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					    Computer Network Design


          CMPE 206
       HOME WORK - 2



       Name: Parth Parikh


       SJSU ID: 007511062


        Class Section: 02




1
                                        R-11

HTTP,FTP,SMTP, & POP3 run on top TCP layer because TCP layer provides three
important factors:-



1)Reliable Data Transfer:- It delivers all data sent without error in the proper
order.



2)Connection-Oriented service:- The client & server exchanges transport layer
control information with each other before the application-level messages begin
to flow.This is called handshaking procedure.After the handshaking phase a TCP
connection is said to exist between the sockets of the two processes.The
connection is full-duplex connection.After the application finishes sending
messages,it must tear down the connection.



3)Congestion-Control Mechanism:- It limits each TCP connection to its fair share
on network bandwidth.



While UDP does not often such services.So HTTP,FTP,POP3 & SMTP runs on TCP.




     2
                                     R-16




POP3 is an extremely simple mail access protocol.POP begins when the user agent
opens a TCP connection to the mail server.With TCP connection established,POP3
progres through three phases: authorization,transaction and update.




     3
Authorization:- the user agent sends a username and a password to authenticate
the user.

Transaction:-user agent retrieves messages,also during this phase the user agent
can mark messages for deletion,remove deletion marks,and obtain mail
statistics.OK and ERR are two possible responses.OK is used by the server to
indicate that the previous command was fine.ERR is used to indicate that
something was wrong with previous command.

Update:-occurs after the client has issued the quit command,ending the POP3
Session;at this time,the mail server deletes the messages that were marked for
deletion.Download and keep,Download and delete are two possible responses.In
download and keep the user can access through anywhere wnd at anytime,while
in download and delete the user can acces only once.

During the session between the user agent and the mail server,pop3 server
maintains some state information.




     4
                                          P-9

                                          (A)



Average object size = 850000 bits

Average request rate = 16/sec

Average access delay = Δ/(1-Δβ)

Δ = average time to send an object over the access link

β = arrival rate of objects to the access link

Δ=L/R

 = 850000 / 1500000

 = 0.56 sec

The traffic intensity on the link Δβ is (1.5 * 0.56) = 0.84

Average access delay Δ/(1-Δβ) is 0.56 sec / (1 – 0.84) = 3.5 sec

Total average response time is therefore 3.5 + 3 = 6.5 sec




     5
                                        (B)



Miss rate = 0.4

Hit rate = 0.6

Therefore response time is

=0.56 / (1- (0.56 * 0.84))

=1.05sec

The response time is approximatelu zero if request is satisfied by the cache.

The avg response response time is 1.21 + 3 =4.21 sec for cache misses

So the total response time is

= 0.6 * ( 0 ) + 0.4 * (4.21)

= 1.68 sec



The average response time is reduced from 6.5 sec to 1.68 sec




      6
                                      P-22



F = 15Gbits

Upload rate Us = 30Mbps

Download rate di = 2Mbps

N = 10,100,1000

u = 300Kbps,700Kbps,2Mbps



for client-server

N=10

Dcs = max (N * F / Us , F / dmin )

   = max (10 * 15 * 109 / 30 * 106 , 15*109 / 2 * 106 )

   = max (5 * 103 , 7.5 * 103 )

   = 7.5 * 103

  =7500

N=100

Dcs = max (N * F / Us , F / dmin )

   = max (100 * 15 * 109 / 30 * 106 , 15*109 / 2 * 106 )

   = max (50 * 103 , 7.5 * 103 )

   = 50 * 103

  =50000

N=1000

     7
Dcs = max (N * F / Us , F / dmin )

   = max (1000 * 15 * 109 / 30 * 106 , 15*109 / 2 * 106 )

   = max (500 * 103 , 7.5 * 103 )

   = 500 * 103

  = 500000



For P2P

N=10

DP2P = max { F / Us , F / dmin , N * F / Us + ∑Ni=1 ui )

     =max {15 * 109 / 30 * 106 , 15 * 109 / 2 * 106 , 10 * 15 * 109 /(33000) }

     =max{0.5 * 103 , 7.5 * 103 ,4.54 * 106 }

     =4.54 * 106

N=100

DP2P = max { F / Us , F / dmin , N * F / Us + ∑Ni=1 ui )

     =max {15 * 109 / 30 * 106 , 15 * 109 / 2 * 106 , 100 * 15 * 109 /(33000) }

     =max{0.5 * 103 , 7.5 * 103 ,45.4 * 106 }

     =45.4 * 106

N=1000

DP2P = max { F / Us , F / dmin , N * F / Us + ∑Ni=1 ui )

     =max {15 * 109 / 30 * 106 , 15 * 109 / 2 * 106 , 1000* 15 * 109 /(33000) }

     =max{0.5 * 103 , 7.5 * 103 ,454.54 * 106 }

     =454.54 * 106

       8
Chart:-

  600000


  500000


  400000


  300000                                         c-s
                                                 p2p
  200000


  100000


       0
           Category 1   Category 2   category3




     9
                                         P-28



As,peer 5 has departed ,since it no longer responds to ping messages.Peer 3 and 4
thus need to update there successor state information.it will do in the following
ways:-

1)peer 3 will kept peer-4 as its first successor.

2)peer 3 then asks its first successor peer 4 for the identifier and IP address of its
immediate successor.Peer 3 then makes peer 8 its second successor

3)peer 3 will now change its second successor .Instead of peer-5 it would be
peer-8.

Thus,peer 4 is first successor and peer 8 is second new successor.




    10

				
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posted:2/10/2011
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