Markov Processes

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Data Mining and Forecast Management

MGMT E - 5070

Machine Operation ProblemA manufacturing firm has developed a transition matrix containingA manufacturing firm has developed a transition matrix containing

the probabilities that a particular machine will operate or break down the probabilities that a particular machine will operate or break down in the following week, given its operating condition in the present week.in the following week, given its operating condition in the present week.

REQUIREMENT:REQUIREMENT:

Assuming that the machine is operating in week 1, that is, the initial state is ( .4 , .6 ) :Assuming that the machine is operating in week 1, that is, the initial state is ( .4 , .6 ) :

1.1. Determine the probabilities that the machine will operate or break down in weeksDetermine the probabilities that the machine will operate or break down in weeks 2, 3, 4, 5, and 6.2, 3, 4, 5, and 6.2.2. Determine the steady-state probabilities for this transition matrix algebraically and Determine the steady-state probabilities for this transition matrix algebraically and indicate the percentage of future weeks in which the machine will break down.indicate the percentage of future weeks in which the machine will break down.

Problem 1

Machine Operation Problem

.16 .24.16 .24

.4 .6.4 .6

.8 .2.8 .2

.48 .12.48 .12

.64 .36 Week No. 2

( .4 , .6 )( .4 , .6 )

Machine Operation Problem

.256 .384.256 .384

.4 .6.4 .6

.8 .2.8 .2

.288 .072.288 .072

.544 .456 Week No. 3

( .64 , .36 )( .64 , .36 )

Machine Operation Problem

.2176 .3264.2176 .3264

.4 .6.4 .6

.8 .2.8 .2

.3648 .0912.3648 .0912

.5824 .4176 Week No. 4

( .544 , .456 )( .544 , .456 )

Machine Operation Problem

.23296 .34944.23296 .34944

.4 .6.4 .6

.8 .2.8 .2

.33408 .08352.33408 .08352

.56704 .43296 Week No. 5

( .5824 , .4176 )( .5824 , .4176 )

Machine Operation Problem

.226816 .340224.226816 .340224

.4 .6.4 .6

.8 .2.8 .2

.346368 .086592.346368 .086592

.57384 .426816 Week No. 6

( .56704 , .43296 )( .56704 , .43296 )

Machine Operation Problem

.4X.4X11 .6X .6X11

.8X.8X22 .2X .2X22

P(O) = 1XP(O) = 1X11 P(B) = 1X P(B) = 1X22

OPERATE BREAKDOWN

P (O) = .4XP (O) = .4X11 + .8X + .8X22 = 1X = 1X11 ( (dependent equation))

P (B) = .6XP (B) = .6X11 + .2X + .2X22 = 1X = 1X2 2 ((dependent equation))

1X1X11 + 1X + 1X22 = 1 ( = 1 (independent equation))

Machine Operation Problem

.6X.6X11 + .2X + .2X22 – 1.0X – 1.0X22 = 0 = 0

becomes……becomes……

.6X.6X11 - .8X - .8X22 = 0 = 0

.4X.4X11 + .8X + .8X22 – 1.0X – 1.0X11 = 0 = 0

becomes……becomes……

- .6X- .6X11 + .8X + .8X22 = 0 = 0

Setdependentequationsequal to

zero

Machine Operation Problem

.6X.6X11 - .8X - .8X22 = 0 = 0.6 ( 1X ( 1X11 + 1X + 1X22 = 1 ) = 1 ) .6X.6X11 + .6X + .6X22 = .6 = .6 -1.4X-1.4X22 = -.6 = -.6 XX22 = = .4285 = P ( = P ( BREAKDOWN BREAKDOWN ))

Since XSince X11 + X + X22 = 1, then: = 1, then:

1 – X1 – X22 = X = X11

1 - .4285 = 1 - .4285 = .5715 = P ( = P ( OPERATION OPERATION ))

STEADY-STATE PROBABILITIES

Newspaper ProblemNewspaper Problem

A city is served by two newspapers – The Tribune and the Daily News. Each Sunday, readers purchase one of the newspapers at a stand. The following transition matrix contains the probabilities of a customer’s buying a particular newspaper in a week, given the newspaper purchased the previous Sunday.

Problem 2

Newspaper ProblemNewspaper Problem

REQUIREMENT:

1. Determine the steady-state probabilities for the transition matrix algebraically, and explain what they mean.

Newspaper Problem

.65 X.65 X11 .35 X .35 X11

.45 X.45 X22 .55 X .55 X22

P(T) = XP(T) = X11 P(DN) = X P(DN) = X22

Tribune Daily News

TribuneDaily News

Newspaper Problem

P ( T ) = .65XP ( T ) = .65X11 + .45X + .45X22 = 1X = 1X11 ( ( dependent equation ))

P ( DN ) = .35XP ( DN ) = .35X11 + .55X + .55X22 = 1X = 1X22 ( ( dependent equation ))

1X1X11 + 1X + 1X22 = 1 ( = 1 ( independent equation ))

Newspaper ProblemNewspaper Problem

.65X1 + .45X2 = 1X1

.65X1 + .45X2 – 1X1 = 0

- .35X1 + .45X2 = 0

.35X1 + .55X2 = 1X2

.35X1 + .55X2 – 1X2 = 0

.35X1 - .45X2 = 0

Newspaper ProblemNewspaper ProblemSTEADY - STATE PROBABILITIES

.35X.35X11 - .45X - .45X22 = 0 = 0

.35 ( 1X( 1X11 + 1X + 1X22 = 1 ) = 1 )

.35X.35X11 + .35X + .35X22 = .35 = .35 - .80X- .80X22 = - .35 = - .35 XX22 = = .4375 = P ( = P ( Daily NewsDaily News ) )

Since XSince X11 + X + X22 = 1, then: = 1, then:

1 – X1 – X22 = X = X11

1 - .4375 = 1 - .4375 = .5625 = P ( = P ( TribuneTribune ) )

Fertilizer ProblemFertilizer Problem

Problem 3

Fertilizer ProblemFertilizer ProblemPROBABILITY TRANSITION MATRIXPROBABILITY TRANSITION MATRIX

This Spring

Next Spring

Fertilizer ProblemFertilizer ProblemThe number of

customers presently

using each brand

of fertilizer is shown below:

Fertilizer ProblemFertilizer Problem

Fertilizer ProblemTransition Matrix

Plant Plus Crop Extra Gro Fast

.4 .3 .3.4 .3 .3 .5 .1 .4.5 .1 .4 .4 .2 .4.4 .2 .4

Fertilizer ProblemTransition Matrix

Plant Plus Crop Extra Gro Fast

.4X.4X11 .3X .3X11 .3X .3X11

.5X.5X2 2 .1X .1X22 .4X .4X22

.4X.4X33 .2X .2X33 .4X .4X33

P (PP) = 1XP (PP) = 1X11 P(CE) = 1X P(CE) = 1X22 P(GF) = 1X P(GF) = 1X33

Fertilizer ProblemTHE EQUATIONS

P (PP) = .4XP (PP) = .4X11 + .5X + .5X22 + .4X + .4X33 = 1X = 1X1 ( 1 ( DEPENDENT ) )

P (CE) = .3XP (CE) = .3X11 + .1X + .1X22 + .2X + .2X33 = 1X = 1X2 ( 2 ( DEPENDENT ))

P (GF) = .3XP (GF) = .3X11 + .4X + .4X22 + .4X + .4X33 = 1X = 1X3 ( 3 ( DEPENDENT ) )

1X1X11 + 1X + 1X22 + 1X + 1X33 = 1 = 1 ( ( INDEPENDENT ) )

Fertilizer Problem

P (PP) = .4X1 + .5X2 + .4X3 - 1.0X1 = 0

P (CE) = .3X1 + .1X2 + .2X3 - 1.0X2 = 0

P (GF) = .3X1 + .4X2 + .4X3 – 1.0X3 = 0

1X1 + 1X2 + 1X3 = 1 ( INDEPENDENT )

Setdependentequationsequal to

zero

Fertilizer ProblemFertilizer Problem

P (PP) = - .6XP (PP) = - .6X11 + .5X + .5X22 + .4X + .4X33 = 0 = 0

P (CE) = .3XP (CE) = .3X11 - .9X - .9X22 + .2X + .2X33 = 0 = 0

P (GF) = .3XP (GF) = .3X11 + .4X + .4X22 - .6X - .6X33 = 0 = 0

Setdependentequationsequal to

zero

Fertilizer Problem

.3X1 - .9X2 + .2X3 = 0

.3X1 + .4X2 - .6X3 = 0

- 1.3X2 + .8X3 = 0

.3 ( 1X1 + 1X2 + 1X3 = 1.0 ) .3X1 + .3X2 + .3X3 = .3 .3X1 + .4X2 - .6X3 = 0

- .1X2 + .9X3 = .3

CANCEL OUT VARIABLE X1

Fertilizer ProblemFertilizer ProblemCANCEL OUT VARIABLE X2

- 1.3X2 + .8X3 = 0 -13 ( .1X2 + .9X3 = .3 ) - 1.3X2 + 11.7X3 = - 3.9

- 10.9X3 = - 3.9 X3 = .357798

Fertilizer Problem

-1.3X2 + .8 ( .358 ) = 0 - 1.3 X2 = - .286 X2 = .220

X1 + .220 + .358 = 1.0 X1 = 1.0 - .578 X1 = .422

SOLVING FOR THE REMAINING VARIABLES

Fertilizer ProblemFertilizer Problem

ΣΣ = 9,000 1.00 = 9,000 1.00 9,0009,000

Markov Processes

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