# Shail C p^n q^120-n n where p= probability

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Last updated: July 19, 2019

Shail Sandip Patel                                         ITCS-6166 HW1                                  Student ID: 801054666 3.            File size F= 4*10^6 bytes               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=120Using Binomial distribution,P(n users are transmitting simultaneously)= 120                                                                                    C       p^n  q^120-n                                                                                       nwhere p= probability of success and q=1-p= probability of failure.    2.         Two ISPs at the same level of hierarchy (eg.

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Tier 2 to Tier 2) will often peer with each other to                avoid paying the fee to intermediate 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.

(Note: For this problem, I referred to class practice solved question)                    (a) For the total average response time:                      The time required to transmit an object of size L over a link or rate R is = L/R.                       The average time = average size of the object / 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 is reduced by 70% .                         Thus the average access delay is = 0.0433/1-(0.3)(0.6928) = 0.

054 seconds                          The response time is zero if the request is satisfied by the cache (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                           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. 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: This network uses a combination of fiber optic and coaxial cable. In HFC, coaxial cable is used between fiber node and subscriber equipment.The bandwidth of this coaxial cable is from 5-750 MHzThe max. data rate achieved in the downstream direction is 1 Gbps and for upstream it is 74 Mbps.

Cable Modem(CM) and cable Modem transmission system(CMTS): Here, the cable company becomes the provider. 