Some Applications of Markov Chain in Python

In this article a few simple applications of Markov chain are going to be discussed as a solution to a few text processing problems. These problems appeared as assignments in a few courses, the descriptions are taken straightaway from the courses themselves.

1. Markov Model of Natural Language

This problem appeared as an assignment in the Princeton course cos126 . This assignment was developed by Prof. Bob Sedgewick and Kevin Wayne, based on the classic idea of Claude Shannon.

Problem Statement

Use a Markov chain to create a statistical model of a piece of English text. Simulate the Markov chain to generate stylized pseudo-random text.

Perspective. In the 1948 landmark paper A Mathematical Theory of Communication, Claude Shannon founded the field of information theory and revolutionized the telecommunications industry, laying the groundwork for today’s Information Age. In this paper, Shannon proposed using a Markov chain to create a statistical model of the sequences of letters in a piece of English text. Markov chains are now widely used in speech recognition, handwriting recognition, information retrieval, data compression, and spam filtering. They also have many scientific computing applications including the genemark algorithm for gene prediction, the Metropolis algorithm for measuring thermodynamical properties, and Google’s PageRank algorithm for Web search. For this assignment, we consider a whimsical variant: generating stylized pseudo-random text.

Markov model of natural language. Shannon approximated the statistical structure of a piece of text using a simple mathematical model known as a Markov model. A Markov model of order 0 predicts that each letter in the alphabet occurs with a fixed probability. We can fit a Markov model of order 0 to a specific piece of text by counting the number of occurrences of each letter in that text, and using these frequencies as probabilities. For example, if the input text is "gagggagaggcgagaaa", the Markov model of order 0 predicts that each letter is 'a' with probability 7/17, 'c' with probability 1/17, and 'g' with probability 9/17 because these are the fraction of times each letter occurs. The following sequence of letters is a typical example generated from this model:

g a g g c g a g a a g a g a a g a a a g a g a g a g a a a g a g a a g ...

A Markov model of order 0 assumes that each letter is chosen independently. This independence does not coincide with statistical properties of English text because there a high correlation among successive letters in a word or sentence. For example, 'w' is more likely to be followed with 'e' than with 'u', while 'q' is more likely to be followed with 'u' than with 'e'.

We obtain a more refined model by allowing the probability of choosing each successive letter to depend on the preceding letter or letters. A Markov model of order k predicts that each letter occurs with a fixed probability, but that probability can depend on the previous k consecutive letters. Let a k-gram mean any k consecutive letters. Then for example, if the text has 100 occurrences of "th", with 60 occurrences of "the", 25 occurrences of "thi", 10 occurrences of "tha", and 5 occurrences of "tho", the Markov model of order 2 predicts that the next letter following the 2-gram "th" is 'e' with probability 3/5, 'i' with probability 1/4, 'a' with probability 1/10, and 'o' with probability 1/20.

A brute-force solution. Claude Shannon proposed a brute-force scheme to generate text according to a Markov model of order 1:

“ To construct [a Markov model of order 1], for example, one opens a book at random and selects a letter at random on the page. This letter is recorded. The book is then opened to another page and one reads until this letter is encountered. The succeeding letter is then recorded. Turning to another page this second letter is searched for and the succeeding letter recorded, etc. It would be interesting if further approximations could be constructed, but the labor involved becomes enormous at the next stage. ”

Our task is to write a python program to automate this laborious task in a more efficient way — Shannon’s brute-force approach is prohibitively slow when the size of the input text is large.


Markov model data type.
 Create an immutable data type MarkovModel to represent a Markov model of order k from a given text string. The data type must implement the following API:

markovapi.png

  • Constructor. To implement the data type, create a symbol table, whose keys will be Stringk-grams. You may assume that the input text is a sequence of characters over the ASCII alphabet so that all char values are between 0 and 127. The value type of your symbol table needs to be a data structure that can represent the frequency of each possible next character. The frequencies should be tallied as if the text were circular (i.e., as if it repeated the first k characters at the end).
  • Order. Return the order k of the Markov Model.
  • Frequency. There are two frequency methods.
    • freq(kgram) returns the number of times the k-gram was found in the original text.
    • freq(kgram, c) returns the number of times the k-gram was followed by the character c in the original text.
  • Randomly generate a character. Return a character. It must be a character that followed the k-gram in the original text. The character should be chosen randomly, but the results of calling rand(kgram) several times should mirror the frequencies of characters that followed the k-gram in the original text.
  • Generate pseudo-random text. Return a String of length T that is a randomly generated stream of characters whose first k characters are the argument kgram. Starting with the argument kgram, repeatedly call rand() to generate the next character. Successive k-grams should be formed by using the most recent k characters in the newly generated text. Use a StringBuilder object to build the stream of characters (otherwise, as we saw when discussing performance, your code will take order of N2 time to generate N characters, which is too slow).

To avoid dead ends, treat the input text as a circular string: the last character is considered to precede the first character.

For example, if k = 2 and the text is the 17-character string "gagggagaggcgagaaa", then the salient features of the Markov model are captured in the table below:

               frequency of   probability that
                next char       next char is 
kgram   freq    a   c   g        a    c    g
----------------------------------------------
 aa      2      1   0   1       1/2   0   1/2  
 ag      5      3   0   2       3/5   0   2/5  
 cg      1      1   0   0        1    0    0
 ga      5      1   0   4       1/5   0   4/5  
 gc      1      0   0   1        0    0    1  
 gg      3      1   1   1       1/3  1/3  1/3  
----------------------------------------------
        17      7   1   9

Taken from an example from the same assignment.

Note that the frequency of "ag" is 5 (and not 4) because we are treating the string as circular.

Markov chain is a stochastic process where the state change depends on only the current state. For text generation, the current state is a k-gram. The next character is selected at random, using the probabilities from the Markov model. For example, if the current state is "ga" in the Markov model of order 2 discussed above, then the next character is 'a' with probability 1/5 and 'g' with probability 4/5. The next state in the Markov chain is obtained by appending the new character to the end of the k-gram and discarding the first character. A trajectory through the Markov chain is a sequence of such states. Below is a possible trajectory consisting of 9 transitions.

trajectory:          ga  -->  ag  -->  gg  -->  gc  -->  cg  -->  ga  -->  ag  -->  ga  -->  aa  -->  ag
probability for a:       1/5      3/5      1/3       0        1       1/5      3/51/5      1/2
probability for c:        0        0       1/3       0        0        0        0        0        0
probability for g:       4/5      2/5      1/3       1        0       4/5      2/5      4/5      1/2

Taken from an example from the same assignment.

Treating the input text as a circular string ensures that the Markov chain never gets stuck in a state with no next characters.

To generate random text from a Markov model of order k, set the initial state to k characters from the input text. Then, simulate a trajectory through the Markov chain by performing T − k transitions, appending the random character selected at each step. For example, if k = 2 and T = 11, the following is a possible trajectory leading to the output gaggcgagaag.

trajectory:          ga  -->  ag  -->  gg  -->  gc  -->  cg  -->  ga  -->  ag  -->  ga  -->  aa  -->  ag
output:              ga        g        g        c        g        a        g        a        a        g

Text generation client. Implement a client program TextGenerator that takes two command-line integers k and T, reads the input text from standard input and builds a Markov model of order k from the input text; then, starting with the k-gram consisting of the first k letters of the input text, prints out T characters generated by simulating a trajectory through the corresponding Markov chain. We may assume that the text has length at least k, and also that T ≥ k.

A Python Implementation

import numpy as np
from collections import defaultdict

class MarkovModel:

    def __init__(self, text, k): 
        '''
        create a Markov model of order k from given text
        Assume that text has length at least k.
        '''
        self.k = k               
        self.tran = defaultdict(float)
        self.alph = list(set(list(text)))
        self.kgrams = defaultdict(int)
        n = len(text)
        text += text[:k]
        for i in range(n):
            self.tran[text[i:i+k],text[i+k]] += 1.
            self.kgrams[text[i:i+k]] += 1

    def order(self):                   # order k of Markov model
        return self.k

    def freq(self, kgram):             # number of occurrences of kgram in text
        assert len(kgram) == self.k    # (check if kgram is of length k)
        return self.kgrams[kgram]

    def freq2(self, kgram, c):   	# number of times that character c follows kgram
        assert len(kgram) == self.k     # (check if kgram is of length k)  
        return self.tran[kgram,c]  

    def rand(self, kgram):             # random character following given kgram
        assert len(kgram) == self.k    # (check if kgram is of length k.
        Z = sum([self.tran[kgram, alph] for alph in self.alph])
        return np.random.choice(self.alph, 1, p=np.array([self.tran[kgram, alph] for alph in self.alph])/Z)

    def gen(self, kgram, T):           # generate a String of length T characters
        assert len(kgram) == self.k    # by simulating a trajectory through the corresponding
        str = ''                       # Markov chain. The first k characters of the newly
        for _ in range(T):             # generated String should be the argument kgram.
             #print kgram, c           # check if kgram is of length k.
             c =  self.rand(kgram)[0]  # Assume that T is at least k.
             kgram = kgram[1:] + c     
             str += c			 
        return str

Some Results

m = MarkovModel('gagggagaggcgagaaa', 2)

generates the following MarkovChain where each state represents a 2-gram.

mc.png

 



Input: news item (taken from the assignment)

Microsoft said Tuesday the company would comply with a preliminary ruling by Federal District Court Judge Ronald H. Whyte that Microsoft is no longer able to use the Java Compatibility Logo on its packaging and websites for Internet Explorer and Software Developers Kit for Java.

“We remain confident that once all the facts are presented in the larger case, the court will find Microsoft to be in full compliance with its contract with Sun,” stated Tom Burt, Associate General Counsel for Microsoft Corporation. “We are disappointed with this decision, but we will immediately comply with the Court’s order.”

Microsoft has been in the forefront of helping developers use the Java programming language to write cutting-edge applications. The company has committed significant resources so that Java developers have the option of taking advantage of Windows features when writing software using the Java language. Providing the best tools and programming options will continue to be Microsoft’s goal.

“We will continue to listen to our customers and provide them the tools they need to write great software using the Java language,” added Tod Nielsen, General Manager for Microsoft’s Developer Relations Group/Platform Marketing.

Markov Model learnt

mc1.png

Generated output: random news item, using input as an order 7 model

 

Microsoft is no longer able to use the Java language,” added Tod Nielsen, General Counsel for Microsoft’s Developers use the Java Compatibility Logo on its packaging and websites for Internet Explorer and Software Developer Relations Group/Platform Marketing.Microsoft to be Microsoft Corporation. “We are disappointed with the Court’s order.”

Microsoft is no longer able to use the Java language. Providing the best tools and provide them the tools and programming options will continue to listen to our customers and provide them the tools and programming option of taking advantage of Windows features when writing software using the Java Compatibility Logo on its packaging and websites for Internet Explorer and Software using the Java programming option of taking advantage of Windows features when writing software Developer Relations Group/Platform Marketing.Microsoft to be in full compliance with its contract with Sun,” stated Tom Burt, Associate General Manager for Microsoft’s goal.

“We will continue to listen to our customers and programming language. Providing the Java language,” added Tod Nielsen, General Manager for Microsoft is no longer able to use the Java language.


Noisy Text Correction

Imagine we receive a message where some of the characters have been corrupted by noise. We represent unknown characters by the ~ symbol (we assume we don’t use ~ in our messages). Add a method replaceUnknown that decodes a noisy message by replacing each ~ with the most likely character given our order k Markov model, and conditional on the surrounding text:

def replaceUnknown(corrupted)  # replace unknown characters with most probable characters

Assume unknown characters are at least k characters apart and also appear at least k characters away from the start and end of the message.  This maximum-likelihood approach doesn’t always get it perfect, but it fixes most of the missing characters correctly.

Here are some details on what it means to find the most likely replacement for each ~. For each unknown character, you should consider all possible replacement characters. We want the replacement character that makes sense not only at the unknown position (given the previous characters) but also when the replacement is used in the context of the k subsequent known characters. For example we expect the unknown character in "was ~he wo" to be 't' and not simply the most likely character in the context of "was ". You can compute the probability of each hypothesis by multiplying the probabilities of generating each of k+1 characters in sequence: the missing one, and the k next ones. The following figure illustrates how we want to consider k+1 windows to maximize the log-likelihood:

correct.png

Using the algorithm described above, here are the results obtained for the following example:

Original : it was the best of times, it was the worst of times.
Noisy :      it w~s th~ bes~ of tim~s, i~ was ~he wo~st of~times.
Corrected (k=4): it was the best of times, it was the worst of times.
Corrected (k=2): it was the best of times, in was the wo st of times.

 

2. Detecting authorship

This problem appeared as an assignment in the Cornell course cs1114 .

The Problem Statement

In this assignment, we shall be implementing an authorship detector which, when
given a large sample size of text to train on, can then guess the author of an unknown
text.

  • The algorithm to be implemented works based on the following idea: An author’s
    writing style can be defined quantitatively by looking at the words he uses. Specifically, we want to keep track of his word flow – that is, which words he tends to use after other words.
  • To make things significantly simpler, we’re going to assume that the author always
    follows a given word with the same distribution of words. Of course, this isn’t true,
    since the words you choose when writing obviously depend on context. Nevertheless, this simplifying assumption should hold over an extended amount of text, where context becomes less relevant.
  • In order to implement this model of an author’s writing style, we will use a Markov
    chain. A Markov chain is a set of states with the Markov property – that is, the
    probabilities of each state are independent from the probabilities of every other state. This behavior correctly models our assumption of word independence.
  • A Markov chain can be represented as a directed graph. Each node is a state (words,
    in our case), and a directed edge going from state Si to Sj represents the probability we will go to Sj when we’re at Si. We will implement this directed graph as a transition matrix. Given a set of words W1, W2, …Wn, we can construct an n by n transition matrix A, where an edge from Wi to Wj of weight p means Aij = p.
  • The edges, in this case, represent the probability that word j follows word i from the
    given author. This means, of course, that the sum of the weights of all edges leaving
    from each word must add up to 1.
  • We can construct this graph from a large sample text corpus. Our next step would be finding the author of an unknown, short chunk of text. To do this, we simply compute the probability of this unknown text occurring, using the words in that order, in each of our Markov chains. The author would likely be the one with the highest probability.
  • We shall implement the Markov chain model of writing style. We are given some sample texts to train our model on, as well as some challenges for you to
    figure out.

 

Constructing the transition matrix

Our first step is to construct the transition matrix representing our Markov chain.
First, we must read the text from a sample file. We shall want to create a sparse array using the scipy csr sparse matrix. Along with the transition matrix, we shall be creating a corresponding vector that contains 2 word frequencies (normalized by the total number of words in the document (including repeated words)).

Calculating likelihood

Once we have our transition matrix, we can calculate the likelihood of an unknown
sample of text. We are given several pieces of literature by various authors,
as well as excerpts from each of the authors as test dataset. Our goal is to identify the authors of each excerpt.

To do so, we shall need to calculate the likelihood of the excerpt occurring in each author’s transition matrix. Recall that each edge in the directed graph that the transition
matrix represents is the probability that the author follows a word with another.
Since we shall be multiplying numerous possibly small probabilities together, our
calculated likelihood will likely be extremely small. Thus, you should compare log(likelihood) instead. Keep in mind the possibility that the author may have used a word he has never used before. Our calculated likelihood should not eliminate an author completely because of this. We shall be imposing a high penalty if a word is missing.

Finding the author with the maximum likelihood

Now that we can compute likelihoods, the next step is to write a routine that takes a set of transition matrices and dictionaries, and a sequence of text, and returns the index of the transition matrix that results in the highest likelihood. You will write this in a function classify text, which takes transition matrices, dictionaries, histograms, and the name of the file containing the test text, and returns a single integer best index.  The following figure shows how to detect an author k (A_k) of the test text t_1..t_n  using the transition matrix P_k with MLE :

ad.png

Python Implementation

from np import log
def log0(x):
return 0 if x <= 0 else log(x)

def compute_text_likelihood(filename, T, dict_rev, histogram, index):

”’
Compute the (log) likelihood L of a given string (in ‘filename’) given
a word transition probability T, dictionary ‘dict’, and histogram
‘histogram’
”’

text = word_tokenize(open(filename).read().replace(‘\n’, ‘ ‘).lower())
num_words = len(text)
text = [word for word in text if word in histogram] # keep only the words that are found in the training dataset
ll = log0(histogram[text[0]]) – log0(sum(histogram.values()))
for i in range(1, len(text)):
ll += log0(T[dict_rev[text[i-1]], dict_rev[text[i]]])
return ll + (num_words – num_matches)*penalty

def classify_text(tmatrices, dict_revs, histograms, filename):

”’
Return the index of the most likely author given the transition matrices, dictionaries, and a test file
”’

for i in range(len(tmatrices)):
ll = compute_text_likelihood(filename, tmatrices[i], dict_revs[i], histograms[i], i)

print i, ll

Training Dataset

The list of authors whose writings are there in the training dataset:

0. Austen
1. Carroll
2. Hamilton
3. Jay
4. Madison
5. Shakespeare
6. Thoreau
7. Twain

 

A few lines of excerpts from the training files (the
literature by several authors, taken from Project Gutenberg), the word clouds and a few states from the corresponding Markov Models Constructed for a few authors

 

Author JANE AUSTEN (Texts taken from Emma, Mansfield Park, Pride and Prejudice)

Emma Woodhouse, handsome, clever, and rich, with a comfortable home and
happy disposition, seemed to unite some of the best blessings of
existence; and had lived nearly twenty-one years in the world with very
little to distress or vex her.

She was the youngest of the two daughters of a most affectionate,
indulgent father; and had, in consequence of her sister’s marriage,
been mistress of his house from a very early period. Her mother had
died too long ago for her to have more than an indistinct remembrance
of her caresses; and her place had been supplied by an excellent woman
as governess, who had fallen little short of a mother in affection.

Sixteen years had Miss Taylor been in Mr. Woodhouse’s family, less as a
governess than a friend, very fond of both daughters, but particularly
of Emma. Between _them_ it was more the intimacy of sisters. Even
before Miss Taylor had ceased to hold the nominal office of governess,
the mildness of her temper had hardly allowed her to impose any
restraint; and the shadow of authority being now long passed away, they
had been living together as friend and friend very mutually attached,
and Emma doing just what she liked; highly esteeming Miss Taylor’s
judgment, but directed chiefly by her own.

The real evils, indeed, of Emma’s situation were the power of having
rather too much her own way, and a disposition to think a little too
well of herself; these were the disadvantages which threatened alloy to
her many enjoyments. The danger, however, was at present so
unperceived, that they did not by any means rank as misfortunes with
her.

0.png

mcAusten.png

 

Author Lewis Carroll (Texts taken from Alice’s Adventures in Wonderland, Sylvie and Bruno)

Alice was beginning to get very tired of sitting by her sister on the
bank, and of having nothing to do: once or twice she had peeped into the
book her sister was reading, but it had no pictures or conversations in
it, ‘and what is the use of a book,’ thought Alice ‘without pictures or
conversation?’

So she was considering in her own mind (as well as she could, for the
hot day made her feel very sleepy and stupid), whether the pleasure
of making a daisy-chain would be worth the trouble of getting up and
picking the daisies, when suddenly a White Rabbit with pink eyes ran
close by her.

There was nothing so VERY remarkable in that; nor did Alice think it so
VERY much out of the way to hear the Rabbit say to itself, ‘Oh dear!
Oh dear! I shall be late!’ (when she thought it over afterwards, it
occurred to her that she ought to have wondered at this, but at the time
it all seemed quite natural); but when the Rabbit actually TOOK A WATCH
OUT OF ITS WAISTCOAT-POCKET, and looked at it, and then hurried on,
Alice started to her feet, for it flashed across her mind that she had
never before seen a rabbit with either a waistcoat-pocket, or a watch
to take out of it, and burning with curiosity, she ran across the field
after it, and fortunately was just in time to see it pop down a large
rabbit-hole under the hedge.

In another moment down went Alice after it, never once considering how
in the world she was to get out again.

The rabbit-hole went straight on like a tunnel for some way, and then
dipped suddenly down, so suddenly that Alice had not a moment to think
about stopping herself before she found herself falling down a very deep
well.

1.png

mcCarroll.png

 

Author William Shakespeare (Texts taken from Henry IV Part 1, Romeo and Juliet, Twelfth Night)

 

KING.
So shaken as we are, so wan with care,
Find we a time for frighted peace to pant,
And breathe short-winded accents of new broils
To be commenced in strands afar remote.
No more the thirsty entrance of this soil
Shall daub her lips with her own children’s blood;
No more shall trenching war channel her fields,
Nor bruise her flowerets with the armed hoofs
Of hostile paces: those opposed eyes,
Which, like the meteors of a troubled heaven,
All of one nature, of one substance bred,
Did lately meet in the intestine shock
And furious close of civil butchery,
Shall now, in mutual well-beseeming ranks,
March all one way, and be no more opposed
Against acquaintance, kindred, and allies:
The edge of war, like an ill-sheathed knife,
No more shall cut his master. Therefore, friends,
As far as to the sepulchre of Christ–
Whose soldier now, under whose blessed cross
We are impressed and engaged to fight–
Forthwith a power of English shall we levy,
To chase these pagans in those holy fields
Over whose acres walk’d those blessed feet
Which fourteen hundred years ago were nail’d
For our advantage on the bitter cross.
But this our purpose now is twelvemonth old,
And bootless ’tis to tell you we will go:
Therefore we meet not now.–Then let me hear
Of you, my gentle cousin Westmoreland,
What yesternight our Council did decree
In forwarding this dear expedience.

WEST.
My liege, this haste was hot in question,
And many limits of the charge set down
But yesternight; when, all athwart, there came
A post from Wales loaden with heavy news;
Whose worst was, that the noble Mortimer,
Leading the men of Herefordshire to fight
Against th’ irregular and wild Glendower,
Was by the rude hands of that Welshman taken;
A thousand of his people butchered,
Upon whose dead corpse’ there was such misuse,
Such beastly, shameless transformation,
By those Welshwomen done, as may not be
Without much shame re-told or spoken of.

 

shakespeare.png

mcShakespeare.png

Author MARK TWAIN (Texts taken from A Connecticut Yankee in King Arthur’s Court, The Adventures of Huckleberry Finn, The Prince and the Pauper)

I am an American. I was born and reared in Hartford, in the State
of Connecticut–anyway, just over the river, in the country. So
I am a Yankee of the Yankees–and practical; yes, and nearly
barren of sentiment, I suppose–or poetry, in other words. My
father was a blacksmith, my uncle was a horse doctor, and I was
both, along at first. Then I went over to the great arms factory
and learned my real trade; learned all there was to it; learned
to make everything: guns, revolvers, cannon, boilers, engines, all
sorts of labor-saving machinery. Why, I could make anything
a body wanted–anything in the world, it didn’t make any difference
what; and if there wasn’t any quick new-fangled way to make a thing,
I could invent one–and do it as easy as rolling off a log. I became
head superintendent; had a couple of thousand men under me.

Well, a man like that is a man that is full of fight–that goes
without saying. With a couple of thousand rough men under one,
one has plenty of that sort of amusement. I had, anyway. At last
I met my match, and I got my dose. It was during a misunderstanding
conducted with crowbars with a fellow we used to call Hercules.
He laid me out with a crusher alongside the head that made everything
crack, and seemed to spring every joint in my skull and made it
overlap its neighbor. Then the world went out in darkness, and
I didn’t feel anything more, and didn’t know anything at all
–at least for a while.

When I came to again, I was sitting under an oak tree, on the
grass, with a whole beautiful and broad country landscape all
to myself–nearly. Not entirely; for there was a fellow on a horse,
looking down at me–a fellow fresh out of a picture-book. He was
in old-time iron armor from head to heel, with a helmet on his
head the shape of a nail-keg with slits in it; and he had a shield,
and a sword, and a prodigious spear; and his horse had armor on,
too, and a steel horn projecting from his forehead, and gorgeous
red and green silk trappings that hung down all around him like
a bedquilt, nearly to the ground.

“Fair sir, will ye just?” said this fellow.

“Will I which?”

“Will ye try a passage of arms for land or lady or for–”

“What are you giving me?” I said. “Get along back to your circus,
or I’ll report you.”

Now what does this man do but fall back a couple of hundred yards
and then come rushing at me as hard as he could tear, with his
nail-keg bent down nearly to his horse’s neck and his long spear
pointed straight ahead. I saw he meant business, so I was up
the tree when he arrived.

twain.png

mcTwain.png

 

 

 

Classifying unknown texts from the Test Dataset

Each of the Markov models learnt from the training texts for each author are used to compute the log-likelihood of the unknown test text, the author with the maximum log-likelihood is chosen to be the likely author of the text.
Unknown Text1 

Against the interest of her own individual comfort, Mrs. Dashwood had
determined that it would be better for Marianne to be any where, at
that time, than at Barton, where every thing within her view would be
bringing back the past in the strongest and most afflicting manner, by
constantly placing Willoughby before her, such as she had always seen
him there. She recommended it to her daughters, therefore, by all
means not to shorten their visit to Mrs. Jennings; the length of which,
though never exactly fixed, had been expected by all to comprise at
least five or six weeks. A variety of occupations, of objects, and of
company, which could not be procured at Barton, would be inevitable
there, and might yet, she hoped, cheat Marianne, at times, into some
interest beyond herself, and even into some amusement, much as the
ideas of both might now be spurned by her.

From all danger of seeing Willoughby again, her mother considered her
to be at least equally safe in town as in the country, since his
acquaintance must now be dropped by all who called themselves her
friends. Design could never bring them in each other’s way: negligence
could never leave them exposed to a surprise; and chance had less in
its favour in the crowd of London than even in the retirement of
Barton, where it might force him before her while paying that visit at
Allenham on his marriage, which Mrs. Dashwood, from foreseeing at first
as a probable event, had brought herself to expect as a certain one.

She had yet another reason for wishing her children to remain where
they were; a letter from her son-in-law had told her that he and his
wife were to be in town before the middle of February, and she judged
it right that they should sometimes see their brother.

Marianne had promised to be guided by her mother’s opinion, and she
submitted to it therefore without opposition, though it proved
perfectly different from what she wished and expected, though she felt
it to be entirely wrong, formed on mistaken grounds, and that by
requiring her longer continuance in London it deprived her of the only
possible alleviation of her wretchedness, the personal sympathy of her
mother, and doomed her to such society and such scenes as must prevent
her ever knowing a moment’s rest.

But it was a matter of great consolation to her, that what brought evil
to herself would bring good to her sister; and Elinor, on the other
hand, suspecting that it would not be in her power to avoid Edward
entirely, comforted herself by thinking, that though their longer stay
would therefore militate against her own happiness, it would be better
for Marianne than an immediate return into Devonshire.

Her carefulness in guarding her sister from ever hearing Willoughby’s
name mentioned, was not thrown away. Marianne, though without knowing
it herself, reaped all its advantage; for neither Mrs. Jennings, nor
Sir John, nor even Mrs. Palmer herself, ever spoke of him before her.
Elinor wished that the same forbearance could have extended towards
herself, but that was impossible, and she was obliged to listen day
after day to the indignation of them all.

log-likelihood values computed for the probable authors

Author  LL
0           -3126.5812874
1           -4127.9155186
2           -7364.15782346
3           -9381.06336055
4           -7493.78440066
5           -4837.98005673
6           -3515.44028659
7           -3455.85716104

As can be seen from above the maximum likelihood value corresponds to the author 0, i.e., Austen.  Hence, the most probable author of the unknown text is Austen.

 

 

Unknown Text2 

Then he tossed the marble away pettishly, and stood cogitating. The
truth was, that a superstition of his had failed, here, which he and
all his comrades had always looked upon as infallible. If you buried a
marble with certain necessary incantations, and left it alone a
fortnight, and then opened the place with the incantation he had just
used, you would find that all the marbles you had ever lost had
gathered themselves together there, meantime, no matter how widely they
had been separated. But now, this thing had actually and unquestionably
failed. Tom’s whole structure of faith was shaken to its foundations.
He had many a time heard of this thing succeeding but never of its
failing before. It did not occur to him that he had tried it several
times before, himself, but could never find the hiding-places
afterward. He puzzled over the matter some time, and finally decided
that some witch had interfered and broken the charm. He thought he
would satisfy himself on that point; so he searched around till he
found a small sandy spot with a little funnel-shaped depression in it.
He laid himself down and put his mouth close to this depression and
called–

“Doodle-bug, doodle-bug, tell me what I want to know! Doodle-bug,
doodle-bug, tell me what I want to know!”

The sand began to work, and presently a small black bug appeared for a
second and then darted under again in a fright.

“He dasn’t tell! So it WAS a witch that done it. I just knowed it.”

He well knew the futility of trying to contend against witches, so he
gave up discouraged. But it occurred to him that he might as well have
the marble he had just thrown away, and therefore he went and made a
patient search for it. But he could not find it. Now he went back to
his treasure-house and carefully placed himself just as he had been
standing when he tossed the marble away; then he took another marble
from his pocket and tossed it in the same way, saying:

“Brother, go find your brother!”

He watched where it stopped, and went there and looked. But it must
have fallen short or gone too far; so he tried twice more. The last
repetition was successful. The two marbles lay within a foot of each
other.

log-likelihood values computed for the probable authors

Author  LL
0             -2779.02810424
1             -2738.09304225
2             -5978.83684489
3             -6551.16571407
4             -5780.39620942
5             -4166.34886511
6             -2309.25043697
7             -2033.00112729

As can be seen from above the maximum likelihood value corresponds to the author 7, i.e., Twain.  Hence, the most probable author of the unknown text is Twain. The following figure shows the relevant states corresponding to the Markov model for Twain trained from the training dataset.

Unknown_2_mc_7.png

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SIR Epidemic model for influenza A (H1N1): Modeling the outbreak of the pandemic in Kolkata, West Bengal, India in 2010 (Simulation in Python & R)

This appeared as a project in the edX course DelftX: MathMod1x Mathematical Modelling Basics and the project report can be found here. This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

Summary

In this report, the spread of the pandemic influenza A (H1N1) that had an outbreak in Kolkata, West Bengal, India, 2010 is going to be simulated. The basic epidemic SIR model will be used, it describes three populations: a susceptible population, an infected population, and a recovered population and assumes the total population (sum of these 3 populations) as fixed over the period of study.

There are two parameters for this model: namely the attack rate (β) per infected person per day through contacts and the recovery rate (α). Initially there will be a small number of infected persons in the population. Now the following questions are to be answered with the simulation / analysis:

1. Whether the number of infected persons increase substantially, producing an epidemic, or the flue will fizzle out.
2. Assuming there is an epidemic, how will it end? Will there still be any susceptibles left when it is over?
3. How long will the epidemic last?

The Euler’s method will be primarily used to solve the system of differential equations for the SIR model and compute the equilibrium points (along with some
analytic solution attempts for a few simplified special cases). Here are the
conclusions obtained from the simulations:

1. When the recovery rate α is ≈ 0 or very very low compared to the attack rate β, the flu will turn out to be an epidemic and the entire population will be infected first (the higher β is the quicker the epidemic outbreak).

2. To be more precise, when the initial susceptible population S(0) is greater than the inverse of the basic reproduction number 1/ R0 = α / β, a proper epidemic will break out.

3. When the initial susceptible population S(0) is less than the inverse of the basic reproduction number 1/R0=α/β, then a proper epidemic will never break out.

4. If the initial susceptible population is non-zero, in the end (at equilibrium) there will always be some susceptible population.

5. When there is an epidemic, it will eventually end in the equilibrium point with 0 infected population, how fast it reaches the equilibrium depends upon the recovery rate (the higher α is the quicker the infection removal).

6. The time to reach the equilibrium can be computed using Euler’s method, it depends on the parameters α (the higher the quicker) and β (the higher the quicker) and the initial infected populated size I(0) (the higher the quicker).

lv.png


Introduction

In 2010, the pandemic influenza A (H1N1) had an outbreak in Kolkata, West Bengal, India. An increased number of cases with influenza like illness (ILI) were reported in Greater Kolkata Metropolitan Area (GKMA) during July and August 2010, as stated in [3]. The main motivation for this research project will be to understand the spread of the pandemic, compute the equilibrium points and find the impact of the initial values of the infected rate and the attack / recovery rate parameters on the spread of the epidemic, using simulations using the basic epidemic SIR model.

The Euler’s method will be primarily used to solve the system of differential equations
for SIR model and compute the equilibrium points. First a few simplified special cases will be considered and both analytic and numerical methods (with Euler method) will be used to compute the equilibrium points. Then the solution for the generic model will be found. As described in [6], the SIR model can also be effectively used (in a broader
context) to model the propagation of computer virus in computer networks, particularly for the networks with Erdos-Renyi type random graph topology.


SIR Epidemic Model

The SIR model is an epidemiological model that computes the theoretical number of people infected with a contagious illness in a closed population over time. One of the basic one strain SIR models is Kermack-McKendrick Model. The Kermack-McKendrick Model is used to explain the rapid rise and fall in the number of infective patients observed in epidemics. It assumes that the population size is fixed (i.e., no births, no deaths due to disease nor by natural causes), incubation period of the infectious agent
is instantaneous, and duration of infectivity is the same as the length of the disease. It also assumes a completely homogeneous population with no age, spatial, or social structure.

The following figure 2.1 shows an electron microscope image of the re-assorted H1N1 influenza virus photographed at the CDC Influenza Laboratory. The viruses are 80 − 120 nm in diameter [1].

virus.png

2.1 Basic Mathematical Model

The starting model for an epidemic is the so-called SIR model, where S stands for susceptible population, the people that can be infected. I is the already infected population, the people that are contagious, and R stands for the recovered population, people who are not contagious any more.

2.1.1 Differential Equations

The SIR model can be defined using the following ordinary differential equations 2.1:

dif1

• The terms dS/dtdI/dtdR/dt in the differential equations indicate the rate of change of the susceptible population size, the infected population size and the recovered population size, respectively.

• The terms β and α indicate the attack rate (number of susceptible persons get infected per day) and the recovery rate of the flu (inverse of the number of days a person remains infected), respectively.

• High value of α means a person will be infected by the flu for less number of days and high value of β means that the epidemic will spread quickly.

• Also, as can be seen from below, from the differential equations it can be shown that the population (S + I + R) is assumed to be constant.

dif2.png

2.1.2 Collected data, units and values for the constants

• As can be seen from the following figure 2.2, the focus of this analysis will be limited to  the population in Kolkata Metropolitan Corporation (KMC, XII) area where the population can be assumed to be ≈ 4.5 million or 4500 thousands, as per [7].

Units

– All population (S, I, R) units will be in thousands persons (so that total population
N = 4500).
– As can be derived from the differential equations 2.1, the unit of β will be in
10^(−6) /persons/day (β = 25 will mean 25 persons in a million gets infected by
susceptible-infected contact per infected person per day).
– Similarly, the units of α will be in 10^(−3) / day (α = 167 will mean 167 × 10^(−3) /
day gets recovered from the flu per day).

• The attack rate is 20-29/100000 and the number of days infected (i.e. the inverse of recovery rate) = 5−7 days on average (with a few exceptions), as per [3].

• Typical values for β and α can be assumed to be 25 /person / day and 10^3/6 ≈ 167 / day, respectively.

f1.jpg

2.2 Simplified Model 1 (with α = 0)

• At first a simplified model is is created assuming that α = 0 (/ day) and that R = 0, so once infected, a person stays contagious for ever. Because S(t) + I(t) + R(t) = S(t) + I(t) = N is constant (since population size N is fixed), S(t) can be eliminated and a single differential equation in just I(t) is obtained as shown in the equation below 2.2.

f2

• Also, let the (fixed) population size N = 4500 = S(0) + I(0), (in thousand persons), initially the number of persons infected = I(0) = 1 (in thousand persons) and number of persons susceptible S(0) = N −I(0) = 4499 (in thousand persons), respectively. Let β = 25 × 10^(−6) /persons / day) to start with.

2.2.1 Analytic Solution

• The analytic solution can be found by following the steps shown in the Appendix A and the final solution is shown in the below equations 2.3:

f3.png
• The following figure 2.3 shows the logistic (bounded) growth in I(t) (in thousands persons) w.r.t. the time (in days) for different values of attack rate β×10^(−6) (/ person / day). As expected, the higher the attack rate, the quicker all the persons in the population become infected.

f4.png

2.2.2 Finding the equilibrium points for I

• The equilibrium points are the points where the rate of change in I is zero, the points that satisfy the following equation

f5

• Considering a small neighborhood of the equilibrium point at I = 0, it can be seen from the figure 2.4 that whenever I > 0, dI/dt > 0, so I increases and goes away from the equilibrium point.

• Hence, the equilibrium point at I = 0 is unstable.

• At I = N = 4500 (in thousand persons) it is a stable equilibrium. As can be seen from the following figure 2.4, in a small neighborhood of the equilibrium point at I = 4500, it always increases / decreases towards the equilibrium point.

• In a small  ε > 0 neighborhood at I = 4500 (in thousand persons),

1. dI/dt > 0, so I increases when I <= 4500 − ε .
2. dI/dt > 0, so I decreases when I >= 4500 +  ε .

• The same can be observed from the direction fields from the figure 2.5.

• Hence, the equilibrium at I = 4500 is stable.

f6.png

f7.png

2.2.3 Numerical Solution with the Euler’s Method

• The algorithm (taken from the course slides) shown in the following figure 2.6 will be used for numerical computation of the (equilibrium) solution using the Euler’s method.

f8.png

• As can be seen from the figure 2.6, then the infection at the next timestep can be (linearly) approximated (iteratively) by the summation of the the infection current timestep with the product of the difference in timestep and the derivative of the infection evaluated at the current timestep.

2.2.3.1 Finding the right step size (with β = 25 × 10^(−6)/person/day)

• In order to decide the best step size for the Euler method, first the Euler’s method is run with different step sizes as shown in the figure 2.7.

• As can be seen from the following table 2.1 and the figure 2.7, the largest differences in the value of I (with two consecutive step sizes) occurs around 78 days:

f9.png

f10.png
• As can be seen from the table in the Appendix B, the first time when the error becomes less than 1 person (in thousands) is with the step size 1/512 , hence this step size will be used for the Euler method.

2.2.3.2 Computing the (stable) equilibrium point

• Now, this timestep will be used to solve the problem to find the equilibrium time teq (in days). Find teq such that N − I(teq) < ε = 10^(−6) , the solution obtained is teq = 272.33398 days ≈ 273 days.

• Now, from the analytic solution 2.3 and the following figure 2.8, it can be verified that the teq solution that the Euler’s method obtained is pretty accurate (to the ε tolerance).

f11.png

2.2.3.3 Results with β = 29 × 10^(−6) / person / day, I(0) = 1 person

• Following the same iterations as above, the steepest error is obtained at t = 67 days in this case, as shown in the figure 2.9.

• The first time when error becomes less than one person for t = 67 days with the Euler ‘s method is with step size 1/512 again.

• The solution obtained is teq = 234.76953125 days ≈ 235 days, so the equilibrium (when all the population becomes infected) is obtained earlier as expected, since the attack rate β is higher.

f12.png

2.2.3.4 Results with β = 25 × 10^(−6) / person / day, with different initial values for infected persons (I(0))

• Following the same iterations as above, the equilibrium point is computed using the Euler’s method with different values of initial infected population I(0), as shown in the figure 2.10.

• The solutions obtained are teq = 272.33, 258.02, 251.85, 248.23, 245.66, 245.66 days for I(0) = 1, 5, 10, 15, 20 days, respectively. So the equilibrium is obtained earlier when the initial infected population size is higher, as expected.

f13.png

2.3 Simplified Model 2 (with β = 0)

• Next, yet another simplified model is considered by assuming that β = 0 and that α > 0, so the flu can no more infect anyone (susceptible, if any, possibly because everyone got infected), an infected person recovers from flu with rate α. This situation can be described again with a single differential equation in just I(t) as shown in the equation below 2.4.

f14

• Also, let the the entire population be infected, N = 4500 = I(0), (in thousand persons), initially the number of persons susceptible = S(0) = 0, respectively. Let α = 167 × 10^(−3)   (/ day) to start with.

2.3.1 Analytic Solution

• The analytic solution can be found by following the steps shown in the below equations 2.5:

f15

• The following figure 2.11 shows the exponential decay in I(t) (in thousand persons)  w.r.t. the time (in days) for different values of recovery rate α × 10^(−3)  (/ day). As expected, the higher the recovery rate, the quicker all the persons in the population get rid of the infection.

f16.png

• Now, I(t) + R(t) = N (since S(t) = 0 forever, since no more infection) and I(0) = N, combining with the above analytic solution I(t) = I(0).exp(−αt) = N.exp(−αt), the following equation is obtained:
f17

• The following figure 2.12 shows the growth in R(t) (in thousand persons) w.r.t. the time (in days) for different values of recovery rate α × 10^(−3) (/ day). As expected, the higher the recovery rate, the quicker all the persons in the population move to the removed state.

f18.png

2.3.2 Numerical Solution with the Euler’s Method

2.3.2.1 Solution with α = 167 × 10−3 / day

• Following the same iterations as above, the steepest error is obtained at t = 6 in this case, as shown in the figure 2.16.

• The first time when error becomes less than one person for t = 67 with the Euler’s method is with step size 1/256 .

• The solution obtained with the Euler’s method is 133.076171875 days ≈ 133 days to remove the infection from population with 10^(−6) tolerance. From the analytic solution,
I(133) = N.exp(−αt) = 1.016478E−06, similar result is obtained.

2.3.2.2 Results

The following figure 2.16 shows the solutions obtained with different step sizes using the Euler’s method.

f19.png

2.4 Generic Model (with α, β > 0)

First, the numeric solution will be attempted for the generic model (using the Euler’s method) and then some analytic insights will be derived for the generic model.

2.4.1 Numerical Solution with the Euler’s  Method

• The following algorithm 2.14 shown in the next figure is going to be used to obtain the solution using Euler method (the basic program for Euler’s method, adapted to include three dependent variables and three differential equations).

f20.png

• As can be seen from the figure 2.14, first the vector X(0) is formed by combining the three variables S, I, R at timestep 0. Then value of the vector at the next timestep can be (linearly) approximated (iteratively) by the (vector) summation of the vector value at the current timestep with the product of the difference in timestep and the derivative of the
vector evaluated at the current timestep.

2.4.1.1 Equilibrium points

• At the equilibrium point,

f21

There will be no infected person at the equilibrium point (infection should get removed).

• As can be seen from the following figure 2.15 also, I = 0 is an equilibrium point, which is quite expected, since in the equilibrium all the infected population will move to the removed state.

• Also, at every point the invariant S + I + R = N holds.

• In this particular case shown in figure 2.15, the susceptible population S also becomes 0 at equilibrium (since all the population got infected initially, all of them need to move to removed state) and R = N = 4500 (in thousand persons).

f22.png

2.4.1.2 Results with Euler method

• As explained in the previous sections, the same iterative method is to find the right stepsize for the Euler method. The minimum of the two stepsizes determined is
∆t = 1/512 day and again this stepsize is going to be used for the Euler’s method.

• The following figures show the solutions obtained with different values of α, β with the initial infected population size I(0) = 1 (in thousand persons). Higher values for the parameter β obtained from the literature are used for simulation, since β = 25 × 10^(−6) /person /day is too small (with the results not interesting) for the growth of the epidemic using the Euler’s method (at least till ∆t = 1/ 2^15), after which the iterative Euler’s
method becomes very slow).

• As can be seen, from the figures 2.16, 2.17 and 2.19, at equilibrium, I becomes zero.

• The solution (number of days to reach equilibrium) obtained at α = 167×10^(−3) /day and β = 25×10^(−5) /person /day is teq = 143.35546875 ≈ 144 days with I(0) = 1 (in thousand persons), the corresponding figure is figure 2.16.

• The solution (number of days to reach equilibrium) obtained at α = 167 × 10^(−3) /day and β = 5 × 10^(−5) /person /day is teq ≈ 542 days with I(0) = 1 (in thousand persons), the corresponding figure is figure 2.17.

• Hence, higher the β value, the equilibrium is reached much earlier.

• The solution obtained at α = 500 × 10^(−3) /day and β = 25 × 10^(−5) /person /day is
teq ≈ 78 days with I(0) = 1 (in thousand persons), the corresponding figure is figure 2.19.

• Hence, higher the α value, the equilibrium is reached earlier.

• The solution obtained at α = 167×10^(−3) /day and β = 25×10^(−5) /person /day is
teq = 140 days with I(0) = 10. Hence, as expected, higher the number of initial infected population size, quicker the equilibrium is reached.

• At equilibrium, S does not necessarily become close to zero, since sometimes the entire  population may not get infected ever, as shown in the figure 2.17, where at equilibrium the susceptible population is non-zero.

f23f24f25f26

• As can be seen from the phase planes from following figure 2.21, at equilibrium, the infected population becomes 0.

f27.png

2.4.2 Analytic Solution and Insights

2.4.2.1 Basic Reproduction Number (R0)

The basic reproduction number (also called basic reproduction ratio) is defined by
R0 = β / α (unit is /day). As explained in [2], this ratio is derived as the expected number of new infections (these new infections are sometimes called secondary infections) from a single infection in a population where all subjects are susceptible. How the dynamics of the system depends on R0 will be discussed next.

2.4.2.2 The dynamics of the system as a function of R0

• By dividing the first equation by the third in 2.1, as done in [2], the following equation is obtained:

f28.png

• Now, at t → ∞, the equilibrium must have been already reached and all infections must have been removed, so that lim (t→∞) I(t) = 0.

• Also, let R∞ = lim (t→∞) R(t).

• Then from the above equation 2.7, R∞ = N − S(0).exp(R0.(R∞−R(0)))
.
• As explained in [2], the above equation shows that at the end of an epidemic, unless
S(0) = 0, not all individuals of the population have recovered, so some must remain  susceptible.

• This means that the end of an epidemic is caused by the decline in the number of infected individuals rather than an absolute lack of susceptible subjects [2].

• The role of the basic reproduction number is extremely important, as explained in [2]. From the differential equation, the following equation can be obtained:

f29

S(t) > 1/R0 ⇒ dI(t)/dt > 0 ⇒ there will be a proper epidemic outbreak with an increase of the number of the infectious (which can reach a considerable fraction of the population).

S(t) < 1 R0 ⇒ dI(t) dt < 0 ⇒ independently from the initial size of the susceptible population the disease can never cause a proper epidemic outbreak.

• As can be seen from the following figures 2.21 and 2.22 (from the simulation results obtained with Euler method), when S(0) > 1/R0 , there is a peak in the infection curve, indicating a proper epidemic outbreak.

• Also, from the figures 2.21 and 2.22, when S(0) > 1/R0 , the higher the the gap between S(0) and 1/R0 , the higher the peak is (the more people get infected) and the quicker the peak is attained.

• Again, from the figure 2.22, when 4490 = S(0) < 1/R0 = 5000, it never causes a proper epidemic outbreak .

f30

f31.png

 

• Again, by dividing the second equation by the first in 2.1, the following equation is obtained:

f32.png

f33.png

 

• As can be noticed from the above figure 2.23 that because the formulas differ only by an additive constant, these curves are all vertical translations of each other.

• The line I(t) = 0 consists of equilibrium points.

• Starting out at a point on one of these curves with I(t) > 0, as time goes on one needs to travel along the curve to the left (because dS/dt < 0), eventually approaching at some positive value of S(t).

• This must happen since on any of these curves, as I(t) → ∞, as S(t) → 0, from equation 2.8.

• So the answer to question (2) is that the epidemic will end as with approaching some positive value and thus there must always be some susceptibles left over.

• As can be seen from the following figure 2.24 (from the simulation results obtained with the Euler’s method), when S(0) > 1/R0 , lesser the the gap between S(0) and 1/R0 , the higher the population remains susceptible at equilibrium (or at t → ∞).

f34.png

Conclusions

In this report, the spread of the pandemic influenza A (H1N1) that had an outbreak in Kolkata, West Bengal, India, 2010 was simulated using the basic epidemic SIR model.Initially there will be a small number of infected persons in the population, most of the population had susceptible persons (still not infected but prone to be infected) and zero removed persons. Given the initial values of the variables and the parameter (attack and recovery rates of the flu) values, the following questions were attempted to be answered with the simulation / analysis:

1. Whether the number of infected persons increase substantially, producing an epidemic, or the flue will fizzle out.

2. Assuming there is an epidemic, how will it end? Will there still be any susceptibles left when it is over?

3. How long will the epidemic last?
The following conclusions are obtained after running the simulations with
different values of the parameters and the initial values of the variables:

1. When the recovery rate α is ≈ 0 or very very low compared to the attack rate β (so that R0 = β / α >> 1) and I(0) > 1, the flu will turn out to be an epidemic and the entire population will be infected first (the higher β is the quicker the epidemic break out).

2. To be more precise, when the initial susceptible population S(0) is greater than the inverse of the basic reproduction number 1/R0 = α / β, a proper epidemic will break out.

3. When the initial susceptible population S(0) is less than the inverse of the basic reproduction number 1/R0 = α/β, then a proper epidemic will never break out.

4. If the initial susceptible population is non-zero, in the end (at equilibrium) there will always be some susceptible population.

5. When there is an epidemic, it will eventually end in the equilibrium point with 0 infected population, how fast it reaches the equilibrium depends upon the recovery rate (the higher α is the quicker the infection removal).

6. The time to reach the equilibrium can be computed using Euler method, it depends on the parameters α (the higher the quicker) and β (the higher the quicker) and the initial infected populated size I(0) (the higher the quicker).

7. Scope of improvement: The SIR model could be extended to The Classic Endemic Model [5] where the birth and the death rates are also considered for the population (this will be particularly useful when a disease takes a long time to reach the equilibrium state).

f35.png

f36.png

f37.png