1 prof. yuan-shyi peter chiu feb. 2011 material management class note # 1-a mrp – capacity...
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Prof. Yuan-Shyi Peter Chiu Feb. 2011
Material ManagementClass Note #1-A
MRP – Capacity Constraints
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§ M1: Push & Pull§ 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.
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§ M2: MRP ~ Push § 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.
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§ M2: MRP ~ Push § M2: MRP ~ Push Production Control System ( p.2 )
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) The master production schedule (MPS)2) The materials requirements planning (MRP)3) 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.
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§ M2: MRP ~ Push § M2: MRP ~ Push Production Control System ( p.3 ) P.405 Fig.8-1
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§ M2: MRP ~ Push § M2: MRP ~ Push Production Control System ( p.4 )
The data sources for determining the MPS include
1) Firm customer orders2) Forecasts of future demand by item3) Safety stock requirements4) Seasonal plans5) Internal orders
Three phases in controlling of the production system
Phase 1: gathering & coordinating info to develop MPSPhase 2: development of MRPPhase 3: development of detailed shop floor and
resource requirements from MRP
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§ M2: MRP ~ Push § M2: MRP ~ Push Production Control System ( p.5 )
How MRP Calculus works:
1. Parent-Child relationships
2. Lead times into Time-Phased requirements
3. Lot-sizing methods result in specific schedules
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§ M3: JIT ~ Pull § M3: JIT ~ Pull Production Control System
Basics :
1. WIP is minimum.2. A Pull system ~ production at each stage is
initiated only when requested.3. JIT extends beyond the plant boundaries.4. The benefits of JIT extend beyond savings of
inventory-related costs.5. Serious commitment from Top mgmt to
workers.
Lean Production ≈ JIT
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§ M4: The Explosion Calculus § M4: The Explosion Calculus (BOM Explosion)
Gross Requirements of one level
Push down
Lower levels
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§ M4: The Explosion Calculus § M4: The Explosion Calculus (page 2)
Eg. 7-1 Fig.7-5 p.353Trumpet
( End Item )
Bell assembly (1)Lead time = 2 weeks
Valve casingassembly (1)
Lead time = 4 weeks
Valves (3) Lead time = 3 weeks
Slide assemblies (3) Lead time = 2 weeks
b-t-14
b-t-15
b-t-13
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§ M4: The Explosion Calculus § M4: The Explosion Calculus (page 3)
=>Steps
1. Predicted Demand (Final Items)2. Net demand (or MPS)
Forecasts Schedule of Receipts Initial Inventory
3. Push Down to the next level (MRP) Lot-for-lot production rule (lot-sizing algorithm)
– no inventory carried over. Time-phased requirements May have scheduled receipts for different parts.
4. Push all the way down
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1 Trumpet1 Bell Assembly1 Valve casing Assembly
3 Slide Assemblies3 Valves
7 weeks to produce a Trumpet ? To plan 7 weeks ahead The Predicted Demands:
Expected schedule of receipts
Week
Demand
Week
Scheduled receipts
8 9 10 11 12 13 14 15 16 17
77 42 38 21 26 112 45 14 76 38
8 9 10 11
12 0 6 9
Eg. 7-1
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Beginning inventory = 23, at the end of week 7 Accordingly the net predicted demands become
MRP calculations for the Bell assembly (one bell assembly for each Trumpet) & Lead time = 2 weeks go-see-10
Master Production Schedule (MPS) for the end product (i.e. Trumpet)
Week
Net Predicted Demands
8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
Week
Gross Requirements
Time-Phased Net Requirements
Planned Order Release (lot for lot)
Net Requirements
6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
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MRP Calculations for the valve casing assembly (1 valves casing assembly for each Trumpet) & Lead time = 4 weeks go-see-10
Week
Gross Requirements
Net Requirements
Time-Phased Net Requirements
Planned Order Release (lot for lot)
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
b-t-20
b-t-38
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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 valvesWeek
Gross Requirements
Net Requirements
Time-Phased Net Requirements
Planned Order Release (lot for lot)
Scheduled Receipts
On-hand inventory
2 3 4 5 6 7 8 9 10 11 12 13
126 126 96 36 78 336 135 42 228 114
96
186 60 30
0 0 66 36 78 336 135 42 228 114
66 36 78 336 135 42 228 114
66 36 78 336 135 42 228 114
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Show the MRP Calculations for the slide assemblies !
§.§. M4.1: M4.1: Class Work Class Work # CW.1# CW.1
Preparation Time : 25 ~ 35 minutesPreparation Time : 25 ~ 35 minutesDiscussion : 20 minutesDiscussion : 20 minutes
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
◆1g-s-62
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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.
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§.§. M4.2: M4.2: Class ProblemsClass Problems Discussion Discussion
Chapter 7 : Chapter 7 : ( # ( # 4, 5,4, 5, 66 ) ) p.356-7
( ( # 9 (b,c,d)# 9 (b,c,d) ) ) p.357
Preparation Time : 25 ~ 35 minutesPreparation Time : 25 ~ 35 minutesDiscussion : 20 minutesDiscussion : 20 minutes
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§ M5: Alternative Lot-sizing schemes§ 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?
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(1) MRP Calculation for the valve casing assembly when applying E.O.Q. lot sizing Technique instead of lot-for-lot (g-s-14)
Week
Net Requirements
Time-Phased Net Requirements
Planned order release (EOQ)
Planned deliveries
Ending inventory
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
139 0 0 0 139 0 139 0 0 139
139 0 0 0 139 0 139 0 0 139
97 55 23 11 124 12 106 92 16 117
? 439/10=43.9
? ($141.82*22%) / 52 $0.6
? 2 3 $22 $132
2139
h per piece
k
kQ
h
(1) EOQ Lot sizing (page 2)
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Ending Inventory
BeginningInventory
Planning Deliveries
Net Requirements
= + -
Total ordering ( times ) = 4 ; cost = $132 * 4 = $528
Total ending inventory = = 653 ;
cost = ($0.6) (653) = $391.80
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8jj
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
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§ M5: Alternative Lot-sizing schemes § 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.
i.e. Once c(j) > c(j-1) stopThem let y1 = r1+r2+…+rj-1 and begin again starting at period j
2
2 3
2 3
(1)
(2) ( ) / 2
(3) ( 2 ) / 3
:
( ) ( 2 ... ( 1) ) /j
c k
c k hr
c k hr hr
c j k hr hr j hr j
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§ M5: § M5: Alternative Lot-sizing schemesAlternative 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) = $80C(2) = [$80+$2(30)] /2 = $70C(3) = [$80+$2(30)+$2(2)(42)] /3 =308/3 = $102.7
∵ C(3) >C(2) STOP ; Set∴ 1 1 2 48y r r
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Starting in Period 3 :
C(1) = 80C(2) = [80+2(5)] /2 = 45C(3) = [80+2(5)+$2(2)(20)] /3 = 170/3 = 56.7
∵ C(3) >C(2) STOP ; Set∴ 53 3 4 47 & 20y r r y ∴ 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)
r = (18, 30, 42, 5, 20)
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§ § M5: Alternative Lot-sizing schemesM5: Alternative Lot-sizing schemes (page 4)
(3) Least Unit Cost (LUC) Similar to the S-M except it divided by total
demanded quantities.
Once c(j) > c(j-1) stop and so on.
1
1 2
1 2
2
2
(1) /
(2) ( ) /
:
( ) [ ..
(
. ( 1) ] /
)
( ... )jj
c k
c k hr
c j k
r
r r
r r rhr j hr
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● LUC example:r = (18, 30, 42, 5, 20)h = $2 K = $80
Solution : in period 1
C(1) = $80 /18 = $4.44C(2) = [80+2(30)] /(18+30) = 140/48 = $2.92C(3) = [80+2(30)+2(2)(42)] /(18+30+42) = 308/90 = $3.42
Starting in period 3
C(1) = $80 /42 = 1.90C(2) = [80+2(5)] /(42+5) = 90 /47 = 1.91
∵ C(3) >C(2) STOP ; Set∴
∵ C(3) >C(2) STOP ; Set∴
1 1 2 48y r r
3 3 42y r
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Starting in period 5
C(1) = $80 /5 = 16C(2) = [80+2(20)] /(5+20) = 120 /25 = 4.8
∴ Set 4 4 5 25y r r
∴ Solution = ( 48, 0, 42, 25, 0) cost = 3(80)+2(30)+2(20) = $340
r = (18, 30, 42, 5, 20)
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§ M5:§ M5: Alternative Lot-sizing schemesAlternative 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.
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● PPB example :r = (18, 30, 42, 5, 20)h = $2K = $80
Starting in Period 1
Order Horizon
Total Holding cost
123
060 (2*30)228 (2*30+2*2*42)
K=80
∵ K is closer to period 2
∴1 1 2 48y r r
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Starting in Period 3 :
Order Horizon
Total Holding cost
123
010 (2*5)90 (2*5+2*2*20)
∵ K is closer to period 3 ∴ 3 3 4 5 67y r r r
∴ Solution = (48, 0, 67, 0, 0)
cost = 2(80)+2(30)+2(5)+2(2)(20) = $310 #
K=80
r = (18, 30, 42, 5, 20)h = $2K = $80
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§.§. M5.1: M5.1: Class ProblemsClass Problems Discussion Discussion
Chapter 7 : Chapter 7 : ( # ( # 14,14, 1717 ) ) p.363
Preparation Time : 25 ~ 40 minutesPreparation Time : 25 ~ 40 minutesDiscussion : 15 minutesDiscussion : 15 minutes
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§ M6:§ M6: Wagner – Whitin Algorithm Wagner – Whitin Algorithm ~~ guarantees an optimal solution to the production
planning problem with time-varying demands.
Eq.
(n-1)2 distinct exact solutions
1 1
1 1, 2
1 1 1 1 2 1 1 2
2 2 2 2 2 3
(52,87,23,56)
52 ; (52 ... 56) 218
[52,218] ~ 167 values ; ( ) ~ 10200 values
~ Enormous ~
y ; ...; ...0; ; ...;
n
r
y y
y y y
r or y r r or y r r ry or y r or y r r
2 2 3 or ...:
0; ~ much smaller set of solutions
n
n n n
y r r r
y or y r
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§ M6:§ M6: Wagner – Whitin Algorithm Wagner – Whitin Algorithm (page 2)(page 2)
Eg. A four periods planning1 1 2 2 3 3 4 4
3 3 4 4
2 2 3 3 4 4
2 2 3 4 3 4
1 1 2 2 3 3 4
, (1)
, 0 (2)
y =r +r y =0, y =r (3)
y =r +r +r y =0, y =0 (4)
y =r +r , y =0 y =r , y
y r y r y r y r
y r r y
1
4
3 3 4 4
1 1 2 3 2 3 4 4
1 2 3 4 2 3 4
=r (5)
y =r +r , y =0 (6)
y =r +r +r , y =0, y =0, y =r (7)
y =r +r +r +r , y =y =y =0 (8)(4-1) 3~ 2 2 8 ◆2g-t-63
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§ M6:§ M6: Wagner – Whitin Algorithm Wagner – Whitin Algorithm (page 3)(page 3)
Enumerating vs. dynamic programming
◆ Dynamic Programming
( 1)min for k = 1, 2, ... , n
j = k, k+1, ... , n
jk k j
j kf c f
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§ M6:§ M6: Wagner – Whitin Algorithm Wagner – Whitin Algorithm (page 4)(page 4)
See ‘ PM00c6-2 ‘ for Example
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§ M6.1: Dynamic Programming§ M6.1: Dynamic Programming
Eq 7.2 r =(18,30,42,5,20) h=$2 k=$805
535
4 44 3 3 54
4 5 33 4
52
42 5
2 32 4
22 3
80 80 10 #$80 40 #
80 10 80 #$160
80 120 200
80 84 20 120 304
80 84 20 80 264
80 84 170 334
80 1
$80
170120
170
2570 #0
c
cc
c c c cc c
c c
c
c cc
c c
c c
51
41 5
31 1 4
21 3
11 2
2 51 3
2 41 3 5
80 60 168 30 160 498
80 60 168 30 80 418
80 60 168 120 428
80 60 170 #
80 250 330
310
(48,0,67,0,0)
(48,0,47,0,20)
c
c c
c c c
c c
c c
c csolution
c c c
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§.§. M M6.2: Class Problems Discussion6.2: Class Problems Discussion
Preparation Time : 10 ~ 15 minutesDiscussion : 10 minutes
4321
300 200 300 200
K=$20C=$0.1h=$0.02
?CCMin C Find 1j14j1
1(j)
#1: Inventory model when demand rate λ is not constant
#2: ( Chapter 7: ( Chapter 7: # 18# 18(a),(b) ) ) p.363
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§ M7: § M7: Incorporating Lot-sizing Algorithms into Incorporating Lot-sizing Algorithms into the Explosion calculusthe Explosion calculus
▓ From Time-phased net requirements applies algorithm p.364
Example 7.6from 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 :
g-s-14
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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)
∴ 4 4 5 6 7 42 42 32 12 128y r r r r
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) ∴ 11
8 8 9 10 118
197ii
y r r r r r
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Starting in week 12 : C(1) = $132
C(2) = [132+(0.6)(38)] /2 = $77.4
12 11 12 76 38 114y r r ∴ ∴ y = (128 , 0 , 0 , 0 , 197 , 0 , 0 , 0 , 114 , 0)
MRP Calculation using Silver-Meal lot-sizing algorithm :
Week
Net Requirements
Time-Phased Net Requirements
Planned Order Release (S-M)
Planned deliveries
Ending inventory
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
128 0 0 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
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S-M : Total cost = 132(3)+(0.6)(86+44+12+171+59+14+38) = $650.4
Lot-For-Lot : $132*10 = $1320
E.O.Q : 4(132)+(0.6)(653) = $919.80
▓ Compute the total costs
for optimal schedule by Wagner-Whitin algorithm it is y4=154 , y9=171 , y12=114 ; Total costs= $ 610.20
▓ push down to lower level…
g-t-20
g-s-14
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§.§. M7.1: Class Work M7.1: Class Work # CW.2# CW.2
Applies Part Period Balancing in MRP Calculation for the valve casing assembly.
Applies Wagner-Whitin algorithm in MRP for the valve casing assembly.
Applies Least Unit Cost in MRP Calculation for the valve casing assembly.
Preparation Time : 25 ~ 35 minutesPreparation Time : 25 ~ 35 minutesDiscussion : 20 minutesDiscussion : 20 minutes
3◆ g-t-64
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§.§. M 7.2: M 7.2: Class ProblemsClass Problems Discussion Discussion
Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutesDiscussion : 10 minutesDiscussion : 10 minutes
Chapter 7 : Chapter 7 : ( ( # 24, 25# 24, 25 ) ) p.365-6
( # 49 ) ( # 49 ) p.393
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§ M8:§ M8: Lot sizing with Capacity ConstraintsLot 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
◇◇
e.g. Demand r = ( 52 , 87 , 23 , 56 ) Total demands = 218
Capacity C= ( 60 , 60 , 60 , 60 ) Total capacity = 240
though total capacity > total demands ; but it is still infeasible (why?)
1 1
1,2, ,j j
i ii i
C for j n
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§ M8:§ M8: Lot sizing with Capacity ConstraintsLot 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’)’]
◇◇
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§ M8:§ M8: Lot sizing with Capacity ConstraintsLot sizing with Capacity Constraints (page 3)
(C-γ’)’ = (20,0,0,5,0,-15,…)
(C-γ’)’ = (10,0,0,0,0,0,…)
γ’ = (50,60,60,60,60,60,25) [γ’=C- (C- γ’)’]
(C-γ’)’ = (10,0,0,0,0,0,35)
(production plan)γ’= (50,60,60,60,60,60,25)
∴ lot-shifting technique solution (backtracking) gives a feasible solution.
▓ ▓ Reasonable improvement rules for capacity constraintsReasonable improvement rules for capacity constraints
◆ Backward lot-elimination rule
◇◇
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§ M8:§ M8: Lot sizing with Capacity ConstraintsLot 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
◆ ◆ ImprovementImprovement 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)
◇◇
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◆ 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)
γ’ = (100,109,200,105,28,50,120,50,30)
∵ 30 units shifts from the 9th period to the 5th period
242 192142
58 108 158
cos $2*4*30 $240 $450 ( . .$ ) '' ''
increases holding t bydecreases setup by i e k okay
§ M8:§ M8: Lot sizing with Capacity ConstraintsLot sizing with Capacity Constraints (page 5) ◇◇
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∵ 50 units shifts from the 8th period to the 5th
∵ 120 units shifts from the 7th period to the 5th [not Okay]
∵ okay to shift 50 from the 6th period to the 5th
Result :
→ γ’ = (100,109,200,105,158,0,120,0,0)
cos $2*3*50 $300 $450 '' ''
increases holding t bydecreases setup by okay
cos $2*2*120 $480 ( $450) " "increases holding t by
not K Not okay
§ M8:§ M8: Lot sizing with Capacity ConstraintsLot sizing with Capacity Constraints (page 6) ◇◇
cos $2*50 $100 $450 '' ''
increases holding t bydecreases setup by okay
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263 0
→ γ’ = (100,109,200,105,158,0,120,0,0)
• (C-γ’) = (20,91,0,295,142,50,0,50,30)
• 137 300Excess capacity
∵ Furthermore, it is okay to shift 158 from the 5th period to the 4th period
cos $2*158 $316 $450 '' ''
increases holding t bydecreases setup by okay
∵ 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)
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◆ 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
§ M8:§ M8: Lot sizing with Capacity ConstraintsLot sizing with Capacity Constraints (page 7)◇◇
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§.§. M 8.1: M 8.1: Class ProblemsClass Problems Discussion Discussion
Chapter 7 : Chapter 7 : # CW.3 ; # CW.3 ; # 28# 28 (a)(a) (b)(b) p.369
# CW.5 ; #CW.4# CW.5 ; #CW.4
Preparation Time : 25 ~ 30 minutesPreparation Time : 25 ~ 30 minutesDiscussion : 15 minutesDiscussion : 15 minutes
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# CW.5# 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 =
1 2 3 4 5 6
335 200 140 440 300 200
Production c=Capacity
600 600 600 400 200 200
(a) Determine the feasible planned order release (b) Determine the optimal production plan
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§ M 9:§ M 9: Shortcoming of MRP Shortcoming of MRP
■ Uncertainty ◆ forecasts for future sales
◆ lead time from one level to another
■ 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.
■ 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.
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§ M 9:§ M 9: Shortcoming of MRP Shortcoming of MRP ( page 2 )
■ Applies the coefficient variation σ/μ
◆ obtain σ, find → ratio = ∴ σ=μx ratio
◆ obtain safety stock σx z (e.g. z = 1.28 → 90% )
◆ obtain (μ+σ*z ) as planned production schedule.
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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
PredictedDemand ( μ )
StandardDeviation ( σ= μ*0.3 )
Mean demandPlus safety stock ( μ+ z σ )
8 9 10 11 12 13 14 15 16 17
77 42 38 21 26 112 45 14 76 38
23.1 12.6 11.4 6.3 7.8 33.6 13.5 4.2 22.8 11.4
107 58 53 29 36 155 62 19 105 53
[ i.e. μ+(1.28) σ ]
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■ 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.
§ M 9:§ M 9: Shortcoming of MRP Shortcoming of MRP (page 3)
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▓ 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
§ M 9:§ M 9: Shortcoming of MRP Shortcoming of MRP (page 3)
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Ultimately, all divisions of the company would worktogether to find a production schedule
consistent with the overall business plan andlong-term financial strategy of the firm.
◇ MRP Ⅱ :~ incorporation of CRP
◆ Imperfect production Process
◆ Data Integrity
§ M 9:§ M 9: Shortcoming of MRP Shortcoming of MRP (page 4)
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§.§. M 9.1: M 9.1: Class ProblemsClass Problems Discussion Discussion
Chapter 7 : Chapter 7 : ( # 33 ) ( # 33 ) p.376
Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutesDiscussion : 10 minutes Discussion : 10 minutes
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§ M 10:§ M 10: J I TJ 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 § M 11: MRP & JIT
36 distinct factors to compare JIT, MRP, & ROP (reorder point) [Krajewski et al 1987]
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The EndThe End
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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
Net Requirements
Time-Phased Net RequirementsPlanned Order Release (lot for lot)
Scheduled Receipts
On-hand inventory
2 3 4 5 6 7 8 9 10 11 12 13
126 126 96 36 78 336 135 42 228 114
78
270 144 96 27
0 0 0 0 51 336 135 42 228 114
63
51 336 135 42 228 114
51 336 135 42 228 114
◆1 # CW.1# CW.1
g-b-16
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4 1 3
(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)
2 2
(1,1,1
8
,1)
◆2
g-b-33
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◆3 Solution: Applies Part Period Balancing in MRP Calculation for the valve casing assembly.
Week
Net Requirements
Time-Phased Net Requirements
Planned Order Release (PPB)
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
?
MRP Calculation using Part Period Balancing lot-sizing algorithm :
Starting in Period 4:
Order Horizon Total Holding cost
12345
0$25.2 (0.6)*(42)$63.6 $25.2+2(0.6)(32)$85.2 $63.6+3(0.6)(12)$147.6 $85.2+4(0.6)(26)
K=132
∵ K is closer to period 5
∴ 4 4 8... 154y r r
# CW.2# CW.2
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◆3
MRP Calculation using Part Period Balancing lot-sizing algorithm :
Week
Net Requirements
Time-Phased Net Requirements
P.O.R. (PPB)
Planned deliveries
Ending inventory
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
154 0 0 0 0 247 0 0 0 38
112 70 38 26 0 135 90 76 0 0
Starting in Period 9:
Order Horizon Total Holding cost
1234
0$27 (0.6)*(45)$43.8 $27+2(0.6)(14)$180.6 $43.8+3(0.6)(76)
K=132 ∵ K is closer to period 4
∴ 9 9 12... 247y r r
13 38y
154 0 0 0 0 247 0 0 0 38
# CW.2# CW.2
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S-M : $650.4
Lot-For-Lot : $132*10 = $1320
E.O.Q : $919.80
▓ Compute the total costs
for optimal schedule by Wagner-Whitin algorithm it is y4=154 , y9=171 , y12=114 ; Total costs= $ 610.20
◆3
PPB : Total cost = $132(3)+(0.6)(547) = $724.2
# CW.2# CW.2
g-s-42
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# CW.3# 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
Net Requirements
Time-Phased Net
Requirements r =
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
Production c=Capacity
50 50 50 50 50 50 50 50 50 50
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Week
Net Requirements
Time-Phased Net
Requirements r =
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
Production c=Capacity
# CW.3# CW.3
excess (c-r) =Capacity
8 8 18 38 24 (62) 5 36 (26) 12
(c-r)’ =
[2] Lot-shifting technique (back-shift demand from rj > cj):
50 50 50 50 50 50 50 50 50 50
8 8 18 38 24 (62) 5 36 (26) 12
8 8 18 0 0 0 5 10 0 12 (c-r)’ =
final r ’ = 42 42 32 50 50 50 45 40 50 38
1 1
1,2, ,j j
i ii i
C for j n
[1]First test for: It is okay!
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# CW.4 # CW.4 ( continues on #CW.3)( 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.Determine the optimal production plan.
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Production c=Capacity
50 50 50 50 50 50 50 50 50 50
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Production c=Capacity (O-T)
120 120 120 120 120 120 120 120 120 120
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# CW.4 # CW.4 ( continues on #CW.3)( 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.Determine the optimal production plan.
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Production c=Capacity
50 50 50 50 50 50 50 50 50 50
final r ’ = 42 42 32 50 50 50 45 40 50 38 (using regular shift)
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
0 0 0 38 62 0 0 26 0 0 Σ= 126 Ending Inventories =
[1] First, the cost for using regular shift is $100(10) + $0.65 (126)
= $1,081.9 [ lot for lot ][ lot for lot ]
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Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Production c=Capacity (O-T)
excess (c-r) =Capacity
120 120 120 120 120 120 120 120 120 120
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
# CW.4# CW.4[1] First, the cost for using regular shift is $100(10) + $0.65 (126)
= $1,081.9 [ lot for lot ][ lot for lot ]
78 78 88 108 94 8 75 106 44 82
42 42 32 12 26 112 45 14 76 38r = 78 78 88 108 94 8 75 106 44 excess (c-r)’=
Capacity 35 77 0 31 43 0 6
42 42 32 12 26 112 45 14 76 38
85 43 120 89 77 120 59 0 114 0
0 0 0 0
85 0 120 0 0 120 0 0 114 0 final r ’ =
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# CW.4# CW.4
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
85 0 120 0 0 120 0 0 114 0 final r ’ =
[2] The cost for using Overtime shift is $205(4) + $0.65(372) = $1061.8
43 1 89 77 51 59 14 0 38 0 Ending Inventories =
Less than the cost for using regular shift $1,081.9, Saved $ 20.10
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# CW.4# CW.4
Ending Inventories =
[3] To think about the following solution:
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Suppose r ’=
74 32 0 38 62 0 0 26 0 0 Σ= 232
116 0 0 50 50 50 45 40 50 38 [ One OT, 7 regular ]
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 $ 26.1Less than the cost for using all Overtime shift $1061.8 Saved $ 6.0
WHY ?
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# CW.4# CW.4
Ending Inventories =
[4] A Better Solution :
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Suppose r ’=
74 32 0 27 1 9 14 0 38 0 Σ= 195
116 0 0 39 0 120 50 0 114 0
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
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§.§. M4.2: M4.2: Class ProblemsClass Problems Discussion Discussion
Chapter 7 : Chapter 7 : ( # ( # 4, 5,4, 5, 66 ) ) p.356-7
( ( # 9 (b,c,d)# 9 (b,c,d) ) ) p.357
Preparation Time : 25 ~ 35 minutesPreparation Time : 25 ~ 35 minutesDiscussion : 20 minutesDiscussion : 20 minutes
p.18p.18
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§.§. M5.1: M5.1: Class ProblemsClass Problems Discussion Discussion
Chapter 7 : Chapter 7 : ( # ( # 14,14, 1717 ) ) p.363
Preparation Time : 25 ~ 40 minutesPreparation Time : 25 ~ 40 minutesDiscussion : 15 minutesDiscussion : 15 minutes
p.31p.31
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§.§. M M6.2: Class Problems Discussion6.2: Class Problems Discussion
Preparation Time : 10 ~ 15 minutesDiscussion : 10 minutes
4321
300 200 300 200
K=$20C=$0.1h=$0.02
?CCMin C Find 1j14j1
1(j)
#1: Inventory model when demand rate λ is not constant
#2: ( Chapter 7: ( Chapter 7: # 18# 18(a),(b) ) ) p.363
p.37p.37
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§.§. M7.1: Class Work M7.1: Class Work # CW.2# CW.2
Applies Part Period Balancing in MRP Calculation for the valve casing assembly.
Applies Wagner-Whitin algorithm in MRP for the valve casing assembly.
Applies Least Unit Cost in MRP Calculation for the valve casing assembly.
Preparation Time : 25 ~ 35 minutesPreparation Time : 25 ~ 35 minutesDiscussion : 20 minutesDiscussion : 20 minutes
3◆ g-t-64
p.42p.42
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§.§. M 7.2: M 7.2: Class ProblemsClass Problems Discussion Discussion
Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutesDiscussion : 10 minutesDiscussion : 10 minutes
Chapter 7 : Chapter 7 : ( ( # 24, 25# 24, 25 ) ) p.365-6
( # 49 ) ( # 49 ) p.393
p.43p.43
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§.§. M 8.1: M 8.1: Class ProblemsClass Problems Discussion Discussion
Chapter 7 : Chapter 7 : # CW.3 ; #CW.5 ; #CW.4 # CW.3 ; #CW.5 ; #CW.4
Preparation Time : 25 ~ 30 minutesPreparation Time : 25 ~ 30 minutesDiscussion : 15 minutesDiscussion : 15 minutes
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# CW.5# 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 =
1 2 3 4 5 6
335 200 140 440 300 200
Production c=Capacity
600 600 600 400 200 200
(a) Determine the feasible planned order release (b) Determine the optimal production plan
p.53p.53
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# CW.5# CW.5 SolutionSolution
Week
Time-Phased Net Requirements r =
1 2 3 4 5 6
335 200 140 440 300 200
Production c=Capacity
600 600 600 400 200 200
(b) Determine the optimal production plan
p.53p.53
(a) Determine the feasible planned order release
265 400 460 -40 -100 0 (c-r) =
265 400 320 0 0 0 Adj.(c-r) =
335 200 280 400 200 200 r’ =
335 200 280 400 200 200 r’ = 265 400 320 0 0 0 (c-r’) =
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Production c=Capacity
600 600 600 400 200 200
(b) Determine the optimal production plan
335 200 280 400 200 200 r’ = 265 400 320 0 0 0 (c-r’) =
# CW.5# CW.5 SolutionSolution (b) (b)
535 0 480 400 0 200 r’’ =
65 600 120 0 200 0
Increase holding cost = $0.3*(200) + $0.3*2*(200)=$180 Saving setup cost = 2*K = 2*$200= $400
Overall savings = $220
535 0 480 400 0 200 Final production plan r’’ =
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§.§. M 9.1: M 9.1: Class ProblemsClass Problems Discussion Discussion
Chapter 7 : Chapter 7 : ( # 33 ) ( # 33 ) p.376
Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutesDiscussion : 10 minutes Discussion : 10 minutes
p.60p.60
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# CW.3# 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
Net Requirements
Time-Phased Net
Requirements r =
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
Production c=Capacity
50 50 50 50 50 50 50 50 50 50
p.68p.68
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# CW.4 # CW.4 ( continues on #CW.3)( 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.Determine the optimal production plan.
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Production c=Capacity
50 50 50 50 50 50 50 50 50 50
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Production c=Capacity (O-T)
120 120 120 120 120 120 120 120 120 120
p.70p.70