1. Markov Processes Homework Solution MGMT E - 5070
Data Mining and Forecast Management
Markov Processes
MGMT E
Homework Solution
Manual
Computer
Based

2. 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.
Problem 1

3. Machine Operation Problem
Week No. 2
( .4 , .6 )

4. Machine Operation Problem
Week No. 3
( .64 , .36 )

5. Machine Operation Problem
Week No. 4
( .544 , .456 )

6. Machine Operation Problem
Week No. 5
( , )

7. Machine Operation Problem
Week No. 6
( , )

8. Machine Operation Problem
OPERATE BREAKDOWN
.4X X1
.8X X2
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)

9. 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

10. Machine Operation Problem
STEADY-STATE PROBABILITIES
.6X1 - .8X2 = 0
.6 ( 1X1 + 1X2 = 1 )
.6X1 + .6X2 = .6
-1.4X2 = -.6
X2 = = P ( BREAKDOWN )
Since X1 + X2 = 1, then:
1 – X2 = X1
= = P ( OPERATION )

15. Newspaper Problem Problem 2
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.

16. Newspaper Problem REQUIREMENT:
Determine the steady-state probabilities for the transition
matrix algebraically, and explain what they mean.

17. Newspaper Problem .65 X1 .35 X1 .45 X2 .55 X2 P(T) = X1 P(DN) = X2
Tribune Daily News
.65 X X1
.45 X X2
P(T) = X1 P(DN) = X2
Tribune
Daily News

18. Newspaper Problem P ( T ) = .65X1 + .45X2 = 1X1 ( dependent equation )
P ( DN ) = .35X X2 = 1X2 ( dependent equation )
1X1 + 1X2 = 1 ( independent equation )

19. Newspaper Problem .65X1 + .45X2 = 1X1 .65X1 + .45X2 – 1X1 = 0

20. Newspaper Problem .35X1 - .45X2 = 0 .35 ( 1X1 + 1X2 = 1 )
STEADY - STATE PROBABILITIES
.35X X2 = 0
.35 ( 1X1 + 1X2 = 1 )
.35X X2 = .35
- .80X2 = - .35
X2 = = P ( Daily News )
Since X1 + X2 = 1, then:
1 – X2 = X1
= = P ( Tribune )

25. Fertilizer Problem
Problem 3

26. Fertilizer Problem PROBABILITY TRANSITION MATRIX Next Spring
This Spring

27. Fertilizer Problem The number of customers presently using each brand
of fertilizer
is shown
below:

28. Fertilizer Problem

29. Fertilizer Problem Plant Plus Crop Extra Gro Fast .4 .3 .3 .5 .1 .4
Transition Matrix
Plant Plus Crop Extra Gro Fast

30. Fertilizer Problem Plant Plus Crop Extra Gro Fast .4X1 .3X1 .3X1
Transition Matrix
Plant Plus Crop Extra Gro Fast
.4X X X1
.5X X X2
.4X X X3
P (PP) = 1X1 P(CE) = 1X2 P(GF) = 1X3

31. Fertilizer Problem P (PP) = .4X1 + .5X2 + .4X3 = 1X1 ( DEPENDENT )
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 = ( INDEPENDENT )

32. Fertilizer Problem P (PP) = .4X1 + .5X2 + .4X3 - 1.0X1 = 0
Set
dependent
equations
equal to
zero
P (PP) = .4X1 + .5X2 + .4X X1 = 0
P (CE) = .3X1 + .1X2 + .2X X2 = 0
P (GF) = .3X1 + .4X2 + .4X3 – 1.0X3 = 0
1X1 + 1X2 + 1X3 = 1 ( INDEPENDENT )

33. Fertilizer Problem P (PP) = - .6X1 + .5X2 + .4X3 = 0
Set
dependent
equations
equal to
zero
P (PP) = - .6X1 + .5X2 + .4X3 = 0
P (CE) = .3X X2 + .2X3 = 0
P (GF) = .3X X X3 = 0

34. Fertilizer Problem .3X1 - .9X2 + .2X3 = 0 .3X1 + .4X2 - .6X3 = 0
CANCEL OUT VARIABLE X1
.3X X2 + .2X3 = 0
.3X1 + .4X X3 = 0
- 1.3X2 + .8X3 = 0
.3 ( 1X1 + 1X2 + 1X3 = 1.0 )
.3X1 + .3X2 + .3X3 = .3
.3X1 + .4X X3 = 0
- .1X2 + .9X3 = .3

35. Fertilizer Problem - 1.3X2 + .8X3 = 0 -13 ( .1X2 + .9X3 = .3 )
CANCEL OUT VARIABLE X2
- 1.3X X3 = 0
-13 ( .1X X3 = .3 )
- 1.3X X3 = - 3.9
- 10.9X3 = - 3.9
X3 =

36. Fertilizer Problem 1.3X2 + .8 ( .358 ) = 0 - 1.3 X2 = - .286 X2 = .220
SOLVING FOR THE REMAINING VARIABLES
1.3X ( .358 ) = 0
- 1.3 X2 =
X2 = .220
X = 1.0
X1 =
X1 = .422

37. Fertilizer Problem
Σ = 9, ,000

42. Markov Processes
Homework Solution
Manual
Computer
Based