Markov processes example 1997 UG exam
In analysing switching by Business Class customers between airlines the following data has been obtained by British Airways (BA):
Next flight by
BA Competition
Last flight by BA 0.85 0.15
Competition 0.10 0.90For example if the last flight by a Business Class customer was by BA the probability that their next flight is by BA is 0.85. Business Class customers make 2 flights a year on average.
Currently BA have 30% of the Business Class market. What would you forecast BA’s share of the Business Class market to be after two years?
Solution
We have the initial system state s1 given by s1 = [0.30, 0.70] and the transition matrix P is given by
P = | 0.85 0.15 |
2
= | 0.7375 0.2625 |
| 0.10 0.90 | | 0.1750 0.8250 |where the square term arises as Business Class customers make 2 flights a year on average.
Hence after one year has elapsed the state of the system s2 = s1P = [0.34375, 0.65625]
After two years have elapsed the state of the system = s3 = s2P = [0.368, 0.632]
and note here that the elements of s2 and s3 add to one (as required).
So after two years have elapsed BA’s share of the Business Class market is 36.8%
Markov processes example 1996 UG exam
An admissions tutor is analysing applications from potential students for a particular undergraduate course at Imperial College (IC). She regards each potential student as being in one of four possible states:
- State 1: has not applied to IC
- State 2: has applied to IC but an accept/reject decision has not yet been made
- State 3: has applied to IC and has been rejected
- State 4: has applied to IC and has been accepted (been made an offer of a place)
At the start of the year (month 1 in the admissions year) all potential students are in state 1.
Her review of admissions statistics for recent years has identified the following transition matrix for the probability of moving between states each month:
To 1 2 3 4
From 1 | 0.97 0.03 0 0 |
2 | 0 0.10 0.15 0.75 |
3 | 0 0 1 0 |
4 | 0 0 0 1 |- What percentage of potential students will have been accepted after 3 months have elapsed?
- Is it possible to work out a meaningful long-run system state or not (and why)?
The admissions tutor has control over the elements in one row of the above transition matrix, namely row 2.
The elements in this row reflect:
- transition from 2 to 2: the speed with which applications are processed each month
- transition from 2 to 3: the proportion of applicants who are rejected each month
- transition from 2 to 4: the proportion of applicants who are accepted each month.
To be more specific, at the start of each month the admissions tutor has to decide the proportion of applicants who should be accepted that month. However she is constrained by a policy decision that, at the end of each month, the total number of rejections should never be more than one-third of the total number of offers, nor should it ever be less than 20% of the total number of offers.
Further analysis reveals that applicants who wait longer than 2 months between applying to IC and receiving a decision (reject or accept) almost never choose to come to IC, even if they get an offer of a place.
Formulate the problem that the admissions tutor faces each month as a linear program. Comment on any assumptions you have made in so doing.
Solution
We have the initial system state s1 given by s1 = [1, 0, 0, 0] and the transition matrix P given by
P = | 0.97 0.03 0 0 |
| 0 0.10 0.15 0.75 |
| 0 0 1 0 |
| 0 0 0 1 |Hence after one month has elapsed the state of the system s2 = s1P = [0.97, 0.03, 0, 0]
After two months have elapsed the state of the system = s3 = s2P = [0.9409, 0.0321, 0.0045, 0.0225]
After three months have elapsed the state of the system = s4 = s3P = [0.912673, 0.031437, 0.009315, 0.046575]
and note here that the elements of s2, s3 and s4 add to one (as required).
Hence 4.6575% of potential students will have been accepted after 3 months have elapsed.
It is not possible to work out a meaningful long-run system state as the admissions year is only (at most) 12 months long. In reality the admissions year is probably shorter than 12 months.
With regard to the linear program we must distinguish within state 2 (those who have applied to IC but an accept/reject decision has not yet been made) how long an applicant has been waiting.
Hence expand state 2 to states:
- 2a – a new application received
- 2b – a new application received one month ago
- In this way we never leave a new application waiting longer than 2 months – applicants in this category almost never come to IC anyway.
Hence we have the new transition matrix
1 2a 2b 3 4
P = 1 | 0.97 0.03 0 0 0 |
2a | 0 0 1-X-Y X Y |
2b | 0 0 0 1-y y |
3 | 0 0 0 1 0 |
4 | 0 0 0 0 1 |Here X is the reject probability each month for a newly received application and Y is be the acceptance probability each month for a newly received application (these are decision variables for the admissions tutor), where X>=0 and Y>=0.
In a similar fashion y is the acceptance probability each month for an application that was received one month ago (again a decision variable for the admissions tutor).
Each month then, at the start of the month, we have a known proportion in each of the states 1, 2a, 2b, 3 and 4.
Hence the equation for the (unknown) proportions [z1,z2a,z2b,z3,z4] at the end of each month is given by:
- [z1,z2a,z2b,z3,z4] = [known proportions at start of month]P
where P is the transition matrix given above involving the variables X,Y and y. If we were to write this matrix equation out in full we would have 5 linear equalities. In addition we must have that:
- z1+z2a+z2b+z3+z4 = 1
- z1,z2a,z2b,z3,z4 >= 0
and the policy conditions are:
- z3 <= z4/3
- z3 >= 0.2z4
Hence we have a set of linear constraints in the variables [X,Y,y,z1,z2a,z2b,z3,z4]
An appropriate objective function might be to maximise the sum of the acceptance probabilities (Y+y) but other objectives could be suggested for this system.
Hence we have an LP that can be solved to decide X,Y and y each month.
Comments are:
- row 1 of the transition matrix constant throughout the year
- does not take into account any information we might have on how applicants respond to the offers made to them
Markov processes example 1993 UG exam
A petrol station owner is considering the effect on his business (Superpet) of a new petrol station (Global) which has opened just down the road. Currently (of the total market shared between Superpet and Global) Superpet has 80% of the market and Global has 20%.
Analysis over the last week has indicated the following probabilities for customers switching the station they stop at each week:
To
Superpet Global
From Superpet 0.75 0.25
Global 0.55 0.45- What will be the expected market share for Superpet and Global after another two weeks have past?
- What would be the long-run prediction for the expected market share for Superpet and Global?
Solution
Letting
- state 1 = Superpet
- state 2 = Global
we have the initial system state s1 given by s1 = [0.80, 0.20] and the transition matrix P given by
P = |0.75 0.25 |
|0.55 0.45 |Hence after one week has elapsed the state of the system s2 = s1P = [0.71, 0.29] and so after two weeks have elapsed the state of the system = s3 = s2P = [0.692, 0.308] and note here that the elements of s2 and s3 add to one (as required).
Hence the market shares after two weeks have elapsed are 69.2% and 30.8% for Superpet and Global respectively.
Assuming that in the long-run the system reaches an equilibrium [x1, x2] where
[x1, x2] = [x1, x2]P and x1 + x2 = 1
we have that
x
1
= 0.75x
1
+ 0.55x
2
(1)
x
2
= 0.25x
1
+ 0.45x
2
(2)
and x
1
+ x
2
= 1 (3)
From (3) we have that x2 = 1-x1
so substituting into (1) we get
x1 = 0.75x1 + 0.55(1-x1)
i.e. (1-0.75+0.55)x1 = 0.55
i.e. x1 = 0.55/0.80 = 0.6875
Hence x2 = 1-x1 = 1-0.6875 = 0.3125
Note that as a check we have that these values for x1 and x2 satisfy equations (1) – (3) (to within rounding errors).
Hence the long-run market shares are 68.75% and 31.25% for Superpet and Global respectively.
Markov processes example 1991 UG exam
A company is considering using Markov theory to analyse brand switching between three different brands of floppy disks. Survey data has been gathered and has been used to estimate the following transition matrix for the probability of moving between brands each month:
To Brand
1 2 3
From Brand 1 | 0.80 0.10 0.10
2 | 0.03 0.95 0.02
3 | 0.20 0.05 0.75The current (month 1) market shares are 45%, 25% and 30% for brands 1, 2 and 3 respectively.
- What will be the expected market shares after two months have elapsed (i.e. in month 3)?
- What is the long-run prediction for the expected market share for each of the three brands?
- Would you expect the actual market share to approach the long-run prediction for the market or not (and why)?
Solution
We have the initial system state s1 given by s1 = [0.45, 0.25, 0.30] and the transition matrix P given by
P = |0.80 0.10 0.10 |
|0.03 0.95 0.02 |
|0.20 0.05 0.75 |Hence after one month has elapsed the state of the system s2 = s1P = [0.4275, 0.2975, 0.2750] and so after two months have elapsed the state of the system = s3 = s2P = [0.4059, 0.3391, 0.2550] and note here that the elements of s2 and s3 add to one (as required).
Hence the market shares after two months have elapsed are 40.59%, 33.91% and 25.50% for brands 1, 2 and 3 respectively.
Assuming that in the long-run the system reaches an equilibrium [x1, x2, x3] where
[x1, x2, x3] = [x1, x2, x3]P and x1 + x2 + x3 = 1
we have that
x1 = 0.80x1 + 0.03x2 + 0.20x3
x2 = 0.10x1 + 0.95x2 + 0.05x3
x3 = 0.10x1 + 0.02x2 + 0.75x3
x1 + x2 + x3 = 1
Rearranging we get
0.20x
1
= 0.03x
2
+ 0.20x
3
(1)
0.05x
2
= 0.10x
1
+ 0.05x
3
(2)
0.25x
3
= 0.10x
1
+ 0.02x
2
(3)
x
1
+ x
2
+ x
3
= 1 (4)
Now subtracting equation (3) from equation (2) we get
0.05x2 – 0.25x3 = 0.05x3 – 0.02x2
i.e. 0.07x2 = 0.30x3
i.e. x
2
= (0.30/0.07)x
3
(5)
Now from equation (4) we have
x1 = 1 – x2 – x3
so substituting in equation (1) we get
0.20(1 – x2 – x3) = 0.03x2 + 0.20x3
i.e. 0.20 = 0.23x2 + 0.40x3
Substituting from equation (5) for x2 we get
0.20 = 0.23(0.30/0.07)x3 + 0.40x3
i.e. x3 = 0.1443
Hence from equation (5)
x2 = (0.30/0.07)x3 = 0.6184
and x1 = 1 – x2 – x3 = 1 – 0.6184 – 0.1443 = 0.2373
Note that as a check we have that these values for x1, x2 and x3 satisfy equations (1) – (4) (to within rounding errors).
Hence the long-run market shares are 23.73%, 61.84% and 14.43% for brands 1, 2 and 3 respectively.
Have that after two months s3 (calculated above) not close to the long-run market shares and would expect changing circumstances to render the transition matrix invalid.
Markov processes example 1990 UG exam
In analysing switching between different brands of copper pipe survey data has been used to estimate the following transition matrix for the probability of moving between brands each month:
To Brand
1 2 3 4
From Brand 1 | 0.95 0.02 0.02 0.01
2 | 0.05 0.90 0.02 0.03
3 | 0.10 0.05 0.83 0.02
4 | 0.13 0.13 0.02 0.72The current (month 1) market shares are 45%, 23%, 20% and 12% for brands 1, 2, 3 and 4 respectively.
- What will be the expected market shares after two months have elapsed (i.e. in month 3)?
- What is the long-run prediction for the expected market share for each of the four brands?
- Would you expect the actual market share to approach the long-run prediction for the market or not (and why)?
- The management of brand 1 are concerned that they should be aiming for a long- run market share of 75% by manipulating the transition probabilities from brand 1 to brands 2, 3 and 4 (as well as the transition probability from brand 1 to brand 1). Formulate the problem of deciding appropriate transition probabilities as a linear program.
Solution
We have the initial system state s1 given by s1 = [0.45, 0.23, 0.20, 0.12] and the transition matrix P given by
P = | 0.95 0.02 0.02 0.01 |
| 0.05 0.90 0.02 0.03 |
| 0.10 0.05 0.83 0.02 |
| 0.13 0.13 0.02 0.72 |Hence after one month has elapsed the state of the system s2 = s1P = [0.4746, 0.2416, 0.1820, 0.1018] and so after two months have elapsed the state of the system = s3 = s2P = [0.494384, 0.249266, 0.16742, 0.08893] and note here that the elements of s2 and s3 add to one (as required).
Hence the market shares after two months have elapsed are 49.44%, 24.93%, 16.74% and 8.89% for brands 1, 2, 3 and 4 respectively.
Assuming that in the long-run the system reaches an equilibrium [x1, x2, x3, x4] where
[x1, x2, x3, x4] = [x1, x2, x3, x4]P and x1 + x2 + x3 + x4 = 1
we have that
x1 = 0.95x1 + 0.05x2 + 0.10x3 + 0.13x4
x2 = 0.02x1 + 0.90x2 + 0.05x3 + 0.13x4
x3 = 0.02x1 + 0.02x2 + 0.83x3 + 0.02x4
x4 = 0.01x1 + 0.03x2 + 0.02x3 + 0.72x4
x1 + x2 + x3 + x4 = 1
Rearranging we get
0.05x
1
= 0.05x
2
+ 0.10x
3
+ 0.13x
4
(1)
0.10x
2
= 0.02x
1
+ 0.05x
3
+ 0.13x
4
(2)
0.17x
3
= 0.02x
1
+ 0.02x
2
+ 0.02x
4
(3)
0.28x
4
= 0.01x
1
+ 0.03x
2
+ 0.02x
3
(4)
x
1
+ x
2
+ x
3
+ x
4
= 1 (5)
Now from equation (3) we have
0.17x3 = 0.02(x1 + x2 + x4)
and from equation (5) we have
x1 + x2 + x4 = 1 – x3
Hence
0.17x3 = 0.02(1-x3)
i.e. 0.19x3 = 0.02
i.e. x3 = (0.2/0.19) = 0.10526
Now subtracting equation (2) from equation (1) we get
0.05x1 – 0.10x2 = 0.05x2 + 0.10x3 – 0.02x1 – 0.05x3
i.e. 0.07x
1
- 0.15x
2
= 0.05x
3
(6)
Also substituting for x4 from equation (5) in equation (4) we have
0.28(1 – x1 – x2 – x3) = 0.01x1 + 0.03x2 + 0.02x3
i.e. 0.28 = 0.29x1 + 0.31x2 + 0.30x3
i.e. 0.29x
1
+ 0.31x
2
= 0.28 - 0.30x
3
(7)
Multiplying equation (6) by 0.31 and equation (7) by 0.15 and adding we get
(0.31)(0.07)x1 + (0.15)(0.29)x1 = (0.31)(0.05)x3 + (0.15)(0.28) – (0.15)(0.30)x3
and since we know x3 = 0.10526 we have x1 = 0.59655
Hence from equation (6) we find that x2 = 0.24330 and from equation (5) that x4 = 0.05489
Note that as a check we have that these values for x1, x2, x3 and x4 satisfy equations (1) – (5) (to within rounding errors).
Hence the long-run market shares are 59.66%, 24.33%, 10.53% and 5.49% for brands 1, 2, 3 and 4 respectively.
Have that after two months s3 (calculated above) not close to the long-run market shares and would expect changing circumstances to render the transition matrix invalid.
We need a long-run system state of [0.75, x2, x3, x4] where x2, x3 and x4 are unknown (but sum to 0.25) and we have a transition matrix given by
P =
| p
1
p
2
p
3
p
4
| | 0.05 0.90 0.02 0.03 | | 0.10 0.05 0.83 0.02 | | 0.13 0.13 0.02 0.72 |
where p1, p2, p3 and p4 are unknown (but sum to one).
Hence using the equation
[0.75, x2, x3, x4] = [0.75, x2, x3, x4]P
we have the equations
0.75 = 0.75p1 + 0.05x2 + 0.10x3 + 0.13x4
x2 = 0.75p2 + 0.90x2 + 0.05x3 + 0.13x4
x3 = 0.75p3 + 0.02x2 + 0.83x3 + 0.02x4
x4 = 0.75p4 + 0.03x2 + 0.02x3 + 0.72x4
together with
x2 + x3 + x4 = 0.25
p1 + p2 + p3 + p4 = 1
Here we have six equations in seven unknowns and so to solve we need an appropriate objective. In order to avoid having to change the transition probabilities too much a suitable objective would be
maximise p1
i.e. find the largest value for the transition probability from brand 1 to itself such that we achieve the long-run market share of 75%.
Hence the LP is
maximise p1
subject to
0.75 = 0.75p1 + 0.05x2 + 0.10x3 + 0.13x4
x2 = 0.75p2 + 0.90x2 + 0.05x3 + 0.13x4
x3 = 0.75p3 + 0.02x2 + 0.83x3 + 0.02x4
x4 = 0.75p4 + 0.03x2 + 0.02x3 + 0.72x4
x2 + x3 + x4 = 0.25
p1 + p2 + p3 + p4 = 1
x2, x3, x4 >= 0
p1, p2, p3, p4 >= 0
Markov processes example 1988 UG exam
An operational researcher is analysing switching between two different products. She knows that in period 1 the market shares for the two products were 55% and 45% but that in period 2 the corresponding market shares were 67% and 33% and in period 3 70% and 30%. The researcher believes that an accurate representation of the market share in any period can be obtained using Markov processes. Assuming her belief is correct:
- Estimate the transition matrix.
- Calculate the market shares in period 4 using the estimated transition matrix.
- If the actual market shares for period 4 were 71% and 29% would you revise your estimate of the transition matrix or not? Give reasons for your decision.
Solution
We have that s1, the state of the system in period 1, is given by s1=[0.55, 0.45] with s2=[0.67, 0.33] and s3=[0.70, 0.30]
Assuming the researcher is correct then s1 and s2 are linked by s2 = s1(P) where P is the transition matrix. We also have that s3 and s2 are linked by s3 = s2(P).
Now we have that P will be a 2 by 2 matrix, and that the elements of each row of P add to one, so that we can write
P = |x
1
1-x
1
|
|x
2
1-x
2
|
where x1 and x2 are unknown.
Using s2 = s1(P) we have
[0.67, 0.33] = [0.55, 0.45] |x
1
1- x
1
|
|x
2
1- x
2
|
and using s3 = s2(P) we have
[0.70, 0.30] = [0.67, 0.33] |x
1
1- x
1
|
|x
2
1- x
2
|
Hence, expanding, we have
0.67 = 0.55x
1
+ 0.45x
2
(1)
0.33 = 0.55(1 - x
1
) + 0.45(1 - x
2
) (2)
0.70 = 0.67x
1
+ 0.33x
2
(3)
0.30 = 0.67(1 - x
1
) + 0.33(1 - x
2
) (4)
Equation (2), when rearranged, becomes equation (1) and similarly equation (4), when rearranged, becomes equation (3). Hence we have two simultaneous equations (equations (1) and (3)) in two unknowns.
From equation (1)
x1 = (0.67 – 0.45x2)/0.55
so substituting for x1 in equation (3) we get
0.70 = 0.67[(0.67 – 0.45x2)/0.55] + 0.33x2
i.e. (0.70)(0.55) = (0.67)(0.67) – (0.67)(0.45)x2 + (0.33)(0.55)x2
i.e. x2 = [(0.67)(0.67)-(0.70)(0.55)]/[(0.67)(0.45)-(0.33)(0.55)]
i.e. x2 = 0.5325 and
x1 = (0.67 – 0.45x2)/0.55 = 0.7825
Hence our estimate of the transition matrix P
= |x
1
1-x
1
|
|x
2
1-x
2
|
= |0.7825 0.2175| |0.5325 0.4675|
Note as a check we can verify that, with this estimated transition matrix, we have s2 = s1(P) and s3 = s2(P).
The market shares for period 4 are given by
s4 = s3(P)
= [0.70, 0.30] |0.7825 0.2175|
|0.5325 0.4675|i.e. s4 = [0.7075, 0.2925]
and note here that the elements of s4 add to one (as required).
Hence the estimated market shares for period 4 are 70.75% and 29.25%.
If the actual market shares are 71% and 29% then this compares well with the shares estimated above and so there would seem no reason to revise the estimate of the transition matrix.
Markov processes example 1987 UG exam
An admissions tutor is analysing (with help from the Management School!) applications from potential students for undergraduate courses at Imperial College (IC). A potential student can (for the purpose of preliminary analysis) be regarded as being in one of five possible states:
- State 1: has not applied to IC
- State 2: has applied to IC
- State 3: has applied to IC and has been interviewed
- State 4: has applied to IC and has been rejected
- State 5: has applied to IC and has been made an offer of a place
At the start of the year (month 1 in the admissions year) all potential students are in state 1.
A review of admissions statistics has identified the following transition matrix for the probability of moving between states each month:
To 1 2 3 4 5
From 1 | 0.95 0.05 0 0 0 |
2 | 0 0.2 0.7 0 0.1 |
3 | 0 0 0.3 0.6 0.1 |
4 | 0 0 0 1 0 |
5 | 0 0 0 0 1 |- What percentage of potential students will have been offered places after 3 months have elapsed?
- Is it possible to work out a long-run system state or not (and why)?
- Formulate the problem of deciding the percentage of potential students who have been offered places in each month up to (and including) the 10th month of the admissions year as a linear program
- Discuss the importance of the objective function of the linear program derived above.
- Discuss what enhancements/changes could be made to the above preliminary analysis in order to enable the admissions tutor to better analyse the admission of students.
Solution
We have the states as defined in the question with
s1=[1,0,0,0,0]
and P = | 0.95 0.05 0 0 0 |
| 0 0.2 0.7 0 0.1 |
| 0 0 0.3 0.6 0.1 |
| 0 0 0 1 0 |
| 0 0 0 0 1 |Hence
s2 = s1P = [0.95, 0.05, 0, 0, 0]
s3 = s2P = [0.9025, 0.0575, 0.035, 0, 0.005]
s4 = s3P = [0.857375, 0.056625, 0.05075, 0.021, 0.01425]
Therefore 1.425% of potential students are offered a place after 3 months.
No long-run system state as absorbing states (4 and 5) are present (or because the admissions year restarts after period 12).
Let sit = the proportion in state i at the end of period t then the matrix equation st = st-1(P) t=2,…,12
becomes the linear equations
s
1t
= 0.95s
1t-1
t=2,.. .,12
s
2t
= 0.05s
1t-1
+ 0.2s
2t-1
t=2,...,12
s
3t
= 0.7s
2t-1
+ 0.3s
3t-1
t=2,...,12
s
4t
= 0.6s
3t-1
+ 1s
4t-1
t=2,...,12
s
5t
= 0.1s
2t-1
+ 0.1s
3t-1
+ 1s
5t-1
t=2,...,12
with initial conditions
s
11
= 1 s
21
= s
31
= s
41
= s
51
= 0
It does not matter about the objective function – any linear function of the variables will do, since there is only one feasible solution to the above system of linear equations.
Points are:
- add more states (e.g. refuses offer, withdraws, etc)
- time alteration of transition matrix e.g. better candidates apply earlier, only interview early in the year, number of applications not spread evenly over the year
- consider departmental pattern
- classes of applicant (e.g. low fee/high fee)
- analysis of results (e.g. how long wait for interview)
Markov processes example 1986 UG exam
A company is considering using Markov theory to analyse brand switching between four different brands of breakfast cereal (brands 1, 2, 3 and 4). An analysis of data has produced the transition matrix shown below for the probability of switching each week between brands.
To brand
1 2 3 4
From brand 1 | 0.5 0.2 0.3 0.0 |
2 | 0.0 0.4 0.4 0.2 |
3 | 0.7 0.0 0.3 0.0 |
4 | 0.8 0.1 0.0 0.1 |- draw the state-transition diagram
- what will be the market shares in week 3 if the current (week 1) market shares for the four brands are 20%, 30%, 15%, 35% for brands 1, 2, 3, 4 respectively?
- what is the long-run prediction for the market shares for each of the four brands?
- would you expect the actual market shares to approach the long-run prediction for the market or not (and why?)
- what advantages and disadvantages can you think of in using Markov theory to forecast market shares in brand switching?
Solution
State-transition diagram easy to draw.
Letting P represent the transition matrix given in the question we have that s1 = [0.2, 0.3, 0.15, 0.35] and so
s2 = s1P = [0.485, 0.195, 0.225, 0.095] and s3 = s2P = [0.476, 0.1845, 0.291, 0.0485]
Note that the elements of s2 and s3 add to one.
Solve the simultaneous equations
[x1, x2, x3, x4] = [x1, x2, x3, x4](P)
and x1 + x2 + x3 + x4 = 1
i.e. x1 = 0.5x1 + 0.7x3 + 0.8x4
x2 = 0.2x1 + 0.4x2 + 0.1x4
x3 = 0.3x1 + 0.4x2 + 0.3x3
x4 = 0.2x2 + 0.1x4
x1 + x2 + x3 + x4 = 1
The solution is x1 = 0.4879 x2 = 0.1689 x3 = 0.3056 x4 = 0.0375
Note, after solution, substitute these values back into the equations above to check that they are consistent.
In general it takes a long time for the system to reach the long-run system state and hence we would not expect that state ever to be reached (due to changing circumstances rendering the transition matrix invalid). However note that in this particular example s3 = the state of the system after just 2 weeks (= [0.476, 0.1845, 0.291, 0.0485]) is close to the long-run system state and so it is likely that we will (effectively) reach the long-run prediction within a few weeks.
Points to make include:
- Markov theory simple to understand and apply
- Markov theory gives us an insight into changes in the system over time
- sensitivity calculations (i.e. “what-if” questions) are easily carried out
- entrants to, and exits from, the market can be modelled
- P may be dependent upon the current state of the system (i.e. the current market share)
- how do we cope with the effect of promotional (advertising) campaigns
- Markov theory is only a simplified model of a complex decision-making process.
Markov processes example 1985 UG exam
British Gas currently has three schemes for quarterly payment of gas bills, namely:
(1) cheque/cash payment
(2) credit card debit
(3) bank account direct debit
Their research department has estimated the following matrix of probabilities for switching between schemes:
Will switch next
quarter to scheme
1 2 3
Currently pays by scheme 1 | 0.85 0.10 0.05 |
2 | 0.04 0.90 0.06 |
3 | 0.02 0.23 0.75 |If 70% currently pay by scheme (1), 20% by scheme (2) and 10% by scheme (3) what will be the corresponding percentages after:
- two quarters; and
- in the long-run.
Solution
Here we have a Markov process with three states where
s1 = [0.7, 0.2, 0.1]
and P = | 0.85 0.10 0.05 |
| 0.04 0.90 0.06 |
| 0.02 0.23 0.75 |The state of the system after one quarter s2 = s1P
= [0.605, 0.273, 0.122]
Note that, as required, the elements of s2 sum to one. The state of the system after 2 quarters s3 = s2P
= [0.52761, 0.33426, 0.13813]
Again note that the elements of s3 add to one.
Hence after two quarters the percentage paying by scheme (1) will be 52.761%, the percentage paying by scheme (2) will be 33.426% and the percentage paying by scheme (3) will be 13.813%.
To calculate the long-run situation let [x1,x2,x3] represent the long-run system state – then we have that
[x1,x2,x3] = [x1,x2,x3](P)
where x1 + x2 + x3 = 1.
This gives us the four equations
x1 = 0.85x1 + 0.04x2 + 0.02x3
x2 = 0.10x1 + 0.90x2 + 0.23x3
x3 = 0.05x1 + 0.06x2 + 0.75x3
x1 + x2 + x3 = 1
Solving these four simultaneous equations gives
x1 = 0.1908, x2 = 0.6218 and x3 = 0.1874
Note that after solution we can substitute these values back into our four equations above to check that these values are consistent with those equations.
Hence in the long-run the percentage paying by scheme (1) will be 19.08%, the percentage paying by scheme (2) will be 62.18% and the percentage paying by scheme (3) will be 18.74%.