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MGMT E-5070. Markov Processes. Homework Solution. Manual Computer-Based. Machine Operation Problem. A manufacturing firm has developed a transition matrix containing the probabilities that a particular machine will operate or break down
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MGMT E-5070 Markov Processes Homework Solution Manual Computer-Based
Machine Operation Problem A manufacturing firm has developed a transition matrix containing the probabilities that a particular machine will operate or break down in the following week, given its operating condition in the present week. • REQUIREMENT: • Assuming that the machine is operating in week 1, that is, the initial state is ( .4 , .6 ) : • Determine the probabilities that the machine will operate or break down in weeks • 2, 3, 4, 5, and 6. • Determine the steady-state probabilities for this transition matrix algebraically and • indicate the percentage of future weeks in which the machine will break down.
Machine Operation Problem .16 .24 .4 .6 .8 .2 .48 .12 .64.36 Week No. 2 ( .4 , .6 )
Machine Operation Problem .256 .384 .4 .6 .8 .2 .288 .072 .544.456 Week No. 3 ( .64 , .36 )
Machine Operation Problem .2176 .3264 .4 .6 .8 .2 .3648 .0912 .5824.4176 Week No. 4 ( .544 , .456 )
Machine Operation Problem .23296 .34944 .4 .6 .8 .2 .33408 .08352 .56704.43296 Week No. 5 ( .5824 , .4176 )
Machine Operation Problem .226816 .340224 .4 .6 .8 .2 .346368 .086592 .57384.426816 Week No. 6 ( .56704 , .43296 )
Machine Operation Problem OPERATEBREAKDOWN .4X1 .6X1 .8X2 .2X2 P(O) = 1X1 P(B) = 1X2 P (O) = .4X1 + .8X2 = 1X1 (dependent equation) P (B) = .6X1 + .2X2 = 1X2 (dependent equation) 1X1 + 1X2 = 1 (independent equation)
Machine Operation Problem SET DEPENDENT EQUATIONS EQUAL TO ZERO .4X1 + .8X2 – 1.0X1 = 0 becomes…… - .6X1 + .8X2 = 0 .6X1 + .2X2 – 1.0X2 = 0 becomes…… .6X1 - .8X2 = 0
Machine Operation Problem STEADY-STATE PROBABILITIES .6X1 - .8X2 = 0 .6 ( 1X1 + 1X2 = 1 ) .6X1 + .6X2 = .6 -1.4X2 = -.6 X2 = .4285 = P ( BREAKDOWN ) Since X1 + X2 = 1, then: 1 – X2 = X1 1 - .4285 = .5715 = P ( OPERATION )
Newspaper 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.
Newspaper Problem REQUIREMENT: • Determine the steady-state probabilities for the transition matrix algebraically, and explain what they mean.
Newspaper Problem Tribune Daily News .65 X1 .35 X1 .45 X2 .55 X2 P(T) = X1 P(DN) = X2 Tribune Daily News
Newspaper Problem P ( T ) = .65X1 + .45X2 = 1X1 ( dependent equation) P ( DN ) = .35X1 + .55X2 = 1X2 ( dependent equation ) 1X1 + 1X2 = 1 ( independent equation )
Newspaper Problem SET DEPENDENT EQUATIONS EQUAL TO ZERO .65X1 + .45X2 = 1X1 .65X1 + .45X2 – 1X1 = 0 - .35X1 + .45X2 = 0 .35X1 + .55X2 = 1X2 .35X1 + .55X2 – 1X2 = 0 .35X1 - .45X2 = 0
Newspaper Problem STEADY - STATE PROBABILITIES .35X1 - .45X2 = 0 .35 ( 1X1 + 1X2 = 1 ) .35X1 + .35X2 = .35 - .80X2 = - .35 X2 = .4375 = P ( Daily News ) Since X1 + X2 = 1, then: 1 – X2 = X1 1 - .4375 = .5625 = P ( Tribune )
Fertilizer Problem In Westville, a small rural town in Maine, virtually all shopping and business is done in the town. The town has one farm and garden center that sells fertilizer 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 three brands. The garden center has 9,000 customers for fertilizer each spring. An extensive market research study has determined that customers switch brands of fertilizer according to the following probability transition matrix.
Fertilizer Problem PROBABILITY TRANSITION MATRIX NEXT SPRING THIS SPRING
Fertilizer Problem The number of customers presently using each brand of fertilizer is shown below:
Fertilizer Problem REQUIREMENT: • Determine the steady-state probabilities for the fertilizer brands. • Forecast the customer demand for each brand of fertilizer in the long run and the changes in customer demand.
Fertilizer Problem Transition Matrix Plant Plus Crop Extra Gro Fast .4 .3 .3 .5 .1 .4 .4 .2 .4
Fertilizer Problem Transition Matrix Plant Plus Crop Extra Gro Fast .4X1 .3X1 .3X1 .5X2 .1X2 .4X2 .4X3 .2X3 .4X3 P (PP) = 1X1 P(CE) = 1X2 P(GF) = 1X3
Fertilizer Problem THE EQUATIONS P (PP) = .4X1 + .5X2 + .4X3 = 1X1 ( DEPENDENT ) P (CE) = .3X1 + .1X2 + .2X3 = 1X2 ( DEPENDENT ) P (GF) = .3X1 + .4X2 + .4X3 = 1X3 ( DEPENDENT ) 1X1 + 1X2 + 1X3 = 1 ( INDEPENDENT )
Fertilizer Problem SET DEPENDENT EQUATIONS EQUAL TO ZERO 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 )
Fertilizer Problem SET DEPENDENT EQUATIONS EQUAL TO ZERO P (PP) = - .6X1 + .5X2 + .4X3 = 0 P (CE) = .3X1 - .9X2 + .2X3 = 0 P (GF) = .3X1 + .4X2 - .6X3 = 0
Fertilizer Problem CANCEL OUT VARIABLE X1 .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
Fertilizer Problem CANCEL 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 SOLVING FOR THE REMAINING VARIABLES • 1.3X2 + .8 ( .358 ) = 0 • - 1.3 X2 = - .286 • X2 = .220 X1 + .220 + .358 = 1.0 X1 = 1.0 - .578 X1 = .422
Fertilizer Problem Σ = 9,000 1.0009,000