1. Introduction to Materials Management
Chapter 4 – Material Requirements Planning

2. Links To Other MPC Functions

3. 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

4. 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

5. 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

6. Sample Bill of Materials

7. 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

8. Sample Product Tree for the BOM

9. Breaking Down the BOM into More Detail – the Multilevel Bill

10. Indented BOM – Use Indentation to Show Parent-Component Relationships

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

12. Sample Planning BOM

13. 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

14. 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

15. 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.

16. 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.

17. 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

18. 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.

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

20. 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)

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

22. 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

23. 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.

24. 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

25. 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)

26. 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

27. Material Requirement Planning MRP

28. 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.

29. 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.

30. 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.

31. 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

32. 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.

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

34. 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

35. 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

36. 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

37. 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

38. 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

39. 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

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

41. Sample Multiproduct MRP Explosion

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

43. 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

44. Solution: Week Gross Requirement Schedule Receipts Project On HandPD 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

45. 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

46. Solution: Week Gross Requirement Schedule Receipts Project On HandPD 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

47. 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

48. 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.

49. The new record reflects those events from week 1:

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

51. 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
Demand

52. 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

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

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

55. Zce
c e= O

56. 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.

57. f0 = 0
f1 = Min (Z11 + f0) = ( )
= 100 for Z11 + f0
f2 = Min (Z12 + f0, Z22 + f1)
= Min ( , )
= 100 for Z12 + f0
f3 = Min (Z13 + f0, Z23 + f1, Z33 + f2)
= Min ( , , )
= 166 for Z13 + f0
f4 = Min (Z14 + f0, Z24 + f1, Z34 + f2, Z44 + f3)
= Min ( , , , )
= 228 for Z34 + f2
f5 = Min (Z15 + f0, Z25 + f1, Z35 + f2, Z45 + f3, Z55 + f4 )
= Min ( , , , , )
= 228 for Z35 + f2
f6 = Min (Z16 + f0, Z26 + f1, Z36 + f2, Z46 + f3, Z56 + f4, Z66 + f5)
= Min ( , , , , , )
= 258 for Z36 + f2

58. fe
c e= fe

59. 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 = 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 units.

60. Result
Period
Demand
Order

61. 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
Demand