Declan bytes. This is 18,874,368 bits. Over a 56-kbs(56,000

DeclanBialowasHassanPeyraviComputerCommunications Network1/27/2018Homework #11.Thereare two reasons for using layered protocols which is that it breaks up theproblem in question smaller and more manageable pieces, as well as thatprotocols can be changed more easily without affecting higher or lower layersinvolved.

Onepossible disadvantage of using layered protocols is the matter of redundancyand overall lower performance comparatively. 2.Message M Layer 1— M + h Layer 2— M + h + h = M + 2h Layer 3— M + 2h + h = M + 3hLayer n— M + (n – 1)h + h = M + n x h   # of header bytes with n-layers and h bytesper header=(n x h). The fraction of the network bandwidth filled with headersis (n x h)/(M+(n x h)). 3.The main difference between TCP and UDP would have to be the fact UDP isconnectionless while TCP is specifically connection oriented.  4.

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Campus                                   Address       Average Time       DistanceStanfordUniversity               stanford.edu         89  ms                2,518 milesUniversityof California, L.A.  ucla.

edu            65 ms             2,386milesUniversityof Utah                 utah.edu              48ms               1,747 milesPrincetonUniversity              princeton.edu      34.6 ms            428 milesMichiganState University     msu.edu               21 ms              255 miles            The findings indeed point to thecorrelation between ping time and physical distance 5.            The image is 1024 x 768 x 3 bytes or 2,359,296 bytes.

This is 18,874,368 bits. Over a 56-kbs(56,000 bits/sec) modem channel, it takes  18,874,368/56,000=337.042 sec. Over a 1-mbs(1,000,000 bits/sec) cable modem, it takes 18,874,368/1,000,000=18.

874 sec. Over a 10-mbs(10,000,000 bits/sec) Ethernet, it takes 18,874,368/10,000,000=1.887 sec. Over a 100-mbs(100,000,000 bits/sec) Ethernet, it takes 18,874,368/100,000,000=0.189 sec.