Question 1

In determining steady-state behavior for a process with absorbing states, the subdivision of the transition matrix yields:

Accepted Answer

A) an identity submatrix, but no zero submatrix. 
B) no identity submatrix, but a zero submatrix. 
C) both an identity submatrix and a zero submatrix. 
D) neither an identity submatrix nor a zero submatrix. 
A) an identity submatrix, but no zero submatrix. 
B) no identity submatrix, but a zero submatrix. 
C) both an identity submatrix and a zero submatrix. 
D) neither an identity submatrix nor a zero submatrix.

Question 2

A simple computer game using a Markov chain starts at the Cave or the Castle with equal probability and ends with Death (you lose) or Treasure (you win).The outcome depends entirely on luck.The transition probabilities are:$\begin{array}{c}\text { Next State }\\begin{array}{lllll}
&&\text { Cave }&\text {Castle }&\text {Death }&\text {Treasure }\
&\text { Cave } & .4 & .3 & .2 & .1 \
\text { Current }&\text { Castle } & .5 & .3 & .1 & .1 \
\text { State }&\text { Death } & 0 & 0 & 1 & 0 \
&\text { Treasure } & 0 & 0 & 0 & 1
\end{array}\end{array}$

A.What is the average number of times you would expect to visit the Cave and the Castle, depending on which state is the starting state?
B.What is the mean time until absorption for the Cave and the Castle?
C.What is the likelihood of winning the game?

Accepted Answer

A.(This is the fundamental matrix:
| 2.

Question 3

If all the rows of a transition matrix are identical, there will be no transient states.

Accepted Answer

A) True 
 B)False

Question 4

Retired people often return to the workforce.If a retired woman returns to work at the same place from which she retired -- even if only part time or for a limited term -- that signifies that retirement is not a:

Accepted Answer

A) transient state. 
B) steady-state. 
C) periodic state. 
D) absorbing state. 
A) transient state. 
B) steady-state. 
C) periodic state. 
D) absorbing state.

Question 5

If we perform the calculations for steady-state probabilities for a Markov process with periodic behavior, what do we get?

Accepted Answer

A) Steady-state probabilities. 
B) An unsolvable set of equations. 
C) The fundamental matrix. 
D) The long run percentage of time the process will be in each state. 
A) Steady-state probabilities. 
B) An unsolvable set of equations. 
C) The fundamental matrix. 
D) The long run percentage of time the process will be in each state.

Question 6

When calculating steady-state probabilities, we multiply the vector of n unknown values times the transition probability matrix to produce n equations with n unknowns.Why do we arbitrarily drop one of these equations?

Accepted Answer

(It is redundant and is replac

Question 7

The transition matrix for customer purchases of alkaline batteries is believed to be as follows:$\begin{array} { c c c c c } 
&{ \text { Next Purchase } } \
&\text { Duracell } & \text { Eveready } & \text { Other } \
\text { Current }  \text { Duracell } & .43 & .35 & .22 \
\text { Purchase Eveready } & .38 & .45 & .17 \
\text { Other } & .13 & .25 & .62
\end{array}$
A.Based on this transition matrix, what is Duracell's market share for the alkaline battery market?
B.Each 1% of the market share of the alkaline battery market is worth $3.2 million in profit.Suppose that Duracell is contemplating an advertising campaign which it believes will result in the transition probabilities for battery purchases to be as follows:
$\begin{array} { c c c c c } 
&{ \text { Next Purchase } } \
&\text { Duracell } & \text { Eveready } & \text { Other } \
\text { Current }  \text { Duracell } & .47 & .31 & .22 \
\text { Purchase Eveready } & .38 & .40 & .22 \
\text { Other } & .22 & .25 & .53
\end{array}$

What is the most that Duracell should be willing to pay for this
campaign?

Accepted Answer

a.(31%)

b.(Market share wi

Question 8

A gambler has an opportunity to play a coin tossing game in which he wins his wager with probability .49 and loses his wager with probability .51.Suppose the gambler's initial stake is $40 and the gambler will continue to make $10 bets until his fortune either reaches $0 or $100 (at which time play will stop).Which of the following statements is true?

Accepted Answer

A) Increasing the amount of each wager from $10 to $20 will increase the expected playing time. 
B) Increasing the initial stake to $50 will increase the expected playing time. 
C) Reducing the initial stake to $20 will increase the expected playing time. 
D) Increasing the probability of winning from .49 to 1.0 will increase the expected playing time. 
A) Increasing the amount of each wager from $10 to $20 will increase the expected playing time. 
B) Increasing the initial stake to $50 will increase the expected playing time. 
C) Reducing the initial stake to $20 will increase the expected playing time. 
D) Increasing the probability of winning from .49 to 1.0 will increase the expected playing time.

Question 9

If we add up the values in the n rows of the fundamental matrix for a Markov process with absorbing states, what is the result?

Accepted Answer

A) The rows each add to 1. 
B) The limiting probability for each state. 
C) A meaningless number. 
D) The mean time until absorption for each state. 
A) The rows each add to 1. 
B) The limiting probability for each state. 
C) A meaningless number. 
D) The mean time until absorption for each state.

Question 10

Consider the transition matrix:
$\begin{array} { l } 
\left| \begin{array} { l l l } 
.3 & .2&.5
\end{array} \right| \
\quad\quad\begin{array} { l l l } 
\mid.1 & .6 & .3 \mid\ 
\mid .2 & .3 & .5\mid 
\end{array} \
\end{array}$
The steady-state probability of being in state 1 is approximately:

Accepted Answer

A) .177 
B) .231 
C) .300 
D) .403 
A) .177 
B) .231 
C) .300 
D) .403

Question 11

What is the steady-state significance, if any, of a zero in the transition matrix?

Accepted Answer

(Not any, necessarily.If there is a zero

Question 12

Is this an identity matrix? Explain your answer.
| 0 1 0 |
| 1 0 0 |
| 0 0 1 |

Accepted Answer

(No.In an identity matrix, th

Question 13

A Markovian system is currently at stage 1.To determine the state of the system at stage 6, we must have, in addition to the transition matrix, the state probabilities at:

Accepted Answer

A) stage 5. 
B) stage 1. 
C) any stage, up to and including 5. 
D) no stage values are needed. 
A) stage 5. 
B) stage 1. 
C) any stage, up to and including 5. 
D) no stage values are needed.

Question 14

Suppose you play a coin flipping game with a friend in which a fair coin is used.If the coin comes up heads you win $1 from yourfriend, if the coin comes up tails, your friend wins $1 from you.
You have $3 and your friend has $4.You will stop the game when one
of you is broke.Determine the probability that you will win all of your friend's money.

Accepted Answer

The answer of Suppose you play a coin flipping game...

Question 15

For Markov processes with absorbing states, steady-state behavior is independent of the initial process state.

Accepted Answer

A) True 
 B)False

Question 16

If a Markov process consists of two absorbing states and two nonabsorbing states, the limiting probabilities for the nonabsorbing states will:

Accepted Answer

A) both equal zero. 
B) be 0.5 and 0.5. 
C) be identical to the transient state probabilities. 
D) depend on the state vector. 
A) both equal zero. 
B) be 0.5 and 0.5. 
C) be identical to the transient state probabilities. 
D) depend on the state vector.

Question 17

For a Markov process with absorbing states, we define Π(j) = state vector at stage j N = fundamental matrix
I = identity matrix
Q = matrix of transition probabilities between non-absorbing states
R = matrix of transition probabilities between non-absorbing states and absorbing states
The limiting state probabilities equal:

Accepted Answer

A) Π(1) * N * R 
B) I * R * Q 
C) (I - Q)-1 
D) Π(1) * N * Q 
A) Π(1) * N * R 
B) I * R * Q 
C) (I - Q)-1 
D) Π(1) * N * Q

Question 18

Markovian transition matrices are necessarily square.That is, there are exactly the same number of rows as there are columns.

Accepted Answer

A) True 
 B)False

Question 19

The transition matrix  
Represents what type of Markov process?

Accepted Answer

A) Periodic. 
B) Absorbing. 
C) Independent. 
D) Nonrecurrent. 
A) Periodic. 
B) Absorbing. 
C) Independent. 
D) Nonrecurrent.

Question 20

In a Markov process, an absorbing state is a special type of transient state.

Accepted Answer

A) True 
 B)False

In determining steady-state behavior for a process with absorbing states, the subdivision of the transition matrix yields:

If all the rows of a transition matrix are identical, there will be no transient states.

Retired people often return to the workforce.If a retired woman returns to work at the same place from which she retired -- even if only part time or for a limited term -- that signifies that retirement is not a:

If we perform the calculations for steady-state probabilities for a Markov process with periodic behavior, what do we get?

When calculating steady-state probabilities, we multiply the vector of n unknown values times the transition probability matrix to produce n equations with n unknowns.Why do we arbitrarily drop one of these equations?

If we add up the values in the n rows of the fundamental matrix for a Markov process with absorbing states, what is the result?

Consider the transition matrix: .3 .2 .5 \mid.1 .6 .3\mid \mid.2 .3 .5\mid The steady-state probability of being in state 1 is approximately:

What is the steady-state significance, if any, of a zero in the transition matrix?

Is this an identity matrix? Explain your answer. | 0 1 0 | | 1 0 0 | | 0 0 1 |

A Markovian system is currently at stage 1.To determine the state of the system at stage 6, we must have, in addition to the transition matrix, the state probabilities at:

For Markov processes with absorbing states, steady-state behavior is independent of the initial process state.

If a Markov process consists of two absorbing states and two nonabsorbing states, the limiting probabilities for the nonabsorbing states will:

Markovian transition matrices are necessarily square.That is, there are exactly the same number of rows as there are columns.

The transition matrix Represents what type of Markov process?

In a Markov process, an absorbing state is a special type of transient state.

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Introduction to Management Science Models

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Exam 12: Markov Process Models

In determining steady-state behavior for a process with absorbing states, the subdivision of the transition matrix yields:

If all the rows of a transition matrix are identical, there will be no transient states.

Retired people often return to the workforce.If a retired woman returns to work at the same place from which she retired -- even if only part time or for a limited term -- that signifies that retirement is not a:

If we perform the calculations for steady-state probabilities for a Markov process with periodic behavior, what do we get?

When calculating steady-state probabilities, we multiply the vector of n unknown values times the transition probability matrix to produce n equations with n unknowns.Why do we arbitrarily drop one of these equations?

If we add up the values in the n rows of the fundamental matrix for a Markov process with absorbing states, what is the result?

Consider the transition matrix: .3 .2 .5 \mid.1 .6 .3\mid \mid.2 .3 .5\mid The steady-state probability of being in state 1 is approximately:

What is the steady-state significance, if any, of a zero in the transition matrix?

Is this an identity matrix? Explain your answer. | 0 1 0 | | 1 0 0 | | 0 0 1 |

A Markovian system is currently at stage 1.To determine the state of the system at stage 6, we must have, in addition to the transition matrix, the state probabilities at:

For Markov processes with absorbing states, steady-state behavior is independent of the initial process state.

If a Markov process consists of two absorbing states and two nonabsorbing states, the limiting probabilities for the nonabsorbing states will:

Markovian transition matrices are necessarily square.That is, there are exactly the same number of rows as there are columns.

The transition matrix Represents what type of Markov process?

In a Markov process, an absorbing state is a special type of transient state.

Introduction to Management Science Models

Introduction to Management Science Models

Applications of Linear and Integer Programming Models

Network Models

Project Scheduling Models

Decision Models

Forecasting

Inventory Models

Queuing Models

Simulation Models

Quality Management Models

Nonlinear Models:dynamic, Goal, and Nonlinear Programming

Filters