An application of Markov chains:
Google - PageRank Algorithm
Problem: list websites in order of rank.
Rank of a page: proportion of time the page
is visited.
Assumption: a page with many links to and
from pages of high rank should also be ranked
high.
Surfer: current page 3 new page:
 Choose a link from the current page randomly:
85%(= p) of the time
 Choose a page from the web randomly:
15%(= 1  p) of the time.
Notation:
 The web consists of n pages.
 Markov chain: each page is a state.
A = [aij] n n transition matrix.
aij is the probability of moving from page
j to page i.
 C = [cij] n  n matrix: describes the page
links.
cij =
8
<
:
1 if there is a link from page j to page i
0 otherwise
 sj = the number of pages to which page j
links:
sj =
nX
i=1
cij
Find aij?
 Page j has no links: sj = 0.
A surfer chooses a page from the web ran-
domly.
aij =
1
n
 Page j has links to other pages: sj T= 0.
2 ways to move from page j to page i:
1. surfer chooses link to page i on page j:
probability is p
cij
sj
2. surfer chooses a page from the web randomly
and the chosen page is page i:
probability (1  p)1
n
.
So if sj T= 0, aij = p
cij
sj
+ 1 p
n
.
Summarizing,
aij =
8
<
:
1
n
if sj = 0
p
cij
sj
+ 1 p
n
if sj T= 0
A has a steady-state vector. The components
of this vector are used to rank the webpages
listed by Google.
Example: suppose the Web consists of ten
pages. The links between the pages are given
by the matrix C:
C =
2
666666666666666664
0 1 1 1 0 0 0 1 1 0
0 0 0 0 1 0 1 0 0 0
0 0 0 1 0 0 0 1 1 0
0 0 1 0 0 0 1 1 0 0
1 0 1 1 0 0 0 0 0 0
0 0 1 0 0 0 0 0 0 0
0 0 1 0 1 0 0 1 0 0
0 1 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
1 0 0 1 1 0 0 0 0 0
3
777777777777777775
Using the formula for aij, you nd that matrix
A is given as follows:
2
666666664
0:015 0:440 0:185 0:2275 0:0150 0:1 0:015 0:2275 0:440 0:015
0:015 0:015 0:015 0:0150 0:2275 0:1 0:440 0:0150 0:015 0:015
0:015 0:015 0:015 0:2275 0:0150 0:1 0:015 0:2275 0:440 0:015
0:015 0:015 0:185 0:0150 0:0150 0:1 0:440 0:2275 0:015 0:015
0:440 0:015 0:185 0:2275 0:0150 0:1 0:015 0:0150 0:015 0:015
0:015 0:015 0:185 0:0150 0:0150 0:1 0:015 0:0150 0:015 0:015
0:015 0:015 0:185 0:0150 0:2275 0:1 0:015 0:2275 0:015 0:015
0:015 0:440 0:015 0:0150 0:2275 0:1 0:015 0:0150 0:015 0:015
0:015 0:015 0:015 0:0150 0:0150 0:1 0:015 0:0150 0:015 0:865
0:440 0:015 0:015 0:2275 0:2275 0:1 0:015 0:0150 0:015 0:015
3
777777775
Steady-state vector:
2
666666666666666664
0:1583
0:0774
0:1072
0:0860
0:1218
0:0363
0:0785
0:0769
0:1282
0:1295
3
777777777777777775
Reference: Spence L., Insel A., & Friedberg S. (2008).
Elementary Linear Algebra: A matrix approach. Upper
Saddle River, New Jersey: Pearson Prentice Hall.