Shail of success and q=1-p= probability of failure.

Shail Sandip Patel                                         ITCS-6166 HW1                                  Student ID: 801054666

 

3.            File size F= 4*10^6 bytes

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               R1=500 *10^3 bps                      R2= 2*10^6 bps                                R3=1*10^6 bps

(1)    The minimum throughput (slowest one)= 500kbps

(2)    Time taken by the file to reach Host B = File size/throughput=8*4*10^6/500*10^3=64 seconds.

 

4.           Link size=3 Mbps

               Transmission speed=150 Kbps

(a)    Number of users supported for ckt switching= Link size/Tx speed=3*10^6/150*10^3=20

(b)    For packet switching,

P(that a given user is transmitting)= 10/100=0.1

 

(c)     Number of users=120

Using Binomial distribution,

P(n users are transmitting simultaneously)= 120

                                                                                    C       p^n  q^120-n

                                                                                       n

where p= probability of success and q=1-p= probability of failure.

 

   2.         Two ISPs at the same level of hierarchy (eg. Tier 2 to Tier 2) will often peer with each other to

               avoid paying the fee to intermediary ISP for exchanging traffic and routing information. This

               reduces the latency as well. Each provider peers with other providers at Internet Exchange

               Point(IXP). At IXP, one or more ISPs can peer with each other. When every ISP that connects to

               the IXP, some amount is charged by an IXP depending on the data traffic send by these ISPs.

               This is how an IXP earns money.

 

5.           (a) For the total average response time:

                   The time to transmit an object of size L over a link or rate R is L/R. The average time is the         

                   average size of the object divided by R:

                   So, ? = 6,50,000 bits/15*10^6 bits/sec

                      =>   ? = 0.0433 seconds

                   Now, the traffic intensity on the link is given by,

                      ?? = 16 req/sec * 0.0433 sec/req

                      Therefore, ?? = 0.6928

                       The average access delay is = ?/1-?? = 0.0433/1-0.6928 = 0.14 seconds.

                       Total average response time = 0.14 sec + 2.4 sec = 2.54 sec

                   (b) For the average response time:

                         Hit rate = 0.7

                         Miss rate = 1- 0.7 = 0.3

                         The traffic intensity on the access link is reduced by 70% since the 70% of the requests are  

                         satisfied within the institutional network.

                         Thus the average access delay is = 0.0433/1-(0.3)(0.6928) = 0.054 seconds

                          The response time is approximately zero if the request is satisfied by the cache (which     

                          happens with probability 0.7).

                          The average response time is 0.054 sec + 2.4 sec = 2.454 sec for cache misses (which  

                          happens 30% of the time).

                          So, the average response time is (0.7)(0 sec) + ( 0.3 )( 2.454 sec) = 0.7362 seconds

                          Thus the average response time is reduced from 2.54 sec to 0.7362 sec.

 

1.       Digital Subscriber Line:

DSL is used for supporting high-speed digital communication over the existing telephone lines and is a distance sensitive technology. These modems contain signal splitters to carry voice signals at low frequencies ranging from 0 to 4KHz.

It is categorized as follows:

(a)    ADSL- Asymmetric DSL : They provide lower rate upstream and higher rate downstream.

(b)    SDSL- Symmetric DSL: They provide similar data rates for both, upstream and downstream.

(c)     VDSL- Very high rate DSL: They work on fiber optic cables and achieves highest data rate and needs cables of shorter lengths.

(d)    HDSL- High data rate DSL: They provide similar bandwidth in both the directions using twisted pair cables and requires two telephone lines for the delivery of basic data rate.

             Cable broadband access networks:

Hybrid Fiber Coaxial (HFC) network:

 

Both, fiber and coaxial cable are used here. It’s a second generation of cable n/w. The coaxial cable has a bandwidth ranging from 5MHz. For the internet access, this BW is divided into three streams making the use of cable modems.

For HFC, the downstream transmission rate is up to 30Mbps.  

 

                         

(Reference: Data communications and networking by BEHROUZ FOROUZAN )

 

 

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