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MGMT E-5070

Machine Operation ProblemMachine 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.

This Week Next Week Operate Next Week Breakdown

Operate .4 .6Break Down .8 .2

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.

Machine Operation ProblemMachine Operation Problem

.16 .24.16 .24

.4 .6.4 .6

.8 .2.8 .2

.48 .12.48 .12

.64.64 .36.36 Week No. 2 Week No. 2

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

Machine Operation ProblemMachine Operation Problem

.256 .384.256 .384

.4 .6.4 .6

.8 .2.8 .2

.288 .072.288 .072

.544.544 .456.456 Week No. 3 Week No. 3

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

Machine Operation ProblemMachine Operation Problem

.2176 .3264.2176 .3264

.4 .6.4 .6

.8 .2.8 .2

.3648 .0912.3648 .0912

.5824.5824 .4176.4176 Week No. 4 Week No. 4

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

Machine Operation ProblemMachine Operation Problem

.23296 .34944.23296 .34944

.4 .6.4 .6

.8 .2.8 .2

.33408 .08352.33408 .08352

.56704.56704 .43296 .43296 Week No. 5 Week No. 5

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

Machine Operation ProblemMachine Operation Problem

.226816 .340224.226816 .340224

.4 .6.4 .6

.8 .2.8 .2

.346368 .086592.346368 .086592

.57384.57384 .426816.426816 Week No. 6 Week No. 6

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

Machine Operation ProblemMachine Operation Problem

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

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

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

OPERATEOPERATE BREAKDOWNBREAKDOWN

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

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

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

Machine Operation ProblemMachine Operation Problem

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

becomes……becomes……

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

SET DEPENDENT EQUATIONS EQUAL TO ZEROSET DEPENDENT EQUATIONS EQUAL TO ZERO

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

becomes……becomes……

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

Machine Operation ProblemMachine Operation Problem

.6X.6X11 - .8X - .8X22 = 0 = 0.6.6 ( 1X ( 1X11 + 1X + 1X22 = 1 ) = 1 ) .6X.6X11 + .6X + .6X22 = .6 = .6 -1.4X-1.4X22 = -.6 = -.6 XX22 = = .4285.4285 = P ( = P ( BREAKDOWN BREAKDOWN ))

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

1 – X1 – X22 = X = X11

1 - .4285 = 1 - .4285 = .5715.5715 = P ( = P ( OPERATION OPERATION ))

STEADY-STATE PROBABILITIESSTEADY-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.

NewspNewspaper Problemaper Problem

( This Sunday )( This Sunday )TribuneTribune

( Next Sunday )( Next Sunday )

Daily NewsDaily News( Next Sunday )( Next Sunday )

TribuneTribune .65 .35

Daily NewsDaily News .45 .55REQUIREMENT:

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

Newspaper ProblemNewspaper 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 Tribune Daily NewsDaily News

TribuneTribuneDaily NewsDaily News

Newspaper ProblemNewspaper Problem

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

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

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

Newspaper ProblemNewspaper ProblemSET DEPENDENT EQUATIONS EQUAL TO ZEROSET DEPENDENT EQUATIONS EQUAL TO ZERO

.65X.65X11 + .45X + .45X22 = 1X = 1X11

.65X.65X11 + .45X + .45X22 – 1X – 1X11 = 0 = 0

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

.35X.35X11 + .55X + .55X22 = 1X = 1X22

.35X.35X11 + .55X + .55X22 – 1X – 1X22 = 0 = 0

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

Newspaper Newspaper PrProblemoblemSTEADY - STATE PROBABILITIESSTEADY - STATE PROBABILITIES

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

.35.35 ( 1X ( 1X11 + 1X + 1X22 = 1 ) = 1 )

.35X.35X11 + .35X + .35X22 = .35 = .35 - .80X- .80X22 = - .35 = - .35 XX22 = = .4375.4375 = P ( = P ( Daily NewsDaily News ) )

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

1 – X1 – X22 = X = X11

1 - .4375 = 1 - .4375 = .5625.5625 = P ( = P ( TribuneTribune ) )

Fertilizer ProblemFertilizer Problem

In Westville, a small rural town in Maine, virtually allshopping and business is done in the town. The town has one farm and garden center that sellsfertilizer to the local farmers and gardeners.The center carries three brands of fertilizer – Plant Plus, Crop Extra, and Gro-fast - so every person in the town who uses fertilizer uses one of the threebrands.The garden center has 9,000 customers for fertilizereach spring. An extensive market research study has determined that customers switch brands of fertilizer according to the following probability transition matrix.

Fertilizer ProblemFertilizer Problem

Plant Plus Crop Extra Gro-Fast

Plant Plus .4 .3 .3

Crop Extra .5 .1 .4

Gro-Fast .4 .2 .4

NEXT SPRINGNEXT SPRING

THIS SPRING

PROBABILITY TRANSITION MATRIXPROBABILITY TRANSITION MATRIX

Fertilizer ProblemFertilizer Problem

The number of customers presently

using each brand of fertilizer is shown

below:

BrandBrand CustomersCustomers

Plant Plus 3,000

Crop Extra 4,000

Gro-Fast 2,000

Fertilizer ProblemFertilizer Problem

REQUIREMENT:

1. Determine the steady-state probabilities for the fertilizer brands.

2. Forecast the customer demand for each brand of fertilizer in the long run and the changes in customer demand.

Fertilizer ProblemFertilizer ProblemTransition MatrixTransition Matrix

Plant Plus Crop Extra Gro FastPlant Plus Crop Extra Gro Fast

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

Fertilizer ProblemFertilizer ProblemTransition MatrixTransition Matrix

Plant Plus Crop Extra Gro FastPlant 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 ProblemFertilizer ProblemTHE EQUATIONSTHE EQUATIONS

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

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

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

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

Fertilizer ProblemFertilizer ProblemSET DEPENDENT EQUATIONS EQUAL TO ZEROSET DEPENDENT EQUATIONS EQUAL TO ZERO

P (PP) = .4XP (PP) = .4X11 + .5X + .5X22 + .4X + .4X33 - 1.0X - 1.0X11 = 0 = 0

P (CE) = .3XP (CE) = .3X11 + .1X + .1X22 + .2X + .2X33 - 1.0X - 1.0X22 = 0 = 0

P (GF) = .3XP (GF) = .3X11 + .4X + .4X22 + .4X + .4X33 – 1.0X – 1.0X33 = 0 = 0

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

Fertilizer ProblemFertilizer Problem

SET DEPENDENT EQUATIONS EQUAL TO ZEROSET DEPENDENT EQUATIONS EQUAL TO ZERO

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

Fertilizer ProblemFertilizer Problem

.3X.3X11 - .9X - .9X22 + .2X + .2X33 = 0 = 0

.3X.3X11 + .4X + .4X22 - .6X - .6X33 = 0 = 0

- 1.3X- 1.3X22 + .8X + .8X33 = 0 = 0

.3.3 ( 1X ( 1X11 + 1X + 1X22 + 1X + 1X33 = 1.0 ) = 1.0 ) .3X.3X11 + .3X + .3X22 + .3X + .3X33 = .3 = .3 .3X.3X11 + .4X + .4X22 - .6X - .6X33 = 0 = 0

- .1X- .1X22 + .9X + .9X33 = .3 = .3

CANCEL OUT VARIABLE XCANCEL OUT VARIABLE X11

Fertilizer ProblemFertilizer ProblemCANCEL OUT VARIABLE XCANCEL OUT VARIABLE X22

- 1.3X- 1.3X22 + .8X + .8X33 = 0 = 0 -13-13 ( .1X ( .1X22 + .9X + .9X33 = .3 ) = .3 ) - 1.3X- 1.3X22 + 11.7X + 11.7X33 = - 3.9 = - 3.9

- 10.9X- 10.9X33 = - 3.9 = - 3.9 XX33 = = .357798.357798

Fertilizer ProblemFertilizer Problem

-1.3X1.3X22 + .8 ( + .8 ( .358.358 ) = 0 ) = 0 - 1.3 X- 1.3 X22 = - .286 = - .286 XX22 = = .220.220

XX11 + + .220.220 + + .358.358 = 1.0 = 1.0 XX11 = 1.0 - .578 = 1.0 - .578 XX11 = = .422.422

SOLVING FOR THE REMAINING VARIABLESSOLVING FOR THE REMAINING VARIABLES

Fertilizer ProblemFertilizer Problem

Fertilizer

Brand

Present

Customers

Long-Term

Market Share

Long-Term

Customers

Plant Plus 3,000 .422 3,798

Crop Extra 4,000 .220 1,980

Gro-Fast 2,000 .358 3,222

ΣΣ = 9,000 1.000 = 9,000 1.000 9,0009,000