CS610 CS610 Computer Network Assignment No 01 Solution & Discussion Spring 2019 Due Date: 14-05-2019
Question No. 1 [Marks: 10]
Create a directory (Folder) and name it as your own student id i.e. mc123456. Go to your own student id directory through command prompt and execute ping and trace route commands for any of these websites:www.google.comwww.vu.edu.pk
Note: Take screenshots for both commands and show them in your solution.Both ping and trace route commands must be clearly visible in the screenshots.You should create the directory with your own student ID. If the name of the directory shown in the screenshot doesn’t match with your student id then the assignment will be considered as copied and will be awarded zero marks.
Question # 2 [Marks: 10]
How much time is required by a machine to transfer a file of 8MB across the network with 64Kbps?How much time is required to send a packet, if file is broken into packets of size 1024 bytes?
Note: You have to show the formula/calculation steps in each part of the numerical.No marks will be given for just the answers without showing the calculations/steps. Deadline: Your assignment must be uploaded/submitted on or before 14 May, 2019.
Note: In Question 1, you should make the directory with your own student id. If the screenshot doesn’t match with your student id then the assignment will be considered as copied and will be awarded zero marks
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Question # 2 [Marks: 10]
a) How much time is required by a machine to transfer a file of 8MB across the network with 64Kbps?
File size =8Mb
Time =8*106 bytes * 8 bits/byte / 64 * 103 * 60 sec/mint
Time= 8*106 * 8 /64 * 103 * 60
b) How much time is required to send a packet, if file is broken into packets of size 1024 bytes?
= 1024 bytes * 8 bit/byte / 64 * 103 bits/sec
=1024*8 / 64 * 103
= 0.128 secs
hy friends tell me about that solution is it true method
8MB=64kbps x T
Here R = 64kbps and I = 8 MB so,
T=8MB/64kbps as 1 MB = 8000kbps 8 x 1 MB = 8 x 8000kbps = 64000kbps
1 m =60 x 1 sec
1 m/60 = 1 sec
1000 m/60 = 1000 sec or kbps 1000kpbs or sec = 16.6666
T=16 minutes and 40 sec
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Solution CS 610 assignment
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CS610 Assignment#01 Solution Spring 2019