Time Analysis for Manufacturing Systems

1

Time Analysis for Manufacturing Systems

2.810

T.Gutowski

reference: Manufacturing Systems Engineering

Prentice Hall 1994 By Stanley B. Gershwin

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0utline

1.  Process Plan 2.  Tools from Operations Research

•  Little’s Law •  Unreliable Machine(s) •  Buffers •  M/M/1 Queue

3.  Applications (Stephanie Proule: submarine manufacturing)

•  Transfer Lines, FMS, TPS Cells, Push Vs Pull, …

3

How would you machine this part?

•  Material; Aluminum bar stock •  Tolerance + 0.005”, except for the two 0.50” radius corners

which have + 0.015”

4

Components of Process Plan

•  Starting material stock: bar stock? extrusion?

•  Machines: tolerances? available? multiple operations?

•  Tools: clamping? Cutting? Form tools? •  Procedures; sequence, feeds and

speeds, tool changes, deburr, inspect

6

Process planning How would you machine this part?

Assumption: 1.  We begin with a stock size of 2.5” X 2.25” X 12” 2.  This will be manufactured in a job shop for very low quantity

We will use: -  A bandsaw to roughly cut the stock to size -  A manual vertical mill to create the planar features and the holes -  A belt sander to sand the radii (tolerance larger here)

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Machines, tools, fixture

8

Process plan Machine Operation

Horizontal band saw Saw stock to ~4.125”

Manual vertical mill

Mill two ends to length 4”

Mill width to 2”

Mill out 2”X1.5”X4”

Drill hole 1” diameter

Bore 1” radius

Belt sender Sand 0.5 radii

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Job shop (Small quantities, Aluminum block)

# Machine

Operation (V = volume, A=Area, P = perimeter)

Fixture Tool Change

Run (R= rough, F =

finish)

Deburr, Inspect, Measure

(Time)


0 0.3 Deburr0.05 Inspect

0.20 2 0.13 0.63 Deburr0.20 0.75 0.05 Inspect

0.13 Measure0.46 0.43 Deburr0.67 0.05 Inspect

0.13 Measure2.19 0.5 Deburr0.93 0.05 Inspect

0.13 Measure0.13 Measure

Drill hole 1" diameter

* Center drill*Pilot drill 1/2" 2 0.05 0.05 Inspect

*Pilot drill 63/64" 2 0.04 0.17 MeasureReam 2 0.01

2 0.96 0.24 Deburr0.1 0.05 Inspect

0.06 Measure0.2 0.1 Deburr

0.05 Inspect0.06 Measure0.06 Measure

Totals 1.31 12 8.75 3.63 25.69

3700.21

00.08Sand 0.5" radii,

V=0.05, A=0.79, P=3.14

250

60 2

0.03 0.21 Deburr

Bore 1" radius, V=0.79, A=1.57,

P=7.280.2

0

0

0.2

2

Mill out 2"*1.5"*4", V=12, A=14, P=15

240 0.00

30 2Mill width to 2",

V=2.5 A=10 P=130.20

20 2Mill two ends to

length 4'', V=0.703, A=11.25, P=19

Saw stock to ~4.125", A=5.625,

2.020.23 110

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CNC production, aluminum block # Machine

Operation (V = volume, A=Area, P = perimeter)

Fixture Tool Change

Run (R= rough, F =

finish)


(Time)


0.3 Deburr0.05 Inspect0.13 Measure

2.19 0.5 Deburr0.93 0.05 Inspect




Drill hole 1" diameter

* Center drill*Pilot drill 1/2" 0.5 0.05 0.05 Inspect

*Pilot drill 63/64" 0.5 0.04 0.17 MeasureReam 0.5 0.01


0.06 MeasureTotals 0.58 3 6.39 2.34 12.31

0

Deburr

50 2Bore 1" radius,

V=0.79, A=1.57, P=7.28

0.13 0.5

0.03

Contour mill 0.5" radii, V=0.05,

A=0.79, P=3.14230

0.21

00

40 2 0

0.5

2.02

20 2Mill out

2"*1.5"*4", V=12, A=14, P=15

0.13

10 1Saw stock to 4", A=5.625, P=9

0.32

0.5

12

Now increase production

Job Shop to large scale production

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Job shop (Large quantities, Aluminum extrusion) # Machine

Operation (V = volume, A=Area,

P = perimeter)Fixture Tool

Change

Run (R= rough, F =

finish)


(Time)


0.3 Deburr

0.05 Inspect

0.13 MeasureDrill hole 1" diameter



2 0.96 0.24 Deburr0.1 0.05 Inspect



Totals 0.71 10 2.32 1.53 14.56

Saw extrusion to 4", A=2, P=9

0.23

Sand 0.5" radii, V=0.05, A=0.79,

P=3.140.0840 3

10 1

0.21

00.21

0 0.72

Deburr

30 2Bore 1" radius, V=0.79,

A=1.57, P=7.280.2

20 2 0.2

2 0.03

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Job shop (Large quantities, Aluminum extrusion)

Form drill

# MachineOperation (V = volume, A=Area,

P = perimeter)Fixture Tool

Change

Run (R= rough, F =

finish)


(Time)


0.3 Deburr

0.05 Inspect

0.13 MeasureDrill hole 1" diameter



2 0.96 0.24 Deburr0.1 0.05 Inspect



Totals 0.71 10 2.32 1.53 14.56

Saw extrusion to 4", A=2, P=9

0.23

Sand 0.5" radii, V=0.05, A=0.79,

P=3.140.0840 3

10 1

0.21

00.21

0 0.72

Deburr

30 2Bore 1" radius, V=0.79,

A=1.57, P=7.280.2

20 2 0.2

2 0.03

- 6 = 8.56

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Flow Line options:

Cold saw Drill press

extrusion

Hole cutter

Form drill

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Flow shop (Unbalanced work load)

# Machine

Ope ration (V = volume , A=Are a, P = pe rime te r)

FixtureTool

Change

Run (R= rough, F =

finish)

De burr, Inspe ct,

Me asure (Time )


Operation Time


0.05 Inspect0.17 Measure

0.2 0.2 0.1 Deburr0.2 0.21 0.05 Inspect

0.06 Measure0.06 Measure0.24 Deburr0.05 Inspect0.06 Measure

Totals 0.96 0 1.41 1.53 3.90

0 0.72

20 2A 0.12

10 1Saw extrusion to 4",

A=2, P=9 0.32

0.21 Deburr

30 3Sand 0.5" radii,

V=0.05, A=0.79, P=3.14

40 2B

Drill with hole cutter 1" radius (use 0.82

in̂ 3/min) V=0.196, P=7.28

0.12 0

Drill hole 1" diameter (R=6.28)

0 0.04

0

0.24

1.52

0.59

1.08

0.71

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Example Flow Line :

Cold saw Drill press

extrusion

Hole cutter

Form drill

Belt sand

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Flow shop (Improved work load) # Machine

Ope ration (V = volume , A=Are a, P = pe rime te r)

FixtureTool

Change

Run (R= rough, F =

finish)

De burr, Inspe ct, Me asure

(Time )


Operation Time


0.05 Inspect0.17 Measure

0.2 0.2 0.1 Deburr0.2 0.21 0.05 Inspect

0.06 Measure0.06 Measure0.24 Deburr0.05 Inspect0.06 Measure

Totals 0.96 0 1.41 1.53 3.90

0.71

20

1.04

0.21

10 1Saw extrusion to

4", A=2, P=9 without D+I+M

0.32

2A

Deburring+ Inspect+Measure of sawing +Drill hole 1" diameter

(R=6.28)1.07

30 3Sand 0.5" radii,

V=0.05, A=0.79, P=3.14

0 1.08

0.12 0

40 2B

Drill with hole cutter 1" radius (use

0.82 in̂ 3/min) V=0.196, P=7.28

0.12

0 N/A

0 0.24

0 0.72

0.04 Deburr

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A Mfg System

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Example Mfg Systems Flow Line(s)

Transfer line

process buffer

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Example Mfg Systems

Toyota Cell(s)

FMS

Machining center with pallets

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A Mfg System

Arrival rate = λ Departure rate = ? Inventory = L (or N) Time in the system = W (or T) Process rate = µ

System boundary

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0utline

1.  Process Plan 2.  Tools from Operations Research

•  Little’s Law (average values) •  Unreliable Machine(s) (operation dependent) •  Buffers (zero buffers & infinite buffers) •  M/M/1 Queue (effects of variation)

3.  Applications •  Transfer Lines, FMS, TPS Cells, Push Vs Pull,

…

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Little’s Law N = λ Τ

N = Average parts in the system λ = Average arrival rate T = Average time in the system

Ref. L. Kleinrock, “Queueing System, Vol 1 Theory, Wiley, 1975

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Queueing Systems

N, T

Observer here Observing arrivals; α(t) = number of arrivals in (0,t)

Observing departures; δ(t) = number of departures in (0,t)

N(t) = α(t) − δ (t) Number in the system, parts or customers

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N(t) = α(t) − δ (t)γ (t) = customer -seconds (shaded area in figure)

average number of customers in the system = N =γ (t)t

average arrival rate = λt =α(t)t

average time per customer = Tt =γ (t)α(t)

=Nλt

this gives N = λt ⋅Ttassuming the limits exist gives N = λ ⋅T

Ref. Kleinrock Vol 1, 1975 See p.16, 17

(or L = λ W)

T = limt→∞

Ttλ = lim

t→∞λt

γ (t) = N(t)dt0

t

∫

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N = λ ⋅T Q. You want a high

rate of production λ, but if you fill too fast the liquid comes out. What do you do?

A. Fill while the bottle is moving making T long enough to avoid losing any liquid.

This results in long lines and large factories N = λ ⋅T

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Ford’s Willow Run Factory Moving assembly line production of B-24s

Ford’s Willow Run plant - 10 mo delay, but in 1944 produced 453 airplanes in 468 hrs About 1 plane every hour!

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How long did they work on assembly?

•  Production rate when fully running was about 1 plane every hour

•  Little’s Law: L = λ W •  λ = 1 plane/hr •  L = ? “Assembly line was over one mile” •  W = ?

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How long did they work on assembly?

•  Production rate when fully running was about 1 plane very hour

•  Little’s Law: L = λ W •  λ = 1 plane/hr •  L = 5280’/68’ = 78 planes,

(if heel to toe for one mile) •  W = L/λ ≈ 78 hours

68’

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Willow Run

Two lines converge into one

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Ford’s Willow Run Factory

Assembly Line, L ~ 81 planes, implies around 81 hrs/plane Ref; Air & Space Aug/Sept 1992

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Applying Little’s Law

•  Boundaries are arbitrary, but you must specify eg. waiting time + service time

•  Internal details are not considered eg. first in first out, flow patterns etc..

(Non-synchronous)

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Unreliable Machine

•  Ref S. B. Gershwin (Ch 2 of his book) •  Preliminaries: conditional probability

and Markov chains - transition probabilities

•  Discrete or continuous time - ODQ •  Probability machine is down -

exponential distribution

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Failure distribution

Note: MTTF = mean time to failure

Probability machine fails at time t = p(1-p)t-1

Geometric distribution

Discrete time model Continuous time model

You can think of this as one half of the “bath tub” curve

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Unreliable Machine with Repair

Note: MTTR = mean time to repair

Steady state solution for probability that machine is up =

MTTFMTTF+MTTR

Discrete time:

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Continuous time

p(0,t)+p(1,t)=1

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Continuous time

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Single unreliable machine

MTTF MTTR

Total working time

Machine up = MTTFMTTF + MTTR

Machine down = MTTRMTTF + MTTR

Average Production rate = 1τ

MTTFMTTF + MTTR

×

Where, τ = operation time = 1/µ

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Operational Dependent Failures

Multiple Machines Zero buffers

Operation dependent e.i. machine can only fail when it is operating

41 Ope

ratio

n de

pend

ent

e.i.

mac

hine

can

onl

y fa

il w

hen

it is

ope

ratin

g Multiple Machine Case: Zero Buffers

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Unreliable Machine(s) Result

•  Multiple identical machines (Transfer line)

•  Single Machine

Buzacott’s formula, µ = 1τ

1×1+ Σ MTTR

MTTF

k

1

1τ

MTTFMTTF + MTTR

×µ = τ = service time without failures

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Time Dependent Estimation of µ

A B“up” PA PB

Assumption: time dependent failure(A and B are two processes with nominal

rate µ=1/τ in series. Their behaviors are not dependent on each other.)

Probability that both A and B are up is A∩ B

Production rate = PA PB =1τ

MTTFB

MTTFB + MTTRB×1

τ

MTTFA

MTTFA + MTTRA

= ×1τ

11 + αA

11 + αB

Where, αi =MTTRi

MTTFi

=1τ

11 + αA + αB + αA αB

A∩ B = PAPB

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Estimation of µ (continued) Note: αA αB << 1

1τ

1

1+ Σ αi2

1

µ ≈Ignoring higher order terms,Same as Buzacott’s result

Note: seems to give the same answer as Buzacott, but second order terms can become important for large systems. Need to differentiate between operation and time dependentfailures

=1τ

11 + αA + αB + αA αB

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Example:Transfer Line

infinite buffer µ0 = (1/τ x p)bottleneck

zero buffer µ∞ = 1/τ x pApB…pN

example;transferline,allp=0.9 µ =(0.9)N x 1/τ

N=1 µ = .9 x 1/τ

N=10 µ = .35 x 1/τ

N=100 µ = .00003 x 1/τ

A B N

µ =(1/(1+0.111N)) x 1/τN=1 µ = .9 x 1/τ

N=10 µ = .47 x 1/τ

N=100 µ = .0825 x 1/τ

Time dependent Operation dependent

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Summary: Production Rates Zero Buffer: 1

τ⋅

1

1+ MTTRiMTTFi1

n

∑

Infinite Buffer: min( 1τ i⋅

MTTFiMTTFi + MTTRi

)

Transfer line

Bottleneck

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Finite Buffer Size R

ate

NN*

Zero Buffer

Infinite Buffer

Buffer Size

How do the two cases connect for finite buffers?

Acts like one big machine Machines are independent rate is controlled by the slowest machine “bottleneck”

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A small amount of buffer space helps a lot, but too much is costly

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Finite buffer approximation

Average Downtime is

N* ≈ 2 to 6 × MTTR × µ

µ1 µ2NMTTR1 + MTTR2

2

For a two machine system :

and, µ1 ≈ µ2, call the rate µ.

Gershwin’s Approximation:Rate

NN*

Zero Buffer

Infinite Buffer Knee

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Simulation of a 20 machine, 19 buffer (cap = 10 parts) Transfer line. Each machine with one minute cycle time

could produce 4800 parts per week. MTTF 3880 minutes, MTTR 120 minutes. See Gershwin p63-64

Zero buffer, 2965/wk

∞ buffer, 4800/week

Ave (3249 sim, 3247 analy)

Perfect machines, ∞buffer

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M/M/1 Queue

λArrival Rate

µService Rate

..how the inventory in the system grows as you approach capacity

(λ & µ vary according to exponential distribution)

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Steady State (λ < µ)

Consider the deterministic case: •  How many people are in the system? A.

λ = 0 L = 0 0 < λ < µ 0 < L < 1 λ = µ L = 1

λ λ λ < µ

Note: From Little’s Law : Time in system, W = L / λ Since L = λ / µ for λ < µ è W = 1 / µ

L

λ

1

µ

L = λ / µ

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When λ > µ

w  What happens at λ > µ ?

w  There is no steady state, parts in the system grow without limit. As t → ∞, L → ∞

L(t)

t

Slope = λ - µ

L

λ

1

µ

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M/M/1 Queue Result Arrival rate = λ , Service rate = µ, where λ ≤ µ

L = “Inventory” = λ / (µ - λ)W = Time in system = 1 / (µ - λ)

λ λL , W

See Notes: Principles needed to derive M/M/1 queue result - on website

L

λ

1

M/M/1

deterministic

µ

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example: two processes

ProcessA:Neverstarved.Outputspartsataveragerateλ withanexponenBaldistribuBon.

ProcessB:withaverageprocessrateµ = (5/4) λ alsowithanexponenBaldistribuBon

Parts in the system: deterministic: L = 4/5; M/M/1: L = 4

?

L

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M/M/1 Queue interpretation

•  Overly simplistic but tractable •  Arrivals (always “on”) vs departures

(stop when the queue is empty) •  Show behavior as you approach

capacity

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G/G/1 Queue result N

ote:

this

resu

lt sh

ows

the

sam

e no

nlin

ear r

ise

in W

and

L

as th

e sy

stem

app

roac

hes

capa

city

as

the

M/M

/1 q

ueue

did

queue Wq=

Note: W = Wq + 1/µ

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For more details take 2.854

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References (optional)

•  Kleinrock (Little’s Law)- handout •  Gershwin, Mfg Systems Engineering

Ch 2 & 3

Time Analysis for Manufacturing Systems

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