 6 7 8 Gross Requirement 10 18 14 Schedule Receipts 20 Project On Hand Net Requirement Planned order Receipts Planned order Releases

Solution: Week Gross Requirement Schedule Receipts Project On Hand PD 1 2 3 4 5 6 7 8 Gross Requirement 10 18 14 Schedule Receipts 20 Project On Hand 15 25 12 Net Requirement Planned order Receipts Planned order Releases

Moving through time – an example This record shows the status of the part Monday morning. The computer is showing the need to release the order of 30

During the week, the following events happen: Only 25 units of the scheduled receipt move into inventory. The balanced is scrapped The gross requirement for week 3 is changed to 10. The gross requirement for week 4 is increased to 50. The gross requirement for week 7 is 15. An inventory count reveals there are 10 more in inventory than the record shows The gross requirement for week 1 is issued from inventory The planned order release of 30 in week 1 is released and becomes a scheduled receipt in week 3.

The new record reflects those events from week 1:

Wagner-Whitin Algorithm Is an exact solution technique for known demand. Developed a dynamic programming procedure to determine the optimum varying order size.

Wagner-Whitin Algorithm An item has a unit purchase cost of $50, an ordering cost per order of $100, and a holding cost fraction per period of $0.02. Determine the order quantities by the Wagner-Whitin algorithm from the demand given below. Assume the initial inventory is zero at the beginning of period 1. Period 1 2 3 4 5 6 Demand 75 0 33 28 0 10

Zce = C + hP Σ (Qce – Qci) for 1≤ c ≤ e ≤ N Where Step 1: Calculate the total variable cost matrix for all possible ordering alternatives. Zce = C + hP Σ (Qce – Qci) for 1≤ c ≤ e ≤ N Where C = ordering cost per order, h = holding cost per period, P = unit purchase cost, Qce = Σ Rk, Rk = demand rate in period k e i = c e k = c

Z11 = 100 + 1 [(75 – 75)] = 100 Z12 = 100 + 1 [(75 – 75) + (75 – 75)] = 100 Z13 = 100 +1 [(108-75) + (108 - 75) + (108 – 108)] = 166 Z14 = 100 +1 [(136-75) + (136-75) + (136 – 108) + (136 - 136)] = 250 Z15 = 100 +1 [(136-75) + (136-75) + (136 – 108) + (136 - 136) + (136 – 136)] = 250 Z16 = 100 +1 [(146 - 75) + (146 - 75) + (146 – 108) + (146 - 136) + (146 – 136) + (146 – 146] = 300 Z22 = 100 + 1 [(0 – 0)] = 100 Z23 = 100 + 1 [(33– 0) + (33 – 33)] = 133 Z24 = 100 +1 [(61- 0) + (61- 33) + (61 – 61)] = 189 Z25 = 100 +1 [(61- 0) + (61- 33) + (61 – 61) + (61 – 61)] = 189 Z26 = 100 +1 [(71 - 0) + (71-33) + (71 – 61) + (71 – 61) + (71 – 71)] = 229

Z33 = 100 + 1 [(33 – 33)] = 100 Z34 = 100 + 1 [(61 – 33) + (61 – 61)] = 128 Z35 = 100 +1 [(61 – 33) + (61 – 61) + (61 – 61)] = 128 Z36 = 100 +1 [(71 - 33) + (71 – 61) + (71 – 61) + (71 – 71)] = 158 Z44 = 100 + 1 [(28 – 28)] = 100 Z45 = 100 + 1 [(28 – 28) + (28 – 28)] = 100 Z46 = 100 +1 [(38 - 28) + (38 - 28) + (38 – 38)] = 120 Z55 = 100 + 1 [(0 – 0)] = 100 Z56 = 100 +1 [(10 - 0) + (10 - 10)] = 110 Z66 = 100 + 1 [(10 – 10)] = 100

Zce c e=1 2 3 4 5 6 1 100 100 166 250 250 300 2 100 133 189 189 229 3 100 128 128 158 4 100 100 120 5 100 11O 6 100

Step 2: Define fe to be the minimum possible cost in period 1 through e, given that the inventory level at the end of period is zero. fe = Min (Zce + fc-1) Lowest cost combination is recorded as the fe strategy to satisfy requirements for periods 1 through e.

f0 = 0 f1 = Min (Z11 + f0) = (100 + 0) = 100 for Z11 + f0 f2 = Min (Z12 + f0, Z22 + f1) = Min (100 + 0, 100 + 100) = 100 for Z12 + f0 f3 = Min (Z13 + f0, Z23 + f1, Z33 + f2) = Min (166 + 0, 133 + 100, 100 + 100 ) = 166 for Z13 + f0 f4 = Min (Z14 + f0, Z24 + f1, Z34 + f2, Z44 + f3) = Min (250 + 0, 189 + 100, 128 + 100, 100 + 166 ) = 228 for Z34 + f2 f5 = Min (Z15 + f0, Z25 + f1, Z35 + f2, Z45 + f3, Z55 + f4 ) = Min (250 + 0, 189 + 100, 128 + 100, 100 + 166, 100 + 228 ) = 228 for Z35 + f2 f6 = Min (Z16 + f0, Z26 + f1, Z36 + f2, Z46 + f3, Z56 + f4, Z66 + f5) = Min (300 + 0, 229 + 100, 158 + 100, 120 + 166, 110 + 228, 100 + 228 ) = 258 for Z36 + f2

fe c e=1 2 3 4 5 6 1 100 100 166 250 250 300 2 200 233 289 289 329 3 200 228 228 258 4 266 266 286 5 328 338 6 328 fe 100 100 166 228 228 258

Step 3: fN = ZwN + fw-1 The final order occurs at period w and is sufficient to satisfy demand in periods w through N. In the example f6 = fN is the combination of Z36 and f2, so the last order will be placed in period 3 and will satisfy the requirement from period 3 through 6, or 33 + 28 + 0 + 10 = 71 units; f2 is the combination of Z12 and f0, so the order will be placed in period 1 and will satisfy requirement from period1 through 2 or 75 + 0 = 75 units.

Result Period 1 2 3 4 5 6 Demand 75 0 33 28 0 10 Order 75 0 71 0 0 0

Try This! An item has a unit purchase cost of $50, an ordering cost per order of $100, and a holding cost fraction per period of $0.02. Determine the order quantities by the Wagner-Whitin algorithm from the demand given below. Assume the initial inventory is zero at the beginning of period 1. Period 1 2 3 4 5 6 7 8 9 10 Demand 10 3 30 100 7 0 80 9 0 90