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G12: Management Science

G12: Management Science. Markov Chains. Outline. Classification of stochastic processes Markov processes and Markov chains Transition probabilities Transition networks and classes of states First passage time probabilities and expected first passage time

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G12: Management Science

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  1. G12: Management Science Markov Chains

  2. Outline • Classification of stochastic processes • Markov processes and Markov chains • Transition probabilities • Transition networks and classes of states • First passage time probabilities and expected first passage time • Long-term behaviour and steady state distribution

  3. Analysing Uncertainty • Computer Models of Uncertainty: • Building blocks: Random number generators • Simulation Models • Static (product launch example) • Dynamic (inventory example and queuing models) • Mathematical Models of Uncertainty: • Building blocks: Random Variables • Mathematical Models • Static: Functions of Random Variables • Dynamic: Stochastic (Random) Processes

  4. Stochastic Processes • Collection of random variables Xt, t in T • Xt’s are typically statistically dependent • State space: set of possible values of Xt’s • State space is the same for all Xt’s • Discrete space: Xt’s are discrete RVs • Continuous space: Xt’s are continuous RVs • Time domain: • Discrete time: T={0,1,2,3,…} • Continuous time: T is an interval (possibly unbounded)

  5. Examples from Queuing Theory • Discrete time, discrete space • Ln: queue length upon arrival of nth customer • Discrete time, continuous space • Wn: waiting time of nth customer • Continuous time, discrete space • Lt: queue length at time t • Continuous time, continuous space • Wt: waiting time for a customer arriving at time t

  6. A gambling example • Game: Flip a coin. You win £ 10 if coin shows head and loose £ 10 otherwise • You start with £ 10 and you keep playing until you are broke • Typical questions • What is the expected amount of money after t flips? • What is the expected length of the game?

  7. A Branching Process 0.5 …. £30 0.5 0.5 £20 …. 0.5 0.5 £10 …. 0.5 10 0.5 0.5 £ 0

  8. Discrete Time - Discrete State Stochastic Processes • Xt: Amount of money you own after t flips • Stochastic Process: X1,X2,X3,… • Each Xt has its own probability distribution • The RVs are dependent: the probability of having £ k after t flips depends on what you had after t’ (<t) flips • Knowing Xt’ changes the probability distribution of Xt (conditional probability)

  9. Outline • Classification of stochastic processes • Markov processes and Markov chains • Transition probabilities • Transition networks and classes of states • First passage time probabilities and expected first passage time • Long-term behaviour and steady state distribution

  10. Markovian Property • Waiting time at time t depends on waiting time at times t’<t • Knowing waiting time at some time t’<t changes the probability distribution of waiting time at time t (Conditional probability) • Knowledge of history generally improves probability distribution (smaller variance) • Generally: The distribution of states at time t depends on the whole history of the process • Knowing states of the system at times t1,…tn<t changes the distribution of states at time t • Markov property: The distribution of states at time t, given the states at times t1<…<tn<t is the same as the distribution of states at time t, given only knowledge of the state at time tn. • The distribution depends only on the last observed state • Knowledge about earlier states does not improve probability distribution

  11. Discrete time, discrete space • P(Xt+1= j | X0=i0,…,Xt=it) = P(Xt+1= j | Xt=it) • In words: The probabilities that govern a transition from state i at time t to state j at time t+1 only depend on the state i at time t and not on the states the process was in before time t

  12. Transition Probabilities • The transition probabilites are P(Xt+1= j | Xt=i) • Transition probabilities are called stationary if P(Xt+1= j | Xt=i) = P(X1= j | X0=i) • If there are only finitely many possible states of the RVs Xt then the stationary transition probabilities are conveniently stored in a transition matrix Pij= P(X1= j | X0=i) • Find the transition matrix for our first example if the game ends if the gambler is either broke or has earned £ 30

  13. Markov Chains • Stochastic process with a finite number, say n, possible states that has the Markov property • Transitions between states in discrete time steps • MC is completely characterised by transition probabilities Pij from state i to state j are stored in an n x ntransition matrix P • Rows of transition matrix sum up to 1. Such a matrix is called a stochastic matrix • Initial distribution of states is given by an initial probability vector p(0)=(p1(0),…,pn(0)) • We are interested in the change of the probability distribution of the states over time

  14. Markov Chains as Modelling Templates • Lawn mower example: • Weekly demand D for lawn mowers has distribution P(D=0)=1/3, P(D=1)=1/2, P(D=2)=1/6 • Mowers can be ordered at the end of each week and are delivered right at the beginning of the next week • Inventory policy: Order two new mowers if stock is empty at the end of the week • Currently (beginning of week 0) there are two lawn mowers in stock • Determine the transition matrix

  15. Market Shares • Two software packages, B and C, enter a market that has so far been dominated by software A • C is more powerful than B which is more powerful than A • C is a big departure from A, while B has some elements in common with both A and C • Market research shows that about 65% of A-users are satisfied with the product and won’t change over the next three months • 30% of A-users are willing to move to B, 5% are willing to move to C….

  16. Transition Matrix • All transition probabilities over the next three months can be found in the following transition matrix • What are the approximate market shares going to be?

  17. Machine Replacement • Many identical machines are used in a manufacturing environment • They deteriorate over time with the following monthly transition probabilities:

  18. Outline • Classification of stochastic processes • Markov processes and Markov chains • Transition probabilities • Transition networks and classes of states • First passage time probabilities and expected first passage time • Long-term behaviour and steady state distribution

  19. 2-step transition probability (graphically) 0 P0j Pi0 i 1 Pi1 P1j j Pi2 P2j 2

  20. 2-step transition probabilities (formally)

  21. Chapman-Kolmogorov Equations • Similarly, one shows that n-step transition probabilities Pij(n)=P(Xn=j | X0=i) obey the following law (for arbitrary m<n:) • The n-step transition probability matrix P(n) is the n-th power of the 1-step TPM P: P(n) =Pn=P…P (n times)

  22. Example see spreadsheet Markov.xls

  23. Distribution of Xn • Given • Markov chain with m states (1,…,m) and transition matrix P • Probability vector for initial state (t=0): p(0)=(p1(0),…, pm(0)) • What is the probability that the process is in state i after n transitions? • Bayes’ formula: P(Xn=i)=P(Xn=i¦X0=1)p1(0)+…+P(Xn=i¦X0=m)pm(0) • Probability vector for Xn: p(n)= p(0)Pn • Iteratively: p(n+1)= p(n)P • Open spreadsheet Markov.xls for lawn mower, market share, and machine replacement examples

  24. Outline • Classification of stochastic processes • Markov processes and Markov chains • Transition probabilities • Transition networks and classes of states • First passage time probabilities and expected first passage time • Long-term behaviour and steady state distribution

  25. An Alternative Representation of the Machine Replacement Example 0.6 OK 0.3 0.9 0.1 New Worn 0.1 1 0.6 0.4 Fail

  26. The transition network • The nodes of the network correspond to the states • There is an arc from node i to node j if Pij > 0 and this arc has an associated value Pij • State i is accessible from state j if there is a path in the network from node i to node j • A stochastic matrix is said to be irreducible if each state is accessible from each other state

  27. Classes of States • State i and j communicate if i is accessible from j and j is accessible from i • Communicating states form classes • A class is called absorbing if it is not possible to escape from it • A class A is said to be accessible from a class B if each state in A is accessible from each state in B • Equivalently: …if some state in A is accessible from some state in B

  28. Find all classes in this example and indicate their accessibility from other classes 1/3 2 3 4 1 1/6 1 1/3 1/2 1/2 1 1 5 2/3 2/3 1/2 1/2 6 7

  29. Return to Gambling Example • Draw the transition network • Find all classes • Is the Markov chain irreducible? • Indicate the accessibility of the classes • Is there an absorbing class?

  30. Outline • Classification of stochastic processes • Markov processes and Markov chains • Transition probabilities • Transition networks and classes of states • First passage time probabilities and expected first passage time • Long-term behaviour and steady state distribution

  31. First passage times • The first passage time from state i to state j is the number of transitions until the process hits state j if it starts at state i • First passage time is a random variable • Define fij(k) = probability that the first passage from state i to state j occurs after k transitions

  32. Calculating fij(k) • Use Bayes’ formula P(A)=P(A|B1)P(B1)+…+ P(A|Bn)P(Bn) • Event A: starting from sate i the process is in state j after n transitions (P(A)=Pij(n)) • Event Bk: first passage from i to j happens after k transitions

  33. Calculating fij(k) (cont.) Bayes’ formula gives: This results in the recursion formula:

  34. Alternative: Simulation • Do a number of simulations, starting from state i and stopping when you have reached state j • Estimate fij(k) = Percentage of runs of length k • BUT: This may take a long time if you want to do this for all state combinations (i,j) and many k’s

  35. Expected first passage time • If Xij = time of first passage from i to j then E(Xij)=fij(1)+2fij(2)+3fij(3)+…. • Use conditional expectation formula E(Xij)=E(Xij|B1)P(B1)+…+ E(Xij|Bn)P(Bn) • Event Bk: first transition goes from i to k • Notice • E(Xij |Bj)=1 and E(Xij|Bk)=1+E(Xkj)

  36. Hence

  37. Example

  38. Outline • Classification of stochastic processes • Markov processes and Markov chains • Transition probabilities • Transition networks and classes of states • First passage time probabilities and expected first passage time • Long-term behaviour and steady state distribution

  39. Long term behaviour • We are interested in distribution of Xn as n tends to infinity: lim p(n)=lim p(0)P(n)= p(0) lim P(n) • If lim P(n) exists then P is called • The limit may not exist, though: • See Markov.xls • Problem: Process has periodic behaviour • Process can only recur to state i after t,2t,3t,… steps • There exists t: if n Not in {t,2t,3t} then Pii(n) = 0 • Period of a state i: maximal such t

  40. Find the periods of the states 1/3 2 3 4 1 1/6 1 1/3 1/2 1/2 1 1 5 2/3 2/3 1/2 1/2 6 7

  41. Aperiodicity • A state with period 1 is called aperiodic • State i is aperiodic if and only if there exists N such that Pii(N) > 0 and Pii(N+1) > 0 • The Chapman-Kolmogorov Equations therefore imply that Pii(n)>0 for every n>=N • Aperiodicity is a class property, i.e. if one state in a class is aperiodic, then so are all others

  42. Regular matrices • A stochastic matrix P is called regular if there exists a number n such that all entries of Pn are positive • A Markov chain with a regular transition matrix is aperiodic (i.e. all states are aperiodic) and irreducible (i.e. all states communicate)

  43. Back to long-term behaviour • Mathematical Fact: If a Markov chain is irreducible and aperiodic then it is ergodic, i.e., all limits exist

  44. Finding the long term probabilities • Mathematical Result: If a Markov chain is irreducible and aperiodic then all rows of its long term transition probability matrix are identical to the unique solution p=(p1,…, pm) of the equations

  45. However,... • …the latter system is of the form pP=p, p1+…+pm=1 and has m+1 equations and m unknowns • It has a solution because P is a stochastic matrix and therefore has 1 as an eigenvalue (with eigenvector x=(1,…,1)). Hence p is just a left eigenvector of P to the eigenvalue 1 and the additional equation normalizes the eigenvector • Calculation: solve the system without the first equation - then check first equation

  46. Example • Find the steady state probabilities for • Solution: (p1,p2)=(0.6,0.4)

  47. Steady state probabilities • The probability vector p with pP=p and p1+..+pm=1 is called the steady state (or stationary) probability distribution of the Markov chain • A Markov chain does not necessarily have a steady state distribution • Mathematical result: an irreducible Markov chain has a steady state distribution

  48. Tending towards steady state • If we start with the steady state distribution then the probability distribution of the states does not change over time • More importantly: If the Markov chain is irreducible and aperiodic then, independently of the initial distribution, the distribution of states gets closer and closer to the steady state distribution • Illustration: see spreadsheet Markov.xls

  49. More on steady state distributions • pj can be interpreted as the long-run proportion of time the process is in state j • Alternatively: pj=1/E(Xjj) where Xjj is the time of the first recurrence to j • E.g. if the expected recurrence time to state j is 2 transitions then, on the long run, the process will be in state j after every 1 out of two transitions,i.e. 1/2 of the time

  50. Average Payoff Per Unit Time • Setting: If process hits state i, a payoff of g(i) is realized (costs = negative payoffs) • Average payoff per period after n transitions Yn=(g(X1)+…+g(Xn))/n • Long-run expected average payoff per time period: lim E(Yn) as n tends to infinity

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