Chapter 2Machine Interference Model

Long Run Analysis

Deterministic Model

Markov Model

IE 512 Chapter 2 2

Problem Description

• Group of m automatic machines

• Operator must change tools or perform minor repairs

• How many machines should be assigned to one operator?

• Performance measures– Operator utilization: = fraction of time the operator is busy

– Production rate: TH = # finished items per unit time

– Machine availability: = TH/G, where G is the gross production rate, or the production rate that would be achieved if each machine were always available

• Note: In this queuing system, the machines are the customers!

IE 512 Chapter 2 3

Long Run Analysis

Each machine has gross production rate h

Pn is the proportion of time that exactly n machines are down:

Then, given Pn,

01

m

nnP

0

0

0

0

1

1

m

nn

mm

nn n

n

TH m n hP

m n hP nPTH

mh mh m

P

IE 512 Chapter 2 4

Eliminate some unknowns

Suppose the mean time to repair a machine is 1/, and the mean time between failures for a single machine is 1/.

= avg. # of repairs in (0,t] = t

= avg. # of failures in (0,t] =

In the long run, assuming the system is stable,

D t

A t 0

m

nnm n P t mt

, so that , and D t A t t mt

mh

TH mh

01 is still unknown...P

IE 512 Chapter 2 5

Queuing Measures of Performance

• = average # of machines waiting for service

• = average number of machines down

• = average downtime duration of a machine

• = average duration of waiting time for repair

L

N

T

01 11 1

1

m m

n nn nn P nP P

m

11Total machine-hrs down in (0,t] 1

m

nnnP t m t

A t mt mt

1

1m

nnnP m

W1 1 1

T

IE 512 Chapter 2 6

Little’s Formula

Observe from the previous equations:

where

is the total average number of failures per unit time

= the arrival rate of customers to the queuing system

Little’s formula relates mean # of customers in system to mean time a customer spends in the system.

1 1

N T m

L m m W

m

IE 512 Chapter 2 7

A Deterministic Model

Suppose each machine spends exactly 1/ time units working followed by exactly 1/ time units in repair. Then if

and we could stagger the failure times, we would have no more than one machine unavailable at any time, so that

(Otherwise,

1 11m

1

1 1, ,

m m h m hTH

m

1, ,h

THm

IE 512 Chapter 2 8

A Markov ModelLet be the time between the (n-1)st repair and the nth failure of machine j, and

be the time duration of the nth repair (indep.)

The time until the first failure is

N(t) = # of machines down at time t follows a CTMC with

S = {0, 1, …, m} and

expjn

expnS 1 21 1 1min , ,..., expm m

0 0 0

1 1 0 0

0 0 0 0 0

0 0

m m

m m

m n m n

Q

IE 512 Chapter 2 9

Steady-State Probabilities

satisfy the balance equations or level-crossing equations

lim Prnt

p N t n

0 1

1 0 2

1 1

1

1

1n n n

m m

m p p

m p m p p

m n p m n p p

p p

0 1

1 2

1

1

1

n n

m m

m p p

m p p

m n p p

p p

IE 512 Chapter 2 10

Solution

1

0 000

! ! and 1

! !

n nm

m

n nnn

m mp p p p

m n m n

1

00

!ˆ ˆ ˆ ˆSay and let , , so that ,

!

k n

n

kG k p G m

k n

ˆ ˆ ˆThen , 1 , 1 , 0,1,...

ˆ , 11 11

ˆ ˆ ˆ ˆ, ,

G k k G k k

G m

m m m G m G m

IE 512 Chapter 2 11

Erlang Distribution

If failure and/or repair times are not exponential, can fit an Erlang distribution by matching moments:

Big advantage: Can still model as a CTMC.

Consider time to machine failure (each machine) as Erlangk. Can think in terms of k phases in the time to failure, where the time the m/c spends in each phase is exponential (k):

Mean time spent in each phase =

Mean total time to failure =

2 2Solve , simultaneously for and X k S k k

1

k

1 1k

k

IE 512 Chapter 2 12

Expanded State Definition

Mi(t) = # of machines operating in phase i at time t

For example, if k = 2, then a single machine without interference follows the CTMC (1 = failed state):

1

1 2

1 2 1

(so )

and , ,..., is a CTMC with

, ,..., ,0 , 1,..., and 0

k

ii

k

k

k i ii

N t m M t

M t M t M t

S l l l l m i k l m

10;1 0;2

IE 512 Chapter 2 13

1 2, Pr machines are operating in phase , 1, 2ip l l l i i

Transitions among States (k=2)

Steady state probabilities:

Rate into state

1 2 1 2 1 2Say : Rate out of state , is 2 2l l l l l l l l

1 2,l l 1 21,l l

1 21, 1l l

1 2, 1l l

2(l1+1)

2(l2+1)

IE 512 Chapter 2 14

Balance Equations

This system of equations (for any k) has the solution:

1 2 1 2 1 1 2

2 1 2

1 2 1 2 1 1 2

0 0,0 2 0,1

2 , 1, 2 1 1, 11 1

2 1 , 1

2 , 1, 2 1 1, 1

l p p

l p l l p l l l p l ll m

l p l l

l m mp l l p l l l p l l

1 21 2

, ,..., 0,0,...0! ! !

l

k

kk

p l l l pl l l

IE 512 Chapter 2 15

SS Number of Machines Working

From the previous equation and

get

Find probabilities by normalizing to 1.

This distribution is independent of k or any other characteristics of the failure time distribution. It can be shown that the same state distribution holds for any failure time distribution!

1 2

1 2, ,...,k

kl l l

p l p l l l

(probability that machines are working)

= 0 (Note: typo in (2.47), p.33 text)!

m l

l

p l P l

pl