MARKOV ANALYSIS Andrei Markov, a Russian Mathematician developed the technique to describe the movement of gas in a closed container in 1940 In 1950s, management scientists began to recognize that the technique could be adapted to decision situations which fit the pattern Markov described

MARKOV ANALYSIS In the early 1960s, the procedure has been used to describe Marketing Strategies Plant maintenance decisions Stock market movements Account Receivable Projections

MARKOV ANALYSIS Markov analysis is used to analyze the current state and movement of a variable to predict the future occurrences and movement of this variable by the use of presently known probabilities.

TERMS USED Market share – The fraction of a population that shops at a particular store or market. When expressed as a fraction, market shares can be used in place of state probabilities.

TERMS USED State probability – The probability of an event occurring at a point in time, say, the probability that a person will purchase a product at a given store during a given month.

TERMS USED Transition probability – The conditional probability that we will be in a future state given a current state.

Characteristics of Markov Process   The following are the characteristics of Markov process as applied in business. The time structure. It is the period of time when a set of outcomes occurs 2. The state variables. These are variables that describe the condition, or state, of the system at each trial or period of the process. 3. At a specific point, often at the beginning or end of a time period in the process, the condition being monitored may changed from one state to another.

Characteristics of Markov Process   4. The decision maker is often uncertain about whether the state variable will change or remain the same; and this uncertainty is reflected in a set of transition probabilities. 5. At a given point in the structure of the process, the set of transition probabilities representing the uncertainty of moving to any future state depend only on the current state.

Assumptions of Markov Analysis   The following are the assumptions of Markov analysis as applied in business. The transition probabilities for a given beginning state of the system equals one. 2. The probabilities apply to all system participants. 3. The transition probabilities are constant over time. 4. The states are independent over time.

Matrix of Transition Probabilities It is the likelihood that the system in a given current state will remain in the same state or move to another state in the next period. It is a proportion of customers which a company retains or loses from period to period.

Matrix of Transition Probabilities

Matrix of Transition Probabilities  Mn = Mn-1(P) where: Mn = matrix of market share at period n Mn-1= matrix of market share at period n-1 P = matrix of transition probabilities

Steady State (Equilibrium) One of the major properties of Markov chains is that, in the long run, the process usually stabilizes. A stabilized system is said to approach steady state or equilibrium when the system’s state probabilities have become independent of time.

Steady State (Equilibrium) One of the major properties of Markov chains is that, in the long run, the process usually stabilizes. A stabilized system is said to approach steady state or equilibrium when the system’s state probabilities have become independent of time.

Matrix of Transition Probabilities  Example: Find the steady state probabilities, [Pp Sp] = [Pp Sp]

Market Share Analysis A community has two gasoline service stations, Petron and Shell. The residents of the community purchase gasoline at the two stations on a monthly basis. The marketing department of the Petron Company surveyed a number of residents and found that the customers were not totally loyal to either brand of gasoline. Customers were willing to change service stations as a result of advertising , service and other factors.

Market Share Analysis The marketing department found that if a customer bought their gasoline from Petron in any given month, there was only a 0.60 probability that the customers would buy from Petron the next month and a .40 probability that the customer would buy their gas from Shell the next month. Likewise, if a customer traded with Shell in a given month, there was an .80 probability that the customer would purchase gasoline from Shell in the next month and a .20 probability that the customer would purchase gasoline from Petron.

Market Share Analysis Required: Set-up the transition probability matrix for this problem. Determine the probability that the customer will trade in a) Petron and b) Shell? Determine the steady-state probabilities and If there are 3,000 customers in the community who purchase gasoline, determine the number of customers that each company can anticipate in the long run

Predicting Future Market Shares Suppose in the canteen of a certain university which offers three types of combo meals, A, B and C, on a given day, 35% of the customers choose A, 30% choose B, and 35% choose C. Based on the canteen staff observations, those ordering A on one day, 40% will reorder A the next day, while 25% will order B and 35% will order C. Of those who order B on a day, 30% will reorder B the next day, 40% will order A and 30% will order C. Of those who order C on a day, 50% will reorder C the next day, 25% will order A and 25% will order B. The transition probabilities of the problem can be illustrated using a tree diagram.

Plant Maintenance Schedule A particular production machine could be assigned the states “operating” and “breakdown”. The transition probabilities could then reflect the probability of a machine’s either breaking down or operating in the next time period. (i.e., month, day, or year)

Plant Maintenance Schedule As an example, consider a machine having the following daily transition matrix. Day 1 Day 2 Operate Breakdown T= Operate .90 .10 .70 .30

Plant Maintenance Schedule The steady-state probabilities for this example are .88 = steady-state probabilities for this example are .12 = steady-state probabilities of the machine’s breaking down If management decides that the long –run probability of .12 for a breakdown is excessive, it might consider increasing preventive maintenance, which would change the transition matrix for this example.

Market Share Analysis The marketing department found that if a customer bought their gasoline from Petron in any given month, there was only a 0.60 probability that the customers would buy from Petron the next month and a .40 probability that the customer would buy their gas from Shell the next month. Likewise, if a customer traded with Shell in a given month, there was an .80 probability that the customer would purchase gasoline from Shell in the next month and a .20 probability that the customer would purchase gasoline from Petron.

Market Share Analysis Required: Set-up the transition probability matrix for this problem. Determine the probability that the customer will trade in a) Petron and b) Shell? Determine the steady-state probabilities and indicate the number of customers that each company can anticipate in the long run.

EXAMPLE #1  For the past twelve months, a computer printer functioned 80% of the time correctly during the current month if it had functioned correctly in the preceding month. This means that 20% of the time the printer did not function correctly for a given month when it was functioning correctly during the preceding month.

EXAMPLE #1  Moreover, it was observed that 85% of the time the printer remained incorrectly adjusted for any given month if it was incorrectly adjusted the preceding month and 15% of the time the printer operated correctly in a given month when it did not operate correctly during the preceding month.

EXAMPLE #1 a. What is the probability that the printer will be functioning correctly one month from now? b. What is the probability that the printer will be functioning correctly two months from now?

APPLICATIONS The mathematical function that describes this distribution, also called the density function, is given by   where: X – is the value of the random variable - standard deviation of the distribution - mean of the distribution = 2.14 e = 2.718