If you think about it, web browsing is basically a Markov chain – the page you will go to next…

If you think about it, web browsing is basically a Markov chain – the page you will go to next depends on the page you are currently at. Suppose our web server has three pages, and we have the following transition probabilities:

If you think about it, web browsing is basically a Markov chain – the page you will go to next...

where Pi,j represents the probability that I will next ask for page j, given that
I am currently at page i. Assume that 0 12 .
Web browsers cache pages so that they can be quickly retrieved later. We will
assume that the cache has enough memory to store two pages. Whenever a
request comes in for a page that is not cached, the browser will store that page
in the cache, replacing the page least likely to be referenced next based on
the current request. For example, if my cache contained pages {2,3} and I
requested page 1, the cache would now store {1,3} (because x 1 − x).
(a) Find the proportion of time that the cache contains the following pages:
(i) {1,2} (ii) {2,3} (iii) {1,3}.
(b) Find the proportion of requests that are for cached pages.

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