10.5: Model Solution10.6 Model Interpretation

10.7 Assumption and Limitation

10.5 Model SolutionThree steps to create a Markov model:1) Construct the state diagram by identifying

all possible states that the modeled system may find itself.

2) Identify the state connections(or transitions).

3)Parameterize the model by specifying the length of time spent in each state once it is entered(or the probability of transitioning from one state to another within the next time period).

10.5 Model SolutionDefinition of “model solution”: To find the long

term(i.e., the “steady state”) probability of being in any particular state.

This steady state solution is the overall probability of being in each system state.

In general, there is one balance equation for each system state.

The balance equation for each system state is:

flows in=flows outGiven N states, there are N desired unknowns

along with N linear equations, which is a straightfoward linear algebra math problem.

Random walk through England: Model Solution

Let Pi represent probability of being in state i. So we have p1,p2,p3,p4 to represent the four states respectively.

So the balance equations for this model are:

0.2*p2+0.1*p3+0.3*p4=0.6*p10.6*p1=p20.2*p4=0.3*p30.8*p2+0.2*p3=0.5*p4

Random walk through England: Model Solution

The final state equations are：0.2*p2+0.1*p3+0.3*p4=0.6*p10.6*p1=p20.2*p4=0.3*p3p1+p2+p3+p4=1The results are:P1=0.2644P2=0.1586P3=0.2308P4=3462

Random walk through England: Model Solution

Database Server Support: Model Solution

The balance equations for the six states are:4*p(1,1,0)+2*(1,0,1)=6*p(2,0,0)3*p(2,0,0)+4*p(0,2,0)+2*p(0,1,1)=10*p(1,1,0

)3*p(2,0,0)+4*p(0,1,1)+2*p(0,0,2)=8*p(1,0,1)3*P(1,1,0)+3*p(1,0,1)=6*p(0,1,1)3*p(1,1,0)=4*p(0,2,0)3*p(1,0,1)=2*p(0,0,2)

Database Server Support: Model Solution

The balance equations for the six states are:4*p(1,1,0)+2*(1,0,1)=6*p(2,0,0)3*p(2,0,0)+4*p(0,2,0)+2*p(0,1,1)=10*p(1,1,0)3*p(2,0,0)+4*p(0,1,1)+2*p(0,0,2)=8*p(1,0,1)3*P(1,1,0)+3*p(1,0,1)=6*p(0,1,1)3*p(1,1,0)=4*p(0,2,0)p(2,0,0)+p(1,1,0)+p(1,0,1)+p(0,2,0)+p(0,1,1)+

p(0,0,2)=1Results:p(2,0,0)=0.1391; p(1,1,0)=0.1043; p(1,0,1)=0.2087;

p(0,2,0)=0.0783; p(0,1,1)=0.1565; p(0,0,2)=0.3131;

Database Server Support: Model Solution

10.6 Model Interpretation

Example 1:Random Walk Through EnglandFather’s question: What percentage of days is the son actually

not drinking in Leeds?Answer: 74%. Interpretation: p1=0.2644=26%. So, the

percentage of days that the son not

drinking in Leeds is: 1-26%=74%

Lake District relative’s question:Once the son finishes a day of kayaking in the Lake

District, how long will it typically be before he returns?

Answer: 3.33 daysInterpretation: The mean time between entering a

particular state (i.e., the state’s “cycle time”)is the inverse of the steady state probability of being in that state.

P3=0.2308; cycle time=1/0.2308=4.33;It takes 1 day for the lad to kayak;The time from when he finishes a day of kayaking

until he typically starts kayaking again is: 4.33-1=3.33

Example 1:Random Walk Through England

Policeman’s question:How many days each month can the bobbies expect to

see the son driving to London after drinking in Leeds?Answer: 4.76 days.Interpretation:p1=0.2644The days will find the lad drinking in a month:

0.2644*30=7.93Since the probability that the lad will go state 2 after

state 1 is 0.6, so the days that the bobbies can expect to find the lad on the road to London is:

7.93*0.6=4.76 days.

Example 1:Random Walk Through England

Kayak renters’ question: How many visits each month does the son typically

visit their shop and typically how long does the son keep their kayak out each visit?

Answer: 2.08 visits per month, keeping the kayak an average of 3.33 days each visit.

Interpretation: The only way to enter state 3 from another state is

from state 4.The days stay in state 4 each month is :

0.3462*30=10.39

Example 1:Random Walk Through England

The probability to go to state 3 after state 4 is 0.2, so the lad typically start a new visit to the Lake District 10.39*0.2=2.08 times each month.

P3=0.2308The days that the lad can be expected to be

kayaking each month is:30*0.2308=6.92The duration of each visit is: 6.92/2.08=3.33

Example 1:Random Walk Through England

Database Server Support: Solution Interpretation

User’s question:What response time can the typical user expect?Answer: 44.24Interactive Response Time Law: R=M/X0-Z ,(Z=0)X0 ,the throughput of the system, measured at the CPU, is

the product of its utilization and its service rate.The CPU is utilized in states(2,0,0),(1,1,0),(1,0,1)The utilization of CPU is:

p(2,0,0)+p(1,1,0)+p(1,0,1)=0.4521The service rate of CPU = 6 transactions/minute.X0 =0.4521*6=2.7126R=M/X0-Z=2/2.7126=0.7373 minutes/ transaction

Example 2:Database Server Support

System administrator’s question:How near capacity (utilization)of each of the

system resources?Answer: Ucpu=0.4521, Ufast=0.3391,

Ulow=0.6783These are found as direct sums of the

relevant steady state probabilities.Ufast=p(1,1,0)+p(0,2,0)+p(0,1,1)=0.3391Ulow=p(1,0,1)+p(0,1,1)+p(0,0,2)=0.6783

Example 2:Database Server Support

Company president’s question : If I capture Company X’s clientele, which will likely double the number of users on my system, I will need to also double the number of active users on my system. What new performance levels should I spin in my speech to the newly acquired customers?

Answer: The throughput is predicted to go from 2.7126 to

3.4768;The response time is predicted to go from 44.24 to

69.03Now we have 4 users and 15 states.

Example 2:Database Server Support

Company pessimist’s question: Since I know that the fast disk is about to fail and all the files will need to be moved to the slow disk, what will the new response time be?

Answer: 65.00 seconds/transactionNow we have 2 devices, and 3 states which are

(1,1),(2,0),(0,2)The above two examples demonstrate how to

use the knowledge of the steady state probabilities to arrive at more meaningful and more useful performance metrics.

Example 2:Database Server Support

10.7 Model Assumptions and LimitationsMarkov Models are quite robust. However, there

are Key assumptions and resulting limitations:Memoryless Assumption: It is assumed that all the important system information

is captured in the state descriptors of a Markov model. That is, simply knowing which state the system is in, uniquely defines all relevant information.

Knowing the current state information alone is sufficient. This is the defining Markov characteristic and any other information is unnecessary as it applies to the system’s future behavior.

That is, previous history can be forgotten.

Resulting Limitation:Because everything must be captured in the

state descriptor, Markov models are susceptible to state space explosion.

Having large state spaces implies additional complexity and a potential loss of accuracy.

10.7 Model Assumptions and Limitations

Exponential Assumption:The exponential distribution is the only continuous

distribution that is memoryless.For example, the service time is 10 seconds, Knowing

that the customer has already received 5 seconds worth of CPU time but not yet finished(previous history, which is irrelevant under the Markov memoryless assumption), the average amount of CPU time still needed is again 10 seconds.

Markov models assume that the time spent between relevant events, such as job arrival times and job service times, is exponentially distributed.

10.7 Model Assumptions and Limitations

Resulting Limitation:To mitigate the limitation imposed by

exponential assumptions, the concept of phases(or stages) can be introduced.

For example, service time can be partitioned into two phases ,each phase being exponentially distributed with an average of five seconds.

However the price is again a potential state space explosion since the state descriptor must now contain this additional phase information.

10.7 Model Assumptions and Limitations