https://www.safaribooksonline.com/library/view/computer-networks-fifth/978013348

Question

https://www.safaribooksonline.com/library/view/computer-networks-fifth/9780133485936/

1. The ping program allows you to send a test packet to a given location and see how long it takes to get there and back. Try using ping to see how long it takes to get from your location to several known locations. From these data, plot the one-way transit time over the Internet as a function of distance. It is best to use universities since the location of their servers is known very accurately. For example, berkeley.edu is in Berkeley, California; mit.edu is in Cambridge, Massachusetts; vu.nl is in Amsterdam; The Netherlands; www.usyd.edu.au is in Sydney, Australia; and www.uct.ac.za is in Cape Town, South Africa.
Does the response times reflect the physical distance in your experiments? Can we conclude it is always the case? (Problem 33, page 88 of the text book.)

2. An image is 1024 x 768 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet? (Problem 23, page 87 of the text.)

Explanation / Answer

1.Yes response times reflect the physical distance, find the detail below:

Microsoft Windows [Version 6.1.7601]
Copyright (c) 2009 Microsoft Corporation. All rights reserved.

C:UsersIBM_ADMIN>ping berkeley.edu

Pinging berkeley.edu [35.163.72.93] with 32 bytes of data:
Reply from 35.163.72.93: bytes=32 time=508ms TTL=45
Reply from 35.163.72.93: bytes=32 time=533ms TTL=45
Reply from 35.163.72.93: bytes=32 time=360ms TTL=45
Request timed out.

7,664 mi
Distance from Berkeley to Delhi

Ping statistics for 35.163.72.93:
    Packets: Sent = 4, Received = 3, Lost = 1 (25% loss),
Approximate round trip times in milli-seconds:
    Minimum = 360ms, Maximum = 533ms, Average = 467ms

C:UsersIBM_ADMIN>ping mit.edu

Pinging mit.edu [184.24.164.227] with 32 bytes of data:
Reply from 184.24.164.227: bytes=32 time=488ms TTL=52
Reply from 184.24.164.227: bytes=32 time=1791ms TTL=52
Reply from 184.24.164.227: bytes=32 time=549ms TTL=52
Reply from 184.24.164.227: bytes=32 time=606ms TTL=52
7,121 mi
Distance from Cambridge to Delhi

Ping statistics for 184.24.164.227:
    Packets: Sent = 4, Received = 4, Lost = 0 (0% loss),
Approximate round trip times in milli-seconds:
    Minimum = 488ms, Maximum = 1791ms, Average = 858ms

C:UsersIBM_ADMIN>ping vu.nl

Pinging vu.nl [37.60.194.64] with 32 bytes of data:
Reply from 37.60.194.64: bytes=32 time=410ms TTL=240
Reply from 37.60.194.64: bytes=32 time=437ms TTL=240
Reply from 37.60.194.64: bytes=32 time=1508ms TTL=240
Reply from 37.60.194.64: bytes=32 time=251ms TTL=240

6,339 km
Distance from Amsterdam to Delhi

Ping statistics for 37.60.194.64:
    Packets: Sent = 4, Received = 4, Lost = 0 (0% loss),
Approximate round trip times in milli-seconds:
    Minimum = 251ms, Maximum = 1508ms, Average = 651ms

C:UsersIBM_ADMIN>ping www.usyd.edu.au

Pinging rp0.ucc.usyd.edu.au [129.78.5.11] with 32 bytes of data:
Reply from 129.78.5.11: bytes=32 time=456ms TTL=242
Reply from 129.78.5.11: bytes=32 time=448ms TTL=242
Reply from 129.78.5.11: bytes=32 time=1704ms TTL=242
Reply from 129.78.5.11: bytes=32 time=476ms TTL=242
Showing results for distance Sydney, Australia to delhi india
Search instead for distiance Sydney, Australia to delhi india
Search Results

10,436 km
Distance from Sydney to Delhi

Ping statistics for 129.78.5.11:
    Packets: Sent = 4, Received = 4, Lost = 0 (0% loss),
Approximate round trip times in milli-seconds:
    Minimum = 448ms, Maximum = 1704ms, Average = 771ms

C:UsersIBM_ADMIN>ping www.uct.ac.za

Pinging ecm-vip-prd.uct.ac.za [137.158.154.230] with 32 bytes of data:
Request timed out.
Request timed out.
Request timed out.
Request timed out.
9,298 km

9,298 km
Distance from Cape Town to Delhi

Ping statistics for 137.158.154.230:
    Packets: Sent = 4, Received = 0, Lost = 4 (100% loss),

C:UsersIBM_ADMIN>

2.Find of image will be

1024*768*3 bytes = 2.359296 megabytes

Since mininum of all decide the bandwidth so, it will have bandwidth of 56 kbps

hence our answer will be (2.359296 megabytes/56kbps)= (2.359296 megabytes/7kBps)

= (2 359.296 kilobytes/7kBps)

=337.042285714 sec

=5.61737142857 minutes

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