Material Requirements Planning
or as we in the business like to call it - MRP
Computer Based Production Planning
and Inventory Control System
Scientific Management meets the Computer
1
Where are we?
Strategic
Tactical
Operational
Purpose
Plan acquisition of
resources
Plan utilization of
resources
Execution of resources
Time horizon
2+ years
6 to 24 months
1-3 months
Time period
Yrs/Months
Months
Days/weeks
Level
Top management
Middle management
Plant management
Questions addressed
What products/Levels
Plant sizes/capacities
Plant/warehouse
locations
What technologies?
Inventory levels
Production rates
Work force sizing
Subcontracting
Batch sizes
Job scheduling
Material control
Machine maintenance
Analysis techniques
Break-even analysis
LP product mix
Distribution models
Supply Chain models
Long range forecasting
Location analysis
Forecasting
Aggregate planning
Production smoothing
Inventory models
Facility layout
Make or buy decisions
Project planning
Job scheduling
Task sequencing
Assembly line balancing
Shift scheduling
Worker assignments
MRP/JIT
Group technology
transfer lines
2
Three Approaches to Production Control
or how much to make, when?
• Traditional
– inventory control and job scheduling
• Material Requirements Planning (MRP)
– along with Manufacturing Resources Planning
(MRP-II)
– capacity requirements planning (CRP),
– and Enterprise Resource Planning (ERP)
• Just-in-Time (JIT) Manufacturing
3
Basic Definitions
• MRP (Materials Requirements Planning). MRP is the
basic process of translating a production schedule for an
end product (MPS or Master Production Schedule) to a set
of requirements for all of the subassemblies and parts
needed to make that item.
• JIT Just-in-Time. Derived from the original Japanese
Kanban system developed at Toyota. JIT seeks to deliver
the right amount of product at the right time. The goal is to
reduce WIP (work-in-process) inventories to an absolute
minimum.
4
Why Push and Pull?
• MRP is the classic push system. The MRP system
computes production schedules for all levels based on
forecasts of sales of end items. Once produced,
subassemblies are pushed to next level whether needed or
not.
• JIT is the classic pull system. The basic mechanism is that
production at one level only happens when initiated by a
request at the higher level. That is, units are pulled through
the system by request.
5
Push versus Pull
•Push – MRP
•Forecast driven
•Production Plan by
period
•Parts – explosion
•Uses lot-sizing
techniques
•Pull – JIT
•Production driven by
demands (in the form of
Kanban) from the next
higher level
•Minimizes work-inprocess
•Requires reduced setup
times
6
Advantages
MRP
reacts to changes in
demands
allows for lot sizing to
reduce setup costs
plans for several time
periods into the future
JIT
reduces work-in-process
quickly identify quality
problems before large
inventories of defect parts
smooth flow of material
through the supply chain
7
What problem does MRP try to solve?
• Independent demand – originates outside the
factory system
– Is subject to uncertainty
– Traditional (stochastic) inventory models work well
• Dependent Demand – demand for components that
make up independent demand products
– No uncertainty – demand is known (at least in
principle) once the final assembly schedule is given
– Traditional (stochastic) inventory models do not work
well
8
MRP versus JIT – an Example
• Assembly of a garden spade requires two screws.
• Spades are assembled in batches of 400 on the
first two days of every month
• The weekly demand pattern for screws therefore
is:
800,0,0,0, 800,0,0,0, 800,0,0,0, 800,0,0,0, etc.
9
The MRP Approach
• Using a weekly demand rate of 200, say, holding
and setup costs are such that the EOQ solution is
Q* = 1,400.
– solution makes no sense since 600 would be stored for
3 weeks and then another order placed for 1,400
– Treating weekly demand as random makes no sense as
well since it is deterministic
• Order 10,400 at the beginning of the year and
incur only one fixed order and delivery cost per
year
• Ideally use dynamic lot sizing based on costs
– i.e. order some multiple of 800
10
The JIT Approach
• Schedule delivery of 800 screws at the beginning
of each month
–
–
–
–
Incur fixed ordering cost 12 times a year
No inventory to store
If usage varies, can modify delivery sizes
If a defective shipment only screwed for 800 screws
• To be economical, fixed order and delivery costs
must be small
11
MRP Objectives
• Insure availability of materials, component,
and products
• Maintain lowest possible inventory level
• Plan manufacturing activities, delivery
schedules, and purchasing requests
An MRP fact:
“MRP deals with two basic dimensions
in production: Quantities and Timing”
12
The MRP
System
Forecast of future demand
Production Planning
Master Production Schedule
Schedule of production quantities by
Product (end item) and time period
Lot sizing
and capacity
planning
planned-order
releases
Customer
orders
Materials requirements planning system
Explode master schedule to obtain requirements
for assemblies, components, and raw material
Time-phased
Purchase orders
Time-phased
Work orders
13
Input 1 - Master Production Schedule
• This is the forecast for the sales of the end
item over the planning horizon. The data
sources for determining the MPS include:
–
–
–
–
–
Firm customer orders
Forecasts of future demand by item
Safety stock requirements
Seasonal variations
Internal orders from other parts of the
organization.
14
Input 2 - Bill of Materials
Bill of materials
• parent-child (hierarchical) relationships
• quantity per application (QPA)
• production or procurement lead-times
15
Input 3 – Inventory Position
• For all items at all levels for each time
period
– On-hand inventory quantities
– On-order quantities (due-in’s)
Let’s see. We have 3
of these with 5 more
arriving on the 23rd.
16
Input - Output
1. Master Production Schedule
2. Bill of Materials and level codes
3. Inventory status of all items – on-hand and on-order
MRP SYSTEM
Compute for all components:
• Gross requirements
• Planned order and work releases
• Time phasing of net requirements
17
A Three-legged Stool
Stool
Final assembly area
End item
Parent level
Child level
1 seat assembly
Department X
1 seat
Dept X
1 cushion
Vendor B
Raw material – lumber
leg assembly
Department Y
3 legs
Dept Y
6 screws
Vendor A
3 supports
Dept Y
- Paul Bunyon Lumber Co.
6 screws
Vendor A
18
A Three-legged Stool
Production
Lead-times
One week
One week
Stool
Final assembly area
1 seat assembly
Department X
leg assembly
Department Y
Vendor
Lead-times
6 screws One week
Vendor A
One week
Two weeks
1 seat
Dept X
1 cushion
Vendor B
3 legs
Dept Y
3 supports
Dept Y
6 screws
Vendor A
Two weeks
Raw material – lumber
- Paul Bunyon Lumber Co.
19
A Three-legged Stool
Independent Demand
Stool
Final assembly area
Dependent Demand
seat assembly
Department X
1 seat
Dept X
1 cushion
Vendor B
Raw material – lumber
leg assembly
Department Y
3 legs
Dept Y
6 screws
Vendor A
3 supports
Dept Y
- Paul Bunyon Lumber Co.
6 screws
Vendor A
20
Lumpy Demands due to lot sizing
time buckets
dependent
demand
Time period
Assembly X
demand
Production
Assembly Y
Production
Assembly Z
Production
Subassembly A
Requirements
1
2
3
4
5
6
10
30
5
10
7
7
10
10
10
10
5
7
7
5
10
7
7
5
7
7
5
10
7
7
10
30
5
57
7
27
37
27
7
7
7
Assemblies X and Z require one unit of assembly A while
Assembly Y requires two.
SubAsbAper 1 = 30 + 2 (10) + 7 = 57
21
Terminology
• bill of materials – all parent-child
relationships
• level coding – the order in which the
requirements must be computed (indenture
level)
• lead-time offset – due date minus the
planned order release date
22
Product Structure
Two end products (1 and 2)
Four assemblies (A, B, C, and D)
Three parts (d, f, and g)
2
1
A(2)
D
C
A
B
B
d(2)
g
C(2)
D
C
d
g(3)
B
f(3)
B(2)
C
( ) – indicates number of components used on next higher assembly
23
Bill of Materials Matrix - B
End
Product
1 2
1
2
A
D
B
C
d
f
g
A
2
subassemblies
D
B
C
1
1
1
1
2
parts
(components)
g
d
f
3
2
1
2
1
1
3
Why this is an upper
triangular matrix isn’t it!
24
use topological sorting
MRP Level Assignments
1
A(2)
2
C
D
A
B
B
g(3)
Level 1
A, D
B
d(2)
g
C
d
Level 0
1, 2
C(2)
Level 3
C, g
D
f(3)
B(2)
Level 2
B
C
Level 4
d, f
25
Topological Sorting
1
rearrange so all
arrows go down
1
4
3
3
7
7
4
6
5
8
9
2
2
5
9
8
note: no loops are present
6
26
Computing Direct Dependent Demands
Let Dn = vector of dependent demand directly resulting
from demand at level n (output vector)
dn = vector of demand at level n (input vector)
Then Dn = dn x B where B is the Bill of Materials Matrix
and D0 = d0 B;
D1 = d1B = D0 B = (d0 x B x B);
D2 = d2 B = (d0 x B x B x B); etc.
Then total demand = D0 + D1 + … + Dm
= d0 x (B + B2 + B3 + … + Bm )
27
Assume independent demand for item 1 = 100 units & item 2 = 200 units
and there is a spare part demand for 20 C’s.
1
D0 = d0 B
= (100, 200, 0, 0, 0, 20, 0, 0, 0) x
1
2 A DB C g d f
0
0

0

0
0

0
0

0
0

2 A D B C g
d f
0
0
0
0
0
0
0
0
0
0
0
2
0
0
1
0
0
0
2
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
2
0
0
0
0
0
1
0
0
1
2
0
0
0
0
0
3
0
0
1
0
0
0
0
0
0 
0

0
0

3
0

0
0 
= (0, 0, 200, 200, 200, 100, 600, 20, 60) = d1
Requirement: A
200
D
200
B
200
C
100
g
600
d
20
f
60
28
Level 2 requirement
1
D1 = d1 B
= (0, 0, 200, 200, 200, 100, 600, 20, 60) x
12 A
D
B C
g d f
0
0

0

0
0

0
0

0
0

2 A D B C g
d f
0
0
0
0
0
0
0
0
0
0
0
2
0
0
1
0
0
0
2
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
2
0
0
0
0
0
1
0
0
1
2
0
0
0
0
0
3
0
0
1
0
0
0
0
0
0 
0

0
0

3
0

0
0 
= (0, 0, 0, 0, 500, 500, 200, 100, 300) = d2
Requirement: A
0
D
0
B
500
C
500
g
200
d
500
f
300
29
Computing Total Requirements
If B is an n x n triangular matrix, then Bk = 0 for any k  n
by matrix multiplication of B x B:
n
bij2   bik bkj  bi1b1 j  bi 2b2 j  ...  bi ,i 1bi 1, j
k 1
If i  j then b2i,j = 0
30
B2 =
=
0
0

0

0
0

0
0

0
0

0
0
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
2
0
0
0
0
0
1
0
0
1
2
0
0
0
0
0
3
0
0
1
0
0
0
0
0
0
2
0
0
1
0
0
0
0
0 
0

0
0

3
0

0
0 
0
0

0

0
0

0
0

0
0

0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
2
0
0
0
0
0
0
0
0
3
2
4
0
0
0
0
0
0
1
1
2
0
0
0
0
0
5
0
0
1
2
0
0
0
0
3
0 
0

3
6

0
0

0
0 
0
0

0

0
0

0
0

0
0

0
0
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
2
0
0
0
0
0
1
0
0
1
2
0
0
0
0
0
3
0
0
1
0
0
0
0
0
0
2
0
0
1
0
0
0
0
0 
0

0
0

3
0

0
0 
Therefore b21,8 = 5 is the
level 1 requirement for
component d in product 1.
31
MRP Level Assignments
1
A(2)
2
C
D
A
B
B
g(3)
Level 1
A, D
B
d(2)
g
C
d
Level 0
1, 2
C(2)
Level 3
C, g
D
f(3)
B(2)
Level 2
B
C
Level 4
d, f
32
B3 =
=
0
0

0

0
0

0
0

0
0

0
0
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
2
0
0
0
0
0
1
0
0
1
2
0
0
0
0
0
3
0
0
1
0
0
0
0
0
0
2
0
0
1
0
0
0
0
0 
0

0
0

3
0

0
0 
0
0

0

0
0

0
0

0
0

0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
4
4
0
0
0
0
0
0
0
2
2
0
0
0
0
0
0
0
0 0
2 9 
2 6

4 12 
0 0

0 0
0 0

0 0
0 0 
0
0

0

0
0

0
0

0
0

0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2
2
0
0
0
0
0
0
0
0
3
2
4
0
0
0
0
0
0
1
1
2
0
0
0
0
0
5
0
0
1
2
0
0
0
0
3
0 
0

3
6

0
0

0
0 
Therefore b32,9 = 9 is the
level 2 requirement for
component f in product 2.
33
Total Requirements matrix (R)
Let ri,j = total number of units of item j required to
produce one unit of item i where ri,i = 1 by definition.
therefore:
R = I + B + B2 + B3 + … + Bn + 0 + …

recall:
1
1
x 
 (1  x)

1 x
n 0
n
Therefore R = (I – B)-1
34
B=
0
0

0

0
0

0
0

0
0

0
0
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
2
0
0
0
0
0
1
0
0
1
2
0
0
0
0
0
3
0
0
1
0
0
0
0
R = (I – B)-1 =
0
0
2
0
0
1
0
0
0
0
0 
0

0
0

3
0

0
0 
1
0

0

0
0

0
0

0
0

0
1
0
0
0
0
0
0
0
2
0
1
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
2
3
1
2
1
0
0
0
0
5
7
2
5
2
1
0
0
0
2
6
1
2
1
0
1
0
0
9 15 
7 21
4 6

5 15 
2 6

1 3
0 0

1 0
0 1 
35
let d = vector of independent demands which includes demands for
end products, spare assemblies, and spare components
Q = total production requirements vector
then
Q = d R = d (I – B)-1
item
1
demand 20
2
A
D
B
C
g
d
f
30
0
10
0
5
0
0
0
Q = d R = (20, 30, 0, 10, 0, 5, 0, 0, 0) x
1
0

0

0
0

0
0

0
0

0
1
0
0
0
0
0
0
0
2
0
1
0
0
0
0
0
0
= (20, 30, 40, 40, 150, 365, 240, 445, 1095)
0
1
0
1
0
0
0
0
0
2
3
1
2
1
0
0
0
0
5
7
2
5
2
1
0
0
0
2
6
1
2
1
0
1
0
0
9 15 
7 21
4 6

5 15 
2 6

1 3
0 0

1 0
0 1 
36
from earlier example:
d R = (100, 200, 0, 0, 0, 20, 0, 0, 0) x
1
0

0

0
0

0
0

0
0

0
1
0
0
0
0
0
0
0
2
0
1
0
0
0
0
0
0
0
1
0
1
0
0
0
0
0
2
3
1
2
1
0
0
0
0
5
7
2
5
2
1
0
0
0
2
6
1
2
1
0
1
0
0
9 15 
7 21
4 6

5 15 
2 6

1 3
0 0

1 0
0 1 
= (100, 200, 200, 200, 800, 1920, 1400, 2320, 5760)
Requirement:
A
D
B
200 200
800
C
1920
g
1400
d
2320
f
5760
37
MRP Procedure
• Netting: Determine net requirements
– gross requirements –[on-hand inventory + scheduled receipts]
• Establish lot sizes
– divide net requirements into production lots or order quantities
• Time Phasing: Determine planned order release dates
– Offset due dates with lead-times
• BOM Explosion: Expand to next level
– Generate gross requirements of all components on next level
• Iterate: Repeat until all levels have been processed.
38
Master Production Scheduling
Qj,t = scheduled receipts (order or lot size quantity)
of item j during period t
Rj,t = gross requirement of item j during period t
Ij,t = expected on-hand inventory of item j at the beginning
of period t
Nj,t = net requirements for item j in period t
Nj,t = max{0, Rj,t – Qj,t – Ij,t}
Note: Rj,t = max{ forecast for period t, customer orders period t}
39
Master Production Scheduling (cont.)
Ij,t = expected on-hand inventory of item j at the beginning
of period t
Qj,t = scheduled receipts (order or lot size quantity)
of item j during period t
Rj,t = gross requirement of item j during period t
I jt  max 0, I j ,t 1  Q j ,t 1  R j ,t 1
40
MRP - Example
CURENT INVENTORY STATUS AND LEADTIMES
level
item
on-hand
inventory
at t=0
L0
L0
L1
L1
L2
L3
L3
L4
L4
1
2
A
D
B
C
g
d
f
120
85
0
10
500
160
0
1200
4000
reorder
lead-time
(weeks)
1
1
2
2
1
1
2
1
2
41
Example - demands
Week
Item
I1
I2
A
D
B
C
g
d
f
1
2
3
4
5
6
7
8
9
50
20
20
30
30
25
40
35
40
10
30
35
25
20
15
25
15
30
30
10
10
20
100
5
Master Production Schedule
42
Example – level 0
Week 1
Item 1
R1
50
Q1
I1
120
N1
OR
2
3
4
5
6
7
8
9
20
30
40
30
25
15
30
70
50
40
120
20
100
60
30
5
10
0
30
Week 1
Item 2
R2
20
Q2
I2
85
N2
OR
2
3
4
5
6
7
8
9
30
25
10
35
20
25
30
65
35
35
100
10
75
65
30
10
15
120
120
100
OR – order releases
(based on lot-sizes of 120 and 100)
0
30
100
43
Gross requirements –lower level period 7
D0 = d0 B
= (120, 100, 0, 0, 0, 0, 0, 0, 0) x
0
0

0

0
0

0
0

0
0

0
0
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
2
0
0
0
0
0
1
0
0
1
2
0
0
0
0
0
3
0
0
1
0
0
0
0
0
0
2
0
0
1
0
0
0
0
0 
0

0
0

3
0

0
0 
= (0, 0, 240, 100, 100, 120, 300, 0, 0)
Gross
Requirement:
A
240
D
100
B
100
C
120
g
300
44
note: lot for lot
re-order policy
OR – order releases
(based on lot-sizing)
Example – level 1
Week
Item A
R
Q
I
N
OR
1
2
3
4
5
6
7
8
9
0
0
0
0
0
240
15
0
0
0
240
240
0
0
0
0
0
240
0
15
0
240
15
Week
Item D
R
Q
I
N
OR
1
2
3
4
5
6
7
8
9
0
10
0
100
0
0
10
10
10
0
0
10
100
100
0
0
0
0
100
0
0
100
10
240
100
lead-time
2 weeks
45
Gross requirements –lower levels period 5
D1 = d1 B
= (0, 0, 240, 100, 0, 0, 0, 0, 0) x
0
0

0

0
0

0
0

0
0

0
0
0
0
0
0
0
0
0
2
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
1
2
0
0
0
0
0
1
0
0
1
2
0
0
0
0
0
3
0
0
1
0
0
0
0
0
0
2
0
0
1
0
0
0
0
0 
0

0
0

3
0

0
0 
= (0, 0, 0, 0, 440, 100, 0, 480, 0)
Gross
Requirement:
B
440
C
100
d
480
46
use topological sorting
MRP Level Assignments
1
A(2)
2
C
D
A
B
B
g(3)
Level 1
A, D
B
d(2)
g
C
d
Level 0
1, 2
C(2)
Level 3
C, g
D
f(3)
B(2)
Level 2
B
C
Level 4
d, f
47
OR – order releases
(based on lot-sizing)
Example – level 2/3
Wagner-Whitin with
A=$150, h=$.60
Week
Item B
R
Q
I
N
OR
1
2
3
4
5
6
7
8
9
440
560
0
20
100
0
440
15
100
0
100
120
100
0
0
440
0
15
0
100
0
0
100
Week
Item C
R
Q
I
N
OR
1
2
- level 3
100
15
115
0
15
555
120
100
3
4
5
6
7
8
9
120
120
0
1150
100
0
120
200
0
0
1150
0
100
0
0
120
0
200
0
1250
320
least unit cost
48
OR – order releases
(based on lot-4-lot)
Week
Item g
R
Q
I
N
OR
Example – level 3/4
1
- level
0
2
3
4
5
6
7
8
9
0
575
0
0
300
100
0
0
0
300
300
0
0
575
0
0
0
300
0
100
0
300
575
300
100
Week 1
2
Item d - level 4
R
480
120
Q
I
1200
720
N
OR
3
4
5
6
7
8
9
1250
0
480
350
0
0
0
600
600
120
0
230
0
0
0
49
Example – level 4
Week
Item f
R
Q
I
N
OR
1
2
- level 4
0
360
4000
110
4000
3
4
5
6
7
8
9
3750
110
3640
0
0
960
0
0
0
0
0
0
960
0
0
0
960
This is a great level.
50
item C
periods 4 - 8
Lot sizing and MRP
least unit cost heuristic:
net
lot
period rqmt size
weeks
in inv
wk 4
wk 5
wk 6
wk 7
wk 8
1150
100
0
120
200
Iteration 2
wk 7
120
wk 8
200
carry-cost
per lot
Inv
$ per
Unit
Unit
Setup
Cost
Total
(unit)
0
60
0
.05
.87
.80
.87
.85
1370
1570
0
1
2
3
4
276
756
.20
.48
.73
.64
.93
1.12
120
320
0
1
0
120
0
.38
8.33
3.13
8.33
2.51
0
1250
setup cost = $1000
holding cost = .60 /unit-period
unit cost = $1
51
Lot Sizing with Capacity Constraints
Known requirements in each period versus capacities
ri = requirement in period i
ci = production capacity in period i
yi = production level in period i
 r1 , r2 ,..., rn 
 c1 , c2 ,..., cn 
Find  y1 , y2 ,..., yn  where
j
j
c  r
i 1
i
i 1
i
yi  ci , for 1  i  n
for j  1,..., n
52
Example 7.7
r = (20, 40, 100, 35, 80, 75, 25)
c = (60, 60, 60, 60, 60, 60, 60)
Check for feasibility:
period
1
2
3
4
5
6
7
additive
rqmt
20
60
160
195
275
350
375
Additive
capacity
60
120
180
240
300
360
420
53
The simple 2-step algorithm
1.
2.
Find next period in which demand > capacity
Back-shift demands to period(s) having excess capacity
0. r = (20, 40, 100, 35, 80, 75, 25)
c = (60, 60, 60, 60, 60, 60, 60)
1. 40 60 60
r = (20, 40, 100, 35, 80, 75, 25)
c = (60, 60, 60, 60, 60, 60, 60)
2. 40 60 60 55 60
r = (20, 40, 100, 35, 80, 75, 25)
c = (60, 60, 60, 60, 60, 60, 60)
3. 50
60
40 60 60 55 60 60 25
r = (20, 40, 100, 35, 80, 75, 25)
c = (60, 60, 60, 60, 60, 60, 60)
Feasible but not necessarily optimal!
Final answer:
y = (50, 60, 60, 60, 60, 60, 25)
54
The Improvement 2-Step
• Start at end and work backwards
• Determine if it is cheaper to shift entire production lot to
prior periods having excess capacity
Assume k = $450 and h = $2:
0. r = (100, 79, 230, 105, 3, 10, 99, 126, 40)
c = (120, 200, 200, 400, 300, 50, 120, 50, 30)
1. r’ = (100, 109, 200, 105, 28, 50, 120, 50, 30)
c = (120, 200, 200, 400, 300, 50, 120, 50, 30)
Cost = 9 x 450 + 432 = $4,482
2. r’= (100, 109, 200, 105, 28, 50, 120, 50, 30)
c = (120, 200, 200, 400, 300, 50, 120, 50, 30)
y = (100, 109, 200, 105, 58, 50, 120, 50, 0)
Cost = 8 x 450 + (432 + 2 x 30 x 4) = $4,272
Find feasible solution
Look for improvement
55
More Improvement
•
•
Start at end and work backwards
Determine if it is cheaper to shift entire production lot to prior periods having
excess capacity
Assume k = $450 and h = $2:
1. r’ = (100, 109, 200, 105, 28, 50, 120, 50, 30)
c = (120, 200, 200, 400, 300, 50, 120, 50, 30)
Cost = 9 x 450 + 432 = $4,482
2. r’= (100, 109, 200, 105, 28, 50, 120, 50, 30)
c = (120, 200, 200, 400, 300, 50, 120, 50, 30)
y = (100, 109, 200, 105, 58, 50, 120, 50, 0)
Cost = 8 x 450 + (432 + 2 x 30 x 4) = $4,272
2. r’= (100, 109, 200, 105, 28, 50, 120, 50, 30)
c = (120, 200, 200, 400, 300, 50, 120, 50, 30)
y = (100, 109, 200, 105, 108, 50, 120, 0, 0)
Cost = 7 x 450 + (432 + 240 + 300) = $4,122
Let’s do another one…
56
Assume k = $450 and h = $2:
Do it again?
2. r’= (100, 109, 200, 105, 28, 50, 120, 50, 30)
c = (120, 200, 200, 400, 300, 50, 120, 50, 30)
y = (100, 109, 200, 105, 108, 50, 120, 0, 0)
Cost = 7 x 450 + (432 + 240 + 300) = $4,122
No! Don’t
do it. It
will cost
more.
2. r’= (100, 109, 200, 105, 28, 50, 120, 50, 30)
c = (120, 200, 200, 400, 300, 50, 120, 50, 30)
y = (100, 109, 200, 105, 228, 50, 0, 0, 0)
Savings = $450
Cost = $2 x 120 x 2 = $480
Look at shifting 50 units in period 6 to period 5
and the resulting 158 units in period 5 to period 4
Final answer: y = (100, 109, 200, 263, 0, 0, 0, 0, 0) with cost = $3,638
57
There are still problems…
The problem is that even if lot sizes at
some level do not exceed production
capacities, there is no guarantee that
when these lot sizes are translated into
gross requirements at a lower level, that
they will not exceed capacities.
58
let
AI = setup cost for item i
hi = holding cost for item i
Iit = inventory level ending period t
Qit = order quantity of item i, period t
zit = 1 if item i is ordered period t; otherwise 0
mi,s(i) = number units of i required to produce s(i)
Rt = end item requirement in period t
Optimal Multilevel Lot Sizing
N
Min
T
 A z
subject to:
i 1 t 1
i it
I1,(t 1)  Q1t  I1t  Rt
 hi I it
for 1  t  T
I i ,(t 1)  Qit  I it  mi , s (i )Qs (i ),t  0 for 1  t  T , 2  i  N
Qit  zit M
for 1  t  T , 2  i  N
Qit , I it  0,
zit  0 or 1
59
Weaknesses and Problems
MRP is based upon a flawed model!
• assumes infinite production capacity
– MRP identifies what is to be done to meet the master
schedule
– capacity planning usually occurs after the fact
– capacity lot-sizing at one level will not solve overall
capacity problem
• deterministic system
– lead-times & demands
– lead-times do not consider plant loading
– lead-times are inflated to compensate creating more
problems
– lead-times may be dependent on lot sizes
60
System Nervousness
Caused by Rolling Horizon
Only 1st period decision of a N-period problem is implemented.
Using the Wagner-Whitin algorithm:
t
Dt
Qt
1
190
190
2
210
400
3
190
0
4
210
400
5
190
0
t
Dt
Qt
1
190
400
2
210
0
3
190
400
4
210
0
5
190
400
A = $400 / order
h = $1 per item
6
210
0
Q1 = 190 if odd number of periods
Q1 = 400 if even number of periods
61
More Problems
• Random numbers of defective items
• Data Integrity
–
–
–
–
machine unavailability
engineering changes
discrepancies in inventory records
changes in demands (forecasts), costs, or lead-times
• Lead-times dependent upon lot sizes
• Reject or rework rates must be addressed
62
After MRP?
• MRP II – Manufacturing Resource Planning
–
–
–
–
includes MRP-I
adds financial, accounting, and marketing functions
MPS becomes a decision variable
capacity resource planning (CRP) an integral feature
• ERP – Enterprise Resource Planning
–
–
–
–
–
includes MRP-II
entire firm operates on the same data
client-server with a single relational database
multi-location (international) support
includes customer orders, distribution, delivery
schedules, etc.
63
More on ERP
• Enterprise systems are commercial software
packages that enable the integration of
transactions-oriented data and business processes
throughout an organization (and perhaps
eventually throughout the entire interorganizational supply chain).
• Enterprise systems include ERP software and
related packages as advanced planning and
scheduling (APS), sales force automation,
customer relationship management, product
configuration, etc.)
64
More of More on ERP
• Most companies have failed to implement ERP
packages successfully or to realize the hoped-for
financial returns on their ERP investment.
• Standish Group Study of ERP Implementations:
– 35% are Cancelled
– 55% overrun their budgets
– Less than 10% are on time and under budget
• Implementation Averages
– Cost: 178% over budget
– Schedule: 230% longer
– Functionality: –59% or: the system will only perform
41% of the functions it was intended to perform.
65
Why Implementations Fail
•
•
•
•
•
•
People Don’t want the systems to succeed
People are comfortable and don’t see the need
for the new system.
People have unrealistic expectations of the new
system.
People don’t understand the basic concepts of
the system.
The basic data is inaccurate.
The system has technical difficulties.
66
What about APS?
• APS is Advance Planning and Scheduling:
The technique that deals with analysis and
planning of logistics and manufacturing over the
short, intermediate and long-term periods. APS
describes any computer program that uses
advanced mathematical algorithms or logic to
perform optimization or simulation on finite
capacity scheduling, sourcing, capital planning,
resource planning, forecasting, demand
management and others.
67
The Five Main Components Of an
APS System
•
•
•
•
•
demand planning
production planning
production scheduling
distribution planning
transportation planning
Gosh there are
five of them.
68
MRP and APS
• APS considers all constraints and capacities
simultaneously using mathematical models (linear
programming or some derivative) and yields a
production plan (either short term or longer term).
• MRP can take this plan and execute the acquisition
of materials.
• MRP is more transaction focused and an older
product than APS
• APS is more modeling oriented.
69
Homework
Hey! This MRP stuff is okay.
Can we have some homework
problems to try it out? Please.
From your textbook: chapter 7: 4,5,6,23,24,25,27,28
+ handout
70