CH 11 Markov analysis Learning objectives After completing
CH – 11 Markov analysis Learning objectives: After completing this chapter , you should be able to: 1. Give examples of systems that may lend themselves to description by a markov model. 2. Explain the meaning of transition probabilities. 3. Use a tree diagram to analyze system behavior. 4. Use matrix multiplication to analyze system behavior. 5. Use algebraic method to solve for steady state values.
Summary Markov analysis can be useful for describing the behavior of a certain class of system that change from state to state on a period – by – period basis according to know transition probabilities. Customers patterns , market share , and equipment breakdowns sometimes lend themselves to description in markov terms.
Glossary Markov process: Steady state: Transition matrix: A closed system that change from state to state according to stable transition probabilities. The long- term tendencies of amarkov system to be in its various states. A matrix that shows the probabilities of markov system changing from is current state to each possible state in the next period.
CH – 11 Markov analysis A markov system has these characteristics 1. It will operate or exist for a number of periods. 2. In each period , the system can assume one of a number of states or conditions. 3. System changes between states from period to period can be described by transition probabilities , which remain constant. Example of system that may be described as markov Brand switching Proportion of customers Who buy brand A Brand B Brand C etc Probability That a customer Will switch from Brand A to brand B , etc
Transition probabilities Which indicate the tendencies of the system to change from one period to the next Example : A car rent agency that has to offices at each of a city’s two airports. Customers are allowed to return a rented car to either airport , regardless of which airport they rented from. Suppose that the manager of the rented agency has made a study of return behavior and has found the following Info :
70% of cars rented from airport A tend to be returned to that airport , and 30% of A tend to returned to B. 10% of cars rented from airport B are returned to airport A. and 90% returned to B. Transition probabilities for car agency Returned to A B A . 70 . 30 = 1. 00 B . 10 . 90 = 1. 00 rented from
The manager has several questions concerning the system : 1. What proportion of cars will be returned to each airport at the short-run , over the next several days. 2. What proportion of cars will be to each location over the long – run. Methods of analysis For a short – run 1. Tree diagram 2. Matrix multiplication For a long – run 3. Algebraic method.
Tree diagram For one period 0 1 A. 70 Strating from A. 30 0. 10 Strating from B 1 A B. 90 At this point A =. 70 B =. 30 B B =. 90 A =. 10
What we are doing for Several periods ?
Example : Use the Info. In the previous example , and prepare a tree diagram for two period. Then compute joint probabilities and use them to determine how many cars will be at location A if A originally has 100 cars and location B has 80 cars. Joint probabilities Period A B 0 1 2. 70 x. 70=. 49. 70 x. 30=. 21. 70 A. 30 x. 10=. 03. 30 x. 90=. 27 A. 70 B. 52. 48. 30 A A. 10 A. 30 B. 90
Period 1 0 . 70. 10 A. 30. 10 A. 90 2 A Joint probabilities A B. 10 x. 70=. 07. 10 x. 30=. 03. 90 x. 10=. 09. 90 x. 90=. 81 B A . 16 . 84 A. 90 B At zero point A=100 cars , B = 80 cars A B 100(. 52) + 80(. 16) = 64 cars in A at the second period.
→ matrix multiplication We need two matrix to use I solution : 1. Shares matrix 2. Transition probabilities matrix. Share’s matrix can be obtained from the calculations period to period. But for the first period ( zero period ) we may assume the shares. for our example : shares for A , B would be this mean that all cars are in A. for period one A = 1(. 70) + 0(. 10) This share will Use in the next period. B 1(. 30) +0(. 90)
→ for the 2 period A =. 70(. 70) =. 49. 30(. 10) =. 03. 52 This share will Use in the 3 period → algebraic Solution The basic Equation is A+B = 1 B. 70(. 30) =. 21. 30(. 90) =. 27. 48
From the transition probability tape we can develop the equation , for A and for B A =. 70 A +. 10 B B =. 30 A +. 90 B A =. 70 A +. 10(1 -A) A =. 70 A +. 10 -. 10 A A =. 60 A +. 10 A -. 60 A =. 10 B = 1 -A B = 1 -. 25 B =. 75 . 40 A =. 10 A= =. 25
→ Analysis for 3 x 3 matrix Example : X Y Z X. 70. 40. 30 Y. 20. 50. 10 Z. 10. 60 Tree diagram solve for 2 period stating from X.
Period 1 0 x x 2 x y z y x y z z x y z The same if you want to solve starting Y or Z ? Joint probabilities x share. 70(. 70) =. 49. 20(. 40) =. 08. 10(. 30) =. 03. 60 y share. 70(. 20) =. 14. 20(. 50) =. 10(. 10) =. 01. 25 z share. 70(. 10) =. 07. 20(. 10) =. 02. 10(. 60) =. 06. 15
Matrix multiplication Solution Period : starting from x x 1(. 70) 0(. 40) 0(. 30). 70 0 2 . 70. 20. 10 . 70(. 70). 20(. 40). 10(. 30). 60 For the 3 ed period , and so on. y 1(. 20) 0(. 50) 0(. 10). 20 z 1(. 10) 0(. 10). 0(. 60). 10 . 70(. 20). 20(. 50). 10(. 10). 25 . 70(. 10). 20(. 10). 10(. 60). 15
Algebraic solution equations 1. X =. 70 x +. 40 y +. 30 z Y =. 20 x +. 50 y +. 10 z Z =. 10 x +. 10 y +. 60 z 1 = x + y +z basic equation 2. Eliminate one the equations , but not the basic equation. Suppose third equation is eliminated. Z=1–X-Y
3. Substitute for Z in the first and second equations X =. 70 x +. 40 y +. 30(1 – x – y) Y =. 20 x +. 50 y +. 10(1 – x – y) X =. 70 x +. 40 y +. 30 -. 30 x -. 30 y Y =. 40 x +. 30 +. 10 y X -. 40 x -. 10 y =. 30. 60 x -. 10 y =. 3 the save for Y -. 10 x +. 60 y =. 10. 60 x -. 10 y =. 30 -. 10 x +. 6 y =. 10
. 60 x -. 10 y =. 30 6 (-. 10 x +. 60 y =. 10 ). 60 x -. 10 y =. 30 -. 60 x + 3. 6 y =. 60 3. 5 y =. 90 Y= =. 257 . 60 x -. 10(. 257) =. 30. 60 x -. 0257 =. 30 . 60 x =. 30 +. 0257 60 x =. 3257 X= =. 543
Z = 1 -. 542 -. 257 =. 200 if the are 900 unit in the system X = 900(. 543) = 488. 7 Y = 900(. 257) = 231. 3 Z = 900(. 200) = 180. 0 900
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