Download SOLUTIONS TO SMS Department of Computer Science and

SOLUTIONS TO SMS
Department of Computer Science and Engineering
1a. Explain Poisson distribution.
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Solution:
In probability theory and statistics, the Poisson distribution is a discrete probability
distribution that expresses the probability of a number of events occurring in a fixed period
of time if these events occur with a known average rate and independently of the time
since the last event.
It is a mathematically simple distribution
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Both the mean and variance of a Poisson distribution are equal to µ.
Poisson probability mass function is
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is the shape parameter which indicates the average number of events in the given time
interval.
The formula for the
The mean and variance are both equal to λ
Standard Deviation:
The formula for the Poisson cumulative probability function is
1b. A computer repair person is beeped each time there is a call for service. The number of
beeps is known to be in accordance with a Poisson distribution with a mean 2 per hour. Find
the probability of 3 beeps in the next hour. Solution:
The probability of three beeps in the next hour is given by the equation:
Substituting x=3 and lamda=2 in above equation, we get p(3,2) =0.18
2a. Explain normal distribution.
• Solution:
• The normal distribution is considered the most prominent probability distribution in
statistics.
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The normal distribution is pattern for the distribution of a set of data which follows a bell
shaped curve. This distribution is sometimes called the Gaussian distribution
Normal distributions are symmetric with scores more concentrated in the middle than in
the tails.
The general formula for the probability density function of the normal distribution is
where
is the location parameter and
is the scale parameter
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The normal cdf is
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The result, p, is the probability that a single observation from a normal distribution with
parameters µ and σ will fall in the interval (-∞ x].
The standard normal distribution has µ = 0 and σ = 1.
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2b. The time to pass through a queue to begin self service at a cafeteria has been found to be
N(15,9). Find the probability that an arriving customer waits b/w 14 to 17 minutes. Solution:
The probability that an arriving customer waits b/w 14 to 17 minutes is computed as follows
:P(14<=X<=17)=F(17) –F(14) = φ((17-5)/3) - φ((14-15)/3)
= φ(0.667)- φ(-0.333)
=0.3780
The probability that an arriving customer waits between 14 and 17 minutes is 0.3780
3.What are the primary Long-run measures of queuing systems? Explain L and Lq with an
example. Solution:
The primary long-run measures of performance of queuing systems are the long-run
time-average number of customers in the system (L) and in the queue (LQ), the long-run average
time spent in system (w) and in the queue (wQ) per customer, and the server utilization, or
proportion of time that a server is busy (ρ). The term "system” usually refers to the waiting line
plus the service mechanism, but, in general, it can refer to any subsystem of the queuing
system; whereas the term "queue" refers to the waiting line alone. Other measures of
performance of interest include the long-run proportion of customers who are delayed in queue
longer than to time units, the long-run proportion of customers turned away because of
capacity constraints, and the long-run proportion of time the waiting line contains more than ko
customers.
4a. Explain with steady state behavior of single server queues with Poisson arrivals and Unlimited
capacity. Solution:
For the infinite population models, the arrivals are assumed to follow a Poisson process with
rate(lamda) arrivals per time unit-that is, he inter arrival time are assumed to be exponentially
distributed with mean (1/lamda).Service times may be exponentially distributed(M) or
arbitrarily(G).The queue discipline will be FIFO. Because of the exponential-distributional of the
arrival process, these models are called Markovian models.
A queuing system is said to be statistical equilibrium, or steady state, if the probability that the
P(L(t)=n)=Pn(t)=Pn
Is independent of time t. Two properties-approaching statistical equilibrium from any starting
state, and remaining in statistical equilibrium once it is reached- are characteristic of many
stochastic models. On the other hand, if the analyst were interested in the transient behavior of
a queue over a relatively short period of time ad were given some specific initial condition, the
result to be presented here would be inappropriate, A transient mathematical analysis or, more
likely, a simulation model would be the chosen tool of analysis.
4b. Widget making machines malfunction apparently at random and then require a mechanic’s
attention. It is assumed that malfunction occur according to a Poisson process at the rate of
(lambda)=1.5 per hour. Repair times of single machine is an avg of 30 minutes with a standard
deviation of 20 minutes. Find the steady state time average number of broken machines. Solution:
The mean service time= 1/µ =1/2 hour The
service rate is µ= 2 per hour.
Also , ơ2=202 minutes2 =1/9 hours2
The customers are the widget makers, and the appropriate model is the M/G/1queue, because
only the mean and variance of service times are known, not their distribution.
The proportion of time the mechanic is busy is =(lamda)/ µ=1.5/2=0.75
Also, the steady state time average number of broken machine therefore is given by,
= 0.75+1.625=2.375 machines.
5a. Define random numbers. Explain the statistical properties of random numbers.
Solution:
A random number is a number chosen as if by chance from some specified distribution such that
selection of a large set of these numbers reproduces the underlying distribution.
The statistical properties of random numbers are:1.
The routine should be fast, individual computations are inexpensive, but simulation may
require many hundreds of thousands of random numbers. The total cost can be managed by
selecting a computationally efficient method-number generation.
2.
The routine should be portable to different computers, and ideally to different
programming languages. This is desirable so that the simulation program produces the same
results wherever it is executed.
3.
The routine should have a sufficiently long cycle. The cycle length or period, represents
the length of the random-number sequence before previous numbers begin to repeat
themselves in an earlier order. Thus, if 10,000 events are to be generated, the period should be
many times that long. A special case of cycling is degenerating. A routine degenerates when the
same random numbers appear repeatedly. Such an occurrence is certainly unacceptable. This
can happen rapidly with some methods.
4.
The random numbers should be replicable. Given the starting point (or conditions), it
should be possible to generate the same set of random numbers, completely independence of
the system that is being simulated. This is helpful for debugging purposes and is a mean of
facilitating comparisons between systems. For the same reasons, it should be possible to easily
specify different starting points, widely separated, within the sequence.
5.
Most important, and as indicated previously, the generated random numbers should
closely approximate the ideal statistical properties of uniformity and independence.
5b. Explain the combined linear congruential generator. Solution:
One fruitful approach is to combine two or more multiplicative congruential generators in such a
way that the combined generator has good statistical properties and a longer period.
If Wi, 1 , Wi , 2. . . , W i,k are any independent, discrete-valued random variables (not necessarily
identically distributed), but one of them, say Wi, 1, is uniformly distributed on the integers 0 to
mi — 2, then
Wi= mod m1 – 1 is uniformly distributed on the integers 0 to mi — 2.
To see how this result can be used to form combined generators, let Xi,1, X i,2,..., X i,k be
the i th output from k different multiplicative congruential generators, where the j th generator
has prime modulus mj, and the multiplier aj is chosen so that the period is mj — 1. Then the j'th
generator is producing integers Xi,j that are approximately uniformly distributed on 1 to mj - 1,
and Wi,j = X i,j
— 1 is approximately uniformly distributed on 0 to mj - 2. L'Ecuyer [1988] therefore suggests
combined generators of the form
k=Σ(-1) j-1 X i,j mod m1 – 1
j=1 with R i = { Xi/m1 , Xi >
0 m1 -1 ⁄m1 Xi = 0
Notice that the " (-1)j-1 "coefficient implicitly performs the subtraction X i,1-1; for example,
if k = 2, then (-1)°(X i 1 - 1) - ( - l ) l ( X i 2 - 1)=Σ2 j=1( -1)
j-1 X i,j
The maximum possible period for such a generator is
P= (m1 -1)(m2 - l ) - - - (mk - 1)
----------------------------------2 k-1
6a. Lead times are found to be exponentially distribute with mean 3.7 days. Generate 4 lead
times from this distribution, given random numbers (0.1306,0.0422,0.6597,0.7965,0.7696)
Solution:
Procedure for generating random variates for the exp dist. Using inverse transform technique.
1. Compute cdf of the desired random variable X.
2. Set F(X) = R on the range of X i.e., [ 1 – e-λx] on the range x >= 0
[Note :- X is a random variable with exp distribution, hence, R is also a random variable with exp
distribution over the interval [0,1] ] 3. Solve the eqn F(X) = R in terms of R. 1 – e-λx = R e-λx =
1 – R -λX = ln(1 – R)
X = -1/λ ln (1 – R)
4. Generate random numbers R1, R2, R3….. and compute random variates X1, X2, X3,….. using
the above eqn.
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1
2
3
4
5
Ri
0.1306
0.0422
0.6597
0.7965
0.7696
Xi
0.1400
0.0431
1.078
1.592
1.468
6b. The sequence of numbers 0.54,0.73,0.98,0.11 and 0.68 has been generated. Use KolmogorovSmirnov test with α=0.05 to learn whether the hypothesis that the numbers are
uniformly distributed on the interval [0,1] can be rejected. Dα=0.565 Solution:
R (i)
0.11
0.54
0.68
0.73
0.98
i/N
0.20
0.4
0.6
0.8
1.0
i/N –R(i)
0.09
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-
0.07
0.02
R(i)-(i-1)/N
0.11
0.34
0.28
0.13
0.18
D+ = max(0.09,0.07,0.02)=0.09
D-=max(0.11,0.34,0.28,0.13,0.18)=0.34
D=max(D+,D-)=max(0.09,0.34)=0.34
Dα=0.565 given
We observe that D<Dα from the above computations. Therefore conclude that no difference has
been detected between the true distribution of (R1,R2…..RN) and the uniform distribution.
7. For the following sequence can the hypothesis that the numbers are independent can be
rejected on the basis of length of runs up and down when α=0.05.
Solution:
Same as in Page 234, Eg.7.8 Its done for 30 samples..Problem requires it for 50 samples. Procedure
remains the same.