Material Management
Class Note #1-A
MRP – Capacity Constraints
Prof. Yuan-Shyi Peter Chiu
Feb. 2011
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
◆1
# CW.1
 Solution: MRP Calculations for the Slide assemblies ( 3 )
 Lead Time = 2 weeks
 On-hand inventory of 270 valves at the end of week 3
 Receipt from an outside supplier of 78 & 63 at the start of week 5 & 7
 MRP Calculations for the valves
Week
Gross
Requirements
2
3
4
126 126
Net
Requirements
Time-Phased
Net Requirements
Planned Order
Release (lot for lot)
6
7
8
10
11
96
36
78 336 135
42
78
63
270 144
96
27
0
0
Scheduled Receipts
On-hand inventory
5
0
51
0
51
336 135
9
336 135
12
13
228 114
42 228 114
42 228 114
g-b-16
51
336 135
42 228 114
63
◆2
  2 41  23  8
(1) (1,0,0,0)
(2) (1,1,0,0)
(3) (1,0,1,0)
(4) (1,0,0,1)
(5) (1,1,1,0)
(6) (1,1,0,1)
(7) (1,0,1,1)
(8) (1,1,1,1)
g-b-33
64
# CW.2
◆3  Solution: Applies Part Period Balancing in MRP Calculation
for the valve casing assembly.
 MRP Calculation using Part Period Balancing lot-sizing algorithm :
Week
4
5
6
7
Net
Requirements
9
42
Time-Phased
Net Requirements 42 42
Planned Order
Release (PPB)
8
32 12
10
11
42
32
12
26 112
45
14
12
26
76
13
112
14
45
15
16 17
14
76 38
38
?
Starting in Period 4:
Order Horizon
1
2
3
4
5
K=132
Total Holding cost
0
$25.2
$63.6
$85.2
$147.6
(0.6)*(42)
∵ K is closer to period 5
$25.2+2(0.6)(32)
∴ y4  r4  ...  r8  154
$63.6+3(0.6)(12)
$85.2+4(0.6)(26)
65
# CW.2
◆3 Starting in Period 9:
Order Horizon
1
2
3
4
Total Holding cost
K=132
0
$27
$43.8
$180.6
∵ K is closer to period 4
(0.6)*(45)
$27+2(0.6)(14) ∴ y9  r9  ...  r12  247
$43.8+3(0.6)(76)
y13  38
 MRP Calculation using Part Period Balancing lot-sizing algorithm :
Week
4
5
6
7
Net
Requirements
8
42
9
10
11
12
15
16 17
45
14
76 38
32
12
26 112
45
14
76
38
0
247
0
0
0
38
Planned deliveries
154
0
0
0
0
247
0
0
0
Ending inventory
112 70
38
26
0
135
90
76
0
P.O.R. (PPB)
154
0
32 12
0
0
112
14
42
Time-Phased
Net Requirements 42 42
26
13
38
66
0
# CW.2
◆3
▓ Compute the total costs
 PPB : Total cost = $132(3)+(0.6)(547) = $724.2
 S-M :
$650.4
 Lot-For-Lot : $132*10 = $1320
 E.O.Q :
$919.80
for optimal schedule by Wagner-Whitin algorithm it is
y4=154 , y9=171 , y12=114 ; Total costs= ＄610.20
g-s-42
67
# CW.3
Consider the example presented previously for the scheduling of
the valve casing assembly. Suppose that the production capacity in
any week is 50 valve casings. Determine the feasible planned order
release for the valve casings. Recall that the time phased net
requirements for the valve casings as followed:
Week
4
5
6
7
Net
Requirements
42
Time-Phased Net
Requirements
Production
Capacity
8
r=
c=
9
10
11
42
32
12
12
26
13
112
42 42
32 12
26 112
45
14
76
38
50 50
50 50
50
50
50
50
50
50
14
45
15
16 17
14
76 38
68
# CW.3
[1]
First test for:
j
j
 C  
i
i 1
i 1
i
for j  1, 2,
,n
It is okay!
[2] Lot-shifting technique (back-shift demand from rj > cj):
Week
4
5
6
7
Net
Requirements
42
Time-Phased Net
Requirements
Production
Capacity
excess (c-r)
Capacity
8
r=
c=
=
(c-r)’ =
(c-r)’ =
final r ’ =
9
10
11
42
32
12
12
13
26
112
42 42
32 12
26 112
45
14
76
38
50 50
50 50
50
50
50
50
50
50
8
8
18 38
24 (62)
5
36
(26)
12
8
8
18 38
24 (62)
5
36
(26)
12
8
8
18
0
0
0
5
10
0
12
32 50
50
50
45
40
50
38
42 42
14
45
15
16 17
14
76 38
69
# CW.4 ( continues on #CW.3)
Suppose that with overtime work on 2nd shifts, the company could
increase the weekly production capacity to 120 valve casings,
however, extra cost per week = $105. Where K=$100, is the regular
setup cost. The holding cost per valve casing per week is estimated
to be $0.65. Determine the optimal production plan.
Week
4
Time-Phased Net
5
6
7
8
9
10
11
12
13
42 42
32 12
26 112
45
14
76
38
50 50
50 50
50
50
50
50
50
120 120 120 120 120 120 120 120
120
Requirements
Production
Capacity
r=
c=
Production
Capacity (O-T)
c= 120
50
14
15
16 17
70
# CW.4 ( continues on #CW.3)
Suppose that with overtime work on 2nd shifts, the company could
increase the weekly production capacity to 120 valve casings,
however, extra cost per week = $105. Where K=$100, is the regular
setup cost. The holding cost per valve casing per week is estimated
to be $0.65. Determine the optimal production plan.
Week
4
Time-Phased Net
Requirements
Production
Capacity
r=
c=
final r ’ =
Ending Inventories =
5
6
7
8
9
10
11
12
13
14
15
16 17
42 42
32 12
26 112
45
14
76
38
50 50
50 50
50
50
50
50
50
50
42 42
32 50
50
50
45
40
50
38 (using regular shift)
0 38
62
0
0
26
0
0
0
0
Σ= 126
[1] First, the cost for using regular shift is $100(10) + $0.65 (126)
= $1,081.9
[ lot for lot ]
71
# CW.4
[1] First, the cost for using regular shift is $100(10) + $0.65 (126)
= $1,081.9
Week
Time-Phased Net
Requirements
Production
Capacity (O-T)
excess (c-r)
Capacity
4
5
42
42
r=
c= 120
=
r=
excess (c-r)’=
Capacity
final r ’ =
6
7
8
12
13
76
38
120 120 120 120 120 120 120 120
120
32 12
9
10
26 112
45
78
78
88 108 94
42
42
32 12
78
78
88 108 94
35
77
0
42
42
32 12
26 112
45
85
43 120 89
77 120
59
85
8
11
[ lot for lot ]
26 112
31 43
0
0
0
0 120
0
0
8
14
75 106
44
82
45
14
76
38
75 106
44
0
14
15
16 17
6
14
76
38
0 114
0
0 114
0
0
120
0
72
# CW.4
Week
Time-Phased Net
r=
final r ’ =
Requirements
Ending Inventories =
4
5
42
42
6
7
32 12
85
0 120
43
1
0
8
9
26 112
10
45
11
14
12
13
76
38
0
120
0
0 114
0
89 77 51
59
14
0
0
[2] The cost for using Overtime shift is $205(4) +
38
14
15
16 17
$0.65(372) = $1061.8
Less than the cost for using regular shift $1,081.9,  Saved $ 20.10
73
# CW.4
[3] To think about the following solution:
Week
Time-Phased Net
4
5
42
42
32 12
0
0 50
r=
Suppose r ’= 116
Requirements
Ending Inventories = 74
6
32
7
8
9
10
11
12
13
26 112
45
14
76
38
50
50
45 40
50
38
0 38 62
0
0 26
0
0
14
15
16 17
[ One OT, 7 regular ]
Σ= 232
The cost for using only one Overtime shift on week 4
is $205(1) + $100(7) + $0.65(232) = $1055.8
Less than the cost for using regular shift $1,081.9,  Saved $ 26.1
Less than the cost for using all Overtime shift $1061.8  Saved $ 6.0
WHY ?
74
# CW.4
[4] A Better Solution :
Week
Time-Phased Net
4
5
42
42
32 12
0
0 39
0
32
0 27
1
r=
Suppose r ’= 116
Requirements
Ending Inventories = 74
6
7
8
9
10
26 112
11
12
13
45
14
76
38
120
50
0
114
0
9
14
0
38
0
14
15
16 17
Σ= 195
The cost for using the above solution
is $205 (3) + $100 (2) + $0.65(195) = $ 941.75
Less than the cost for using regular shift $1,081.9,  Saved $ 140.05
Wow !
42
113
0
0
0
120
50
0
114
0
$944.35
75
§. 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
p.18
76
§. M5.1: Class Problems
Discussion
Chapter 7 :
( # 14, 17 )
p.363
Preparation Time : 25 ~ 40 minutes
Discussion : 15 minutes
p.31
77
§. 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) )
Preparation Time : 10 ~ 15 minutes
Discussion : 10 minutes
p.363
p.37
78
§. 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
p.42
79
§. M 7.2: Class Problems
Discussion
Chapter 7 :
( # 24, 25 )
( # 49 )
p.365-6
p.393
Preparation Time : 15 ~ 20 minutes
Discussion : 10 minutes
p.43
80
§. M 8.1: Class Problems
Discussion
Chapter 7 :
# CW.3 ; #CW.5 ; #CW.4
Preparation Time : 25 ~ 30 minutes
Discussion : 15 minutes
81
# 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
p.53
82
# CW.5 Solution
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
(c-r) =
265 400 460 -40 -100 0
Adj.(c-r)
=
265
400
320
0
0
0
r’ =
335
200
280
400
200
200
(b) Determine the optimal production plan
r’ =
(c-r’) =
335
200
280
400
200
200
265
400
320
0
0
0
p.53
83
# CW.5 Solution (b)
Production
Capacity
c=
600
600
600
400
200
200
(b) Determine the optimal production plan
r’ =
(c-r’) =
r’’ =
335
200
280
400
200
200
265
400
320
0
0
0
65
600
120
0
200
0
0
480
400
0
535
200
Increase holding cost = $0.3*(200) + $0.3*2*(200)=$180
Saving setup cost = 2*K = 2*$200= $400
Overall savings = $220
Final production plan r’’ =
535
0
480
400
0
200
84
§. M 9.1: Class Problems
Discussion
Chapter 7 :
( # 33 )
Preparation Time : 15 ~ 20 minutes
Discussion : 10 minutes
p.376
p.60
85
# CW.3
Consider the example presented previously for the scheduling of
the valve casing assembly. Suppose that the production capacity in
any week is 50 valve casings. Determine the feasible planned order
release for the valve casings. Recall that the time phased net
requirements for the valve casings as followed:
Week
4
5
6
7
Net
Requirements
42
Time-Phased Net
Requirements
Production
Capacity
8
r=
c=
9
10
11
42
32
12
12
26
13
112
42 42
32 12
26 112
45
14
76
38
50 50
50 50
50
50
50
50
50
50
14
45
15
16 17
14
76 38
p.68
86
# CW.4 ( continues on #CW.3)
Suppose that with overtime work on 2nd shifts, the company could
increase the weekly production capacity to 120 valve casings,
however, extra cost per week = $105. Where K=$100, is the regular
setup cost. The holding cost per valve casing per week is estimated
to be $0.65. Determine the optimal production plan.
Week
4
Time-Phased Net
5
6
7
8
9
10
11
12
13
42 42
32 12
26 112
45
14
76
38
50 50
50 50
50
50
50
50
50
120 120 120 120 120 120 120 120
120
Requirements
Production
Capacity
r=
c=
Production
Capacity (O-T)
c= 120
50
14
15
16 17
p.70
87