Material Management
Class Note #1-A
MRP – Capacity Constraints
Prof. Yuan-Shyi Peter Chiu
Feb. 2012
1
§ M1: Push & Pull
Production Control System

MRP: Materials Requirements Planning (MRP) ~ PUSH

JIT: Just-in-time (JIT) ~ PULL

Definition (by Karmarkar, 1989)
A pull system initiates production as a reaction to
present demand, while
A push system initiates production in anticipation of
future demand
Thus, MRP incorporates forecasts of future demand
while JIT does not.
2
§ M2: MRP ~ Push
Production Control System


We determine lot sizes based on forecasts
of future demands and possibly on cost
considerations
A top-down planning system in that all
production quantity decisions are derived
from demand forecasts.
Lot-sizing decisions are found for every
level of the production system. Item are
produced based on this plan and pushed
to the next level.
3
§ M2: MRP ~ Push

A production plan is a complete spec. of
The amounts of final product produced
The exact timing of the production lot sizes
The final schedule of completion




The production plan may be broken down into
several component parts
1)
2)
3)

Production Control System ( p.2 )
The master production schedule (MPS)
The materials requirements planning (MRP)
The detailed Job Shop schedule
MPS - a spec. of the exact amounts and timing of
production of each of the end items in a production
system.
4
§ M2: MRP ~ Push
Production Control System ( p.3 )
P.405 Fig.8-1
5
§ M2: MRP ~ Push

The data sources for determining the MPS
include
1)
2)
3)
4)
5)

Production Control System ( p.4 )
Firm customer orders
Forecasts of future demand by item
Safety stock requirements
Seasonal plans
Internal orders
Three phases in controlling of the production
system
Phase 1: gathering & coordinating info to develop MPS
Phase 2: development of MRP
Phase 3: development of detailed shop floor and
resource requirements from MRP
6
§ M2: MRP ~ Push

Production Control System ( p.5 )
How MRP Calculus works:
1.
2.
3.
Parent-Child relationships
Lead times into Time-Phased requirements
Lot-sizing methods result in specific schedules
7
§ M3: JIT ~ Pull
Production Control System
Basics :
1.
2.
3.
4.
5.
WIP is minimum.
A Pull system ~ production at each stage is
initiated only when requested.
JIT extends beyond the plant boundaries.
The benefits of JIT extend beyond savings of
inventory-related costs.
Serious commitment from Top mgmt to
workers.
Lean Production ≈ JIT
8
§ M4: The Explosion Calculus
(BOM Explosion)
Gross Requirements of one level
Push down
Lower levels
9
§ M4: The Explosion Calculus
Eg. 7-1
Valve casing
assembly (1)
Lead time = 2 weeks
b-t-13
Fig.7-5 p.353
Trumpet
( End Item )
Bell assembly (1)
(page 2)
Lead time = 4 weeks
b-t-14
Slide assemblies (3)
Valves (3)
Lead time = 2 weeks
Lead time = 3 weeks
b-t-15
10
§ M4: The Explosion Calculus
(page 3)
=>Steps
1. Predicted Demand (Final Items)
2. Net demand (or MPS)



3.
Push Down to the next level (MRP)



4.
Forecasts
Schedule of Receipts
Initial Inventory
Lot-for-lot production rule (lot-sizing algorithm)
– no inventory carried over.
Time-phased requirements
May have scheduled receipts for different parts.
Push all the way down
11
Eg. 7-1

1
Trumpet
1 Bell Assembly
1 Valve casing Assembly
3 Slide Assemblies
3 Valves
 7 weeks to produce a Trumpet ?
 To plan 7 weeks ahead
 The Predicted Demands:
Week
Demand
8
9
77 42
10 11 12
13 14 15 16 17
38 21 26 112 45 14 76 38
 Expected schedule of receipts
Week
Scheduled receipts
8
9
10
11
12
0
6
9
12
 Beginning inventory = 23, at the end of week 7
 Accordingly the net predicted demands become
Week
Net Predicted
Demands
8
9
10
11
12
13
14
15
16
17
42
42
32
12
26
112
45
14
76
38
Master Production Schedule (MPS) for the end product (i.e. Trumpet)
 MRP calculations for the Bell assembly (one bell assembly
for each Trumpet) & Lead time = 2 weeks go-see-10
Week
6
7
8
9
10
11
12
Gross
Requirements
42
42
32
12
Net
Requirements
42
42
32
12
26
112
45
26
13
14
15
16
17
112
45
14
76
38
14
76
38
Time-Phased
Net Requirements
42
42 32 12
26
112
45
14
76
38
Planned Order
Release (lot for lot)
42
42 32 12
26
112
45
14
76
38
13
 MRP Calculations for the valve casing assembly (1 valves
casing assembly for each Trumpet) & Lead time = 4 weeks go-see-10
Week
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Gross
Requirements
42 42
32
12
26
112
45
14
76
38
Net
Requirements
42 42
32
12
26
112
45
14
76
38
Time-Phased
Net Requirements 42 42
32 12
26 112 45
14
76
38
Planned Order
Release (lot for lot) 42 42
32 12
26 112 45
14
76
38
b-t-20
b-t-38
14
 MRP Calculations for the valves ( 3 valves for each valve casing
assembly) go-see-10
 Lead Time = 3 weeks
 On-hand inventory of 186 valves at the end of week 3
 Receipt from an outside supplier of 96 valves at the start of week 5
 MRP Calculations for the valves
Week
2
3
Gross
Requirements
4
126 126
Scheduled Receipts
On-hand inventory
6
7
8
9
10
11
96
36
78 336 135
42
12
13
228 114
96
186
Net
Requirements
Time-Phased
Net Requirements
Planned Order
Release (lot for lot)
5
60
30
0
0
66
36
78
336 135
66
36 78
336 135
42 228 114
66
36 78
336 135
42 228 114
42 228 114
15
§. M4.1: Class Work
# CW.1
 What is the MRP Calculations for the slide assemblies ?
( 3 slide assemblies for each valve casing )
 Lead Time = 2 weeks
 Assume On-hand inventory of 270 slide assemblies at the end of week
3 & Scheduled receipts of 78 & 63 at the beginning of week 5 & 7
Show the MRP Calculations for the slide assemblies !
Preparation Time : 25 ~ 35 minutes
Discussion : 20 minutes
◆1g-s-62
16

To Think about …
 Lot-for-Lot may not be feasible ?!
e.g. 336 Slide assemblies required at week 9 may exceeds
plant’s capacity of let’s say 200 per week.
 Lot-for-Lot may not be the best way in production !?
 Why do we have to produce certain items (parts) every week?
why not in batch ? To minimize the production costs.
17
§. M4.2: Class Problems
Discussion
Chapter 7 :
( # 4, 5, 6 )
( # 9 (b,c,d) )
p.356-7
p.357
Preparation Time : 25 ~ 35 minutes
Discussion : 20 minutes
18
§ M5: Alternative Lot-sizing schemes

Log-for-log : in general, not optimal

If we have a known set of time-varying demands
and costs of setup & holding, what production
quantities will minimize the total holding & setup
costs over the planning horizon?
19
(1) EOQ Lot sizing   ? 
439/10=43.9
(page 2)
h  ?  ($141.82* 22%) / 52  $0.6 per piece
k  ?  2  3  $22  $132

2k 
 139
h
(1) MRP Calculation for the valve casing assembly when applying E.O.Q.
lot sizing Technique instead of lot-for-lot (g-s-14)
Q
Week
4
Net
Requirements
Time-Phased
Net Requirements 42
Planned order
release (EOQ)
Planned deliveries
Ending inventory
139
5
6
42
32
0
0
7
8
9
10
11
12
13
42
42
32
12
26
112 45
26
112
45
14
76
38
0 139
0
139
0
0
139
139
0
0
0
139
97
55
23
12
11 124
14
0 139
15 16 17
14
76 38
0
0 139
12 106 92 16 117
20
 Ending =
Inventory
Beginning
Inventory
+
Planning
Deliveries
 Total ordering ( times ) = 4 ;
 Total ending inventory =
Net
Requirements
cost = $132 * 4 = $528
17
= 653 ;

 j
j 8
cost = ($0.6) (653) = $391.80
Total Costs
= Setup costs + holding costs
= 4*132+$0.6*653 = $919.80
vs. lot-for-lot 10*132 = $1320 (setup costs)
g-b-41
21
§ M5: Alternative Lot-sizing schemes
(page 3)
(2) The Silver-Meal Heuristic (S-M)


Forward method ~ avg. cost per period (to span)
Stop when avg. costs increases.
c(1)  k
c(2)  (k  hr2 ) / 2
c(3)  (k  hr2  2hr3 ) / 3
:
c( j )  (k  hr2  2hr3  ...  ( j  1)hrj ) / j

i.e. Once c(j) > c(j-1) stop
Them let y1 = r1+r2+…+rj-1 and begin again starting at period j
22
§ M5: Alternative Lot-sizing schemes
The
silver-meal heuristic Will Not Always result in an
optimal solution (see eg.7.3; p.360)
Computing
Technology enables heuristic solution
● S-M example 1 :
 Suppose demands for the casings are r = (18, 30, 42, 5, 20)
 Holding cost = $2 per case per week
 Production setup cost = $80
Starting in Period 1 :
C(1) = $80
C(2) = [$80+$2(30)] /2 = $70
C(3) = [$80+$2(30)+$2(2)(42)] /3 =308/3 = $102.7
∵ C(3) >C(2)
∴ STOP ; Set
y1  r1  r2  48
23
Starting in Period 3 :
r = (18, 30, 42, 5, 20)
C(1) = 80
C(2) = [80+2(5)] /2 = 45
C(3) = [80+2(5)+$2(2)(20)] /3 = 170/3 = 56.7
∵ C(3) >C(2)
∴ STOP ; Set
y3  r3  r4  47 & y5  20
∴ Solution = (48, 0, 47, 0, 20) cost = $310
● S-M example 2 : (counterexample)
Let r = (10, 40, 30) , k=50 & h=1
Silver-Meal heuristic gives the solution y=(50,0,30)
but the optimal solution is (10,70,0)
Conclusion of Silver-Meal heuristic
 It will not always result in an optimal solution
 The higher the variance (in demand) , the better the
improvement the heuristic gives (versus EOQ)
24
§ M5: Alternative Lot-sizing schemes (page 4)
(3) Least Unit Cost (LUC)

Similar to the S-M except it divided by total
demanded quantities.
c(1)  k / r1
c(2)  ( k  hr2 ) /( r1  r2 )
:
c( j )  [k  hr2  ...  ( j  1)hrj ]/( r1  r2  ...  rj )

Once c(j) > c(j-1) stop and so on.
25
● LUC example:
r = (18, 30, 42, 5, 20)
h = $2
K = $80
Solution : in period 1
C(1) = $80 /18 = $4.44
C(2) = [80+2(30)] /(18+30) = 140/48 = $2.92
C(3) = [80+2(30)+2(2)(42)] /(18+30+42) = 308/90 = $3.42
∵ C(3) >C(2)
∴ STOP ; Set
y1  r1  r2  48
Starting in period 3
C(1) = $80 /42 = 1.90
C(2) = [80+2(5)] /(42+5) = 90 /47 = 1.91
∵ C(3) >C(2)
∴ STOP ; Set
y3  r3  42
26
r = (18, 30, 42, 5, 20)
Starting in period 5
C(1) = $80 /5 = 16
C(2) = [80+2(20)] /(5+20) = 120 /25 = 4.8
∴ Set
y4  r4  r5  25
∴ Solution = ( 48, 0, 42, 25, 0)
cost = 3(80)+2(30)+2(20) = $340
27
§ M5: Alternative Lot-sizing schemes (page 5)
(4) Part Period Balancing (PPB)



More popular in practice
Set the order horizon equal to “# of periods”
~ closely matches total holding cost closely with the setup
cost over that period.
Closer rule
Eg. 80 vs. (0, 10, 90) then choose 90
Last three : S-M, LUC, and PPB are heuristic methods
~ means reasonable but not necessarily give
the optimal solution.
28
● PPB example :
r = (18, 30, 42, 5, 20)
h = $2
K = $80
Starting in Period 1
Order
Horizon
Total Holding
cost
1
2
3
0
60 (2*30)
228 (2*30+2*2*42)
K=80
∵ K is closer to period 2
∴ y  r  r  48
1
1
2
29
Starting in Period 3 :
Order
Horizon
1
2
3
Total Holding
cost
0
10
90
r = (18, 30, 42, 5, 20)
h = $2
K = $80
(2*5)
(2*5+2*2*20)
K=80
∵ K is closer to period 3
∴ y3  r3  r4  r5  67
∴ Solution = (48, 0, 67, 0, 0)
cost = 2(80)+2(30)+2(5)+2(2)(20) = $310 #
30
§. M5.1: Class Problems
Discussion
Chapter 7 :
( # 14, 17 )
p.363
Preparation Time : 25 ~ 40 minutes
Discussion : 15 minutes
31
§ M6: Wagner – Whitin Algorithm
~ guarantees an optimal solution to the production
planning problem with time-varying demands.
Eq.
r  (52,87, 23,56)
y1  52 ;
y1   (52  ...  56)  218
y1  [52, 218] ~ 167 values ;
( y1, y2 ) ~ 10200 values
~ Enormous ~
y1  r1 ; or y1  r1  r2  ...; or y1  r1  r2  ...  rn
y2  0; or y2  r2 ; or y2  r2  r3  ...;
or y2  r2  r3...  rn
:
yn  0; or yn  rn ~ much smaller set of solutions
 2(n-1) distinct exact solutions
32
§ M6: Wagner – Whitin Algorithm
(page 2)
Eg. A four periods planning
y1  r1 y2  r2 y3  r3 , y4  r4
(1)
y3  r3  r4 , y4  0 (2)
y 2 =r2 +r3  y3 =0, y 4 =r4
(3)
y 2 =r2 +r3 +r4 y3 =0, y 4 =0 (4)
y1 =r1 +r2 , y 2 =0 y3 =r3 ,
y 4 =r4 (5)
y3 =r3 +r4 , y 4 =0
(6)
y1 =r1 +r2 +r3 , y 2 =0, y3 =0, y 4 =r4
(7)
y 1 =r1 +r2 +r3 +r4 ,
(8)
y 2 =y3 =y 4 =0
~2
(4-1)
2 8
3
◆2g-t-63
33
§ M6: Wagner – Whitin Algorithm

(page 3)
Enumerating vs. dynamic programming
◆ Dynamic Programming
f k  min c  f ( j 1)  for k = 1, 2, ... , n
j k
j
k
j = k, k+1, ... , n
34
§ M6: Wagner – Whitin Algorithm
See ‘ PM00c6-2 ‘
(page 4)
for Example
35
§ M6.1: Dynamic Programming
Eq 7.2
c5  $80
r =(18,30,42,5,20) h=$2
k=$80
c35  80  80  10  170 #
c45  $80  40  120 #
 4
c4   4
c3  c3  c5  80  10  80  170 #
 3
c4  c5  $160
c3  c4  80  120  200
5

c
1  80  60  168  30  160  498
5
c2  80  84  20  120  304
 4
 4
c1  c5  80  60  168  30  80  418
c2  c5  80  84  20  80  264

c2   3
c1  c13  c4  80  60  168  120  428
c2  c4  80  84  170  334
 2
c 2  c  80  170  250 #
c1  c3  80  60  170  310 #
 2 3
c1  c  80  250  330
1 2
c12c35  (48, 0, 67, 0, 0)
solution  2 4
c1 c3 c5  (48, 0, 47, 0, 20)
36
§. M6.2: Class Problems Discussion
#1: Inventory model when demand rate λ is
not constant
1  2  3  4
300 200 300 200
K=$20
C=$0.1
h=$0.02

Find C1  Min
C
1  Cj  1  ?
1 j 4
(j)
#2:
( Chapter 7:
# 18(a),(b) )
p.363
Preparation Time : 10 ~ 15 minutes
Discussion : 10 minutes
37
§ M7: Incorporating Lot-sizing Algorithms into
the Explosion calculus
▓ From Time-phased net requirements applies algorithm
Example 7.6
p.364
g-s-14
from the time-phased net requirements for the valve casing assembly :
Week
Time-Phased
Net Requirements
4
5
6
7
8
9
10
11
12
13
42
42
32
12
26
112
45
14
76
38
 Setup cost = $132 ; h= $0.60 per assembly per week
 Silver-Meal heuristic :
38
Starting in week 4 :
C(1) = $132
C(2) = [132+(0.6)(42)] /2 = 157.2/2 = $78.6
C(3) = [132+(0.6)(42)+(0.6)(2)(32)] /3= 195.6/3 =$65.2
C(4) = [195.6+(0.6)(3)(12)] /4 = 217.2/4 = $54.3
C(5) = [217.2+(0.6)(4)(26)] /5 = 279.6/5 = $55.9 (STOP)
∴
y4  r4  r5  r6  r7  42  42  32  12  128
Starting in week 8 :
C(1) = $132
C(2) = [132+(0.6)(112)] /2 = 199.2/2 = $99.6
C(3) = [199.2+(0.6)(2)(45)] /3= 253.2/3 =$84.4
C(4) = [253.2+(0.6)(3)(14)] /4 = 278.4/4 = $69.6
C(5) = [278.4+(0.6)(4)(76)] /5 = 460.8/5 = $92.2 (STOP)
∵ C(5) >C(4)
∴
y8 
11
r
i 8
i
 r8  r9  r10  r11  197
39
Starting in week 12 : C(1) = $132
C(2) = [132+(0.6)(38)] /2 = $77.4
y12  r11  r12  76  38  114
∴ y = (128 , 0 , 0 , 0 , 197 , 0 , 0 , 0 , 114 , 0)
∴
 MRP Calculation using Silver-Meal lot-sizing algorithm :
Week
4
5
6
7
Net
Requirements
42
Time-Phased
Net Requirements 42 42
Planned Order
Release (S-M)
Planned deliveries
Ending inventory
8
128
0
32 12
0
9
10
11
12
26
13
42
32
12
112
26 112
45
14
76
38
14
15
16 17
45
14
76 38
0 197
0
0
0
114
0
128
0
0
0
197
0
0
0
114
0
86
44
12
0
171
59
14
0
38
0
40
▓ Compute the total costs
 S-M : Total cost = 132(3)+(0.6)(86+44+12+171+59+14+38) = $650.4
 Lot-For-Lot : $132*10 = $1320
g-s-14
 E.O.Q : 4(132)+(0.6)(653) = $919.80
g-t-20
for optimal schedule by Wagner-Whitin algorithm it is
y4=154 , y9=171 , y12=114 ; Total costs= ＄610.20
▓ push down to lower level…
41
§. M7.1: Class Work # CW.2
  Applies Least Unit Cost in MRP Calculation for
the valve casing assembly.
  Applies Part Period Balancing in MRP Calculation
for the valve casing assembly.
◆3g-t-64
 Applies Wagner-Whitin algorithm in MRP for the
valve casing assembly.
Preparation Time : 25 ~ 35 minutes
Discussion : 20 minutes
42
§. M 7.2: Class Problems
Discussion
Chapter 7 :
( # 24, 25 )
( # 49 )
p.365-6
p.393
Preparation Time : 15 ~ 20 minutes
Discussion : 10 minutes
43
§ M8:
Lot sizing with Capacity Constraints
▓ Requirements vs production capacities.
’’realistic’’~more complex.
◇
▓ True optimal is difficult, time-consuming and probably not practical.
▓ Even finding a feasible solution may not be obvious.
▓ Feasibility condition must be satisfied
j
j
 C  
i 1
i
i 1
i
for j  1, 2,
e.g. Demand r = ( 52 , 87 , 23 , 56 )
Capacity C= ( 60 , 60 , 60 , 60 )
,n
Total demands = 218
Total capacity = 240
though total capacity > total demands ;
but it is still infeasible (why?)
44
§ M8:
Lot sizing with Capacity Constraints
◇
(page 2)
▓ Lot-shifting technique to find initial solution
▓ Eg. #7.7 (p.376) γ=(20,40,100,35, 80,75,25)
C =(60,60, 60,60, 60,60,60)
◆ First tests for Feasibility condition → satisfied
◆ Lot-shifting
C = (60,60, 60,60,60,60,60)
γ = (20,40,100,35,80,75,25)  demand
(C-γ) = (40,20,-40,…)
(C-γ)’ =(20, 0, 0,…)
(production plan) γ’= (40,60,60,35,80,75,25)
[γ’=C- (C- γ)’]
(C-γ’)’ = (20,0,0,25,-20,…)
(C-γ’)’ = (20,0,0,5,0,…)
γ’ = (40,60,60,55,60,75,25) [γ’=C- (C- γ’)’]
45
§ M8:
Lot sizing with Capacity Constraints
(C-γ’)’ = (20,0,0,5,0,-15,…)
(C-γ’)’ = (10,0,0,0,0,0,…)
γ’ = (50,60,60,60,60,60,25)
(C-γ’)’ = (10,0,0,0,0,0,35)
(production plan)
(page 3)
◇
[γ’=C- (C- γ’)’]
γ’= (50,60,60,60,60,60,25)
∴ lot-shifting technique solution (backtracking)
gives a feasible solution.
▓ Reasonable improvement rules for capacity constraints
◆ Backward lot-elimination rule
46
§ M8:
Lot sizing with Capacity Constraints
(page 4)
◆ Eg. 7.8
◇
Assume k=$450 , h=$2
C = (120,200,200,400,300,50,120, 50,30)
γ= (100, 79,230,105, 3,10, 99,126,40)
from lot-shifting γ’=？
γ’ = (100,109,200,105,28,50,120,50,30) [ How ? ]
costs = (9*$450)+2*(216)=$4482
◆ Improvement
Find Excess capacity first.
C = (120,200,200,400,300,50,120, 50,30)
γ’ = (100,109,200, 105, 28, 50,120, 50,30)
(C - γ’) = ( 20, 91, 0, 295,272, 0, 0, 0, 0)
47
§ M8:
Lot sizing with Capacity Constraints
(page 5)
◇
◆ Is there enough excess capacity in prior periods to
consider shifting this lot?
excess capacity: (C –γ’) = (20,91,0,295,272,0,0,0,0)
242
192
142
γ’ = (100,109,200,105,28,50,120,50,30)
58
108
158
∵ 30 units shifts from the 9th period to the 5th period
increases holding cos t by $2*4*30  $240
 decreases setup by $450 (i.e.$k ) '' okay ''

48
§ M8:
Lot sizing with Capacity Constraints
(page 6)
◇
∵ 50 units shifts from the 8th period to the 5th
increases holding cos t by $2*3*50  $300
 decreases setup by $450 '' okay ''

∵ 120 units shifts from the 7th period to the 5th [not Okay]
increases holding cos t by $2* 2*120  $480

not  K (  $450) " Not okay "
∵ okay to shift 50 from the 6th period to the 5th
increases holding cos t by $2 *50  $100
 decreases setup by $450 '' okay ''

Result :
→ γ’ = (100,109,200,105,158,0,120,0,0)
49
∵ Furthermore, it is okay to shift 158 from the 5th period to
the 4th period
increases holding cos t by $2 *158  $316
decreases setup by $450 '' okay ''

263 0
→ γ’ = (100,109,200,105,158,0,120,0,0)
•
(C-γ’) = (20,91,0,295,142,50,0,50,30)
•
Excess capacity
137 300
∵ 158 units shifts from the 5th period to the 4th
increase holding cost by $2*158=$316 < $K “ okay ’’
→ final γ’ = (100,109,200,263,0,0,120,0,0)
50
§ M8:
Lot sizing with Capacity Constraints
(page 7)
◇
◆ after improvement;
total cost = [ 5*$450+ $2*(694) ] = $3638
vs { $4482 (before improvement)
where γ’ = (100,109,200,105,28,50,120,50,30) }
◆ improvement save 20% of costs
51
§. M 8.1: Class Problems
Discussion
Chapter 7 :
# CW.3 ; # 28 (a) (b)
# CW.5 ; #CW.4
p.369
Preparation Time : 25 ~ 30 minutes
Discussion : 15 minutes
52
# CW.5
Consider problem #28 (a), suppose the setup cost for the
construction of the base assembly is $200, and the holding cost is
$0.30 per assembly per week, and the time-phased net requirements
and production capacity for the base assembly in a table lamp over
the next 6 weeks are:
Week
Time-Phased Net
Requirements r
Production
Capacity
=
c=
1
2
3
4
5
6
335
200
140
440
300
200
600
600
600
400
200
200
(a) Determine the feasible planned order release
(b) Determine the optimal production plan
53
§ M 9: Shortcoming of MRP
■ Uncertainty
◆ forecasts for future sales
◆ lead time from one level to another
■ Two implication in MRP
 all of the lot-sizing decisions could be incorrect.
 former decisions that are currently being implemented
in the production process may be incorrect.
■ Safety stock to protect against the uncertainty of
demand
◆ not recommended for all levels
◆ recommended for end products only, they will be
transmitted down thru the explosion calculus.
54
§ M 9: Shortcoming of MRP
( page 2 )
■ Applies the coefficient variation σ/μ
◆ obtain σ, find → ratio =
◆ obtain safety stock σx z

 ∴ σ=μx ratio
(e.g. z = 1.28 → 90％)
◆ obtain (μ+σ*z ) as planned production schedule.
55
 Example 7.9
(p.381)
[ Using a Type 1 service lever of 90 %]
 Consider example 7.1 (p.362) Demands for Trumpets
 If analyst finds that the ratio σ/μ (coefficient of variation) is 0.3
 Harmon co. decided to produce enough Trumpets to meet
all weekly demand with probability 0.90
 0.90 for Normally Distributed demand has a Z = 1.28
Week
8
9
10
11
12
13
14
15
16
17
77
42
38
21
26
112
45
14
76
38
Standard
23.1 12.6 11.4
Deviation ( σ= μ*0.3 )
Mean demand
107 58 53
Plus safety stock
( μ+ z σ )
[ i.e. μ+(1.28) σ ]
6.3
7.8
33.6 13.5
4.2 22.8 11.4
29
36
155
19 105
Predicted
Demand ( μ )
62
53
56
§ M 9: Shortcoming of MRP
(page 3)
■ Capacity Planning
◆ Feasible solution at one level may result in an
‘’ infeasible ’’ requirements schedule at a lower level.
◆ CRP – Capacity requirements planning by using MRP
planned order releases.
~ If CRP results in an ‘’ infeasible ’’ case then to
correct it by
◇ schedule overtime, outsourcing
◇ revise the MPS
~ Trial & Error between CRP and MRP until fitted.
57
§ M 9: Shortcoming of MRP
(page 3)
▓ Rolling Horizons and System Nervousness
◆ MRP is not always treated as a static system.
~ may need to rerun each period for
1st period decision
▓ Other considerations
◆ Lead times is not always dependent on lot sizes
~ sometimes lead time increases
when lot size increases
◆ MRP Ⅱ：Manufacturing Resource Planning
◇ MRP converts an MPS into planned order releases.
◇ MRP Ⅱ：Incorporate Financial , Accounting ,
& Marketing functions into the production
planning process
58
§ M 9: Shortcoming of MRP
(page 4)
Ultimately, all divisions of the company would work
together to find a production schedule
consistent with the overall business plan and
long-term financial strategy of the firm.
◇ MRP Ⅱ：～ incorporation of CRP
◆ Imperfect production Process
◆ Data Integrity
59
§. M 9.1: Class Problems
Discussion
Chapter 7 :
( # 33 )
p.376
Preparation Time : 15 ~ 20 minutes
Discussion : 10 minutes
60
§ M 10: J I T
◆ Kanban
◆ SMED (Single minute exchange of dies)
‧IED (inside exchange of dies )
‧OID (out side exchange of dies )
◆ Advantages vs. Disadvanges (See Table 6-1)
§ M 11: MRP & JIT
36 distinct factors to compare JIT, MRP, & ROP
(reorder point) [Krajewski et al 1987]
61
The End
62