prof. yuan-shyi peter chiu feb. 2011

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Prof. Yuan-Shyi Peter Chiu Feb. 2011

Material ManagementClass Note #1-A

MRP – Capacity Constraints

2

§ 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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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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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



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



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


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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


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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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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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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Chapter 7 : Chapter 7 : ( # ( # 14,14, 1717 ) ) p.363


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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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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

35


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

38

§ 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


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

40

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


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.


3◆ g-t-64

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§.§. M 7.2: M 7.2: Class ProblemsClass Problems Discussion Discussion


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

45

§ 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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(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

◇◇

47


◆ 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)

◇◇

48

◆ 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) ◇◇

49

∵ 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 '' ''


50

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 '' ''


∵ 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)

51

◆ 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)◇◇

52


Chapter 7 : Chapter 7 : # CW.3 ; # CW.3 ; # 28# 28 (a)(a) (b)(b) p.369

# CW.5 ; #CW.4# CW.5 ; #CW.4


53

# 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

54

§ 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.

55

§ 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.

56

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 ]

57

■ 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)

58

▓ 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


59

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


60


Chapter 7 : Chapter 7 : ( # 33 ) ( # 33 ) p.376

Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutesDiscussion : 10 minutes Discussion : 10 minutes

61

§ 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]

62

The EndThe End

63

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

64

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

65

◆3 Solution: Applies Part Period Balancing in MRP Calculation for the valve casing assembly.

Week

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

66

◆3

MRP Calculation using Part Period Balancing lot-sizing algorithm :

Week

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

67

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

68

# 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


50 50 50 50 50 50 50 50 50 50

69

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


# 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!

70

# 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


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

71



Time-Phased Net

Requirements r =42 42 32 12 26 112 45 14 76 38


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 ]

72

Time-Phased Net

Requirements r =42 42 32 12 26 112 45 14 76 38


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 ’ =

73

# 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

74

# 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 ?

75

# 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

76


Chapter 7 : Chapter 7 : ( # ( # 4, 5,4, 5, 66 ) ) p.356-7

( ( # 9 (b,c,d)# 9 (b,c,d) ) ) p.357


p.18p.18

77


Chapter 7 : Chapter 7 : ( # ( # 14,14, 1717 ) ) p.363


p.31p.31

78

§.§. 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

79

§.§. 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.


3◆ g-t-64

p.42p.42

80



Chapter 7 : Chapter 7 : ( ( # 24, 25# 24, 25 ) ) p.365-6

( # 49 ) ( # 49 ) p.393

p.43p.43

81


Chapter 7 : Chapter 7 : # CW.3 ; #CW.5 ; #CW.4 # CW.3 ; #CW.5 ; #CW.4


82

# 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


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.53p.53

83

# CW.5# CW.5 SolutionSolution

Week


1 2 3 4 5 6

335 200 140 440 300 200


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’) =

84


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’’ =

85


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

86

# 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


50 50 50 50 50 50 50 50 50 50

p.68p.68

87



Time-Phased Net

Requirements r =42 42 32 12 26 112 45 14 76 38


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


120 120 120 120 120 120 120 120 120 120

p.70p.70

prof. yuan-shyi peter chiu feb. 2011

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