A packet of 10 Kb is to be downloaded from a web server.
Find the time needed to download this packet using:
a) A dial up telephone connection at 28 Kbps
b) A cable modem at 28 Mbps.
Answers were Sorted based on User's Feedback
Answer / david tilley
The straight maths answer is: Ti = Fs/Sp
Where:
Ti...Time
Fs..FileSize
Sp..Speed
a) Ti = 10,000 / 28,000
= 357ms
b) Ti = 10,000/ 28,000,000
= 357uS
However.... In practice transfer times would be longer than
this depending on the overhead added to handle error
checking/recovery.
Is This Answer Correct ? | 30 Yes | 6 No |
Answer / vaibhav
the correct answer is below
10kb=10*1024bytes=10240 bytes
28Kilobitsps=28/8KiloBytesps=3.5KBps=3.5*1024Bytesps=3584Bps
time=10240/3584=2.85 s
do same with 28.8Mbps=28.8*1024*1024.
Is This Answer Correct ? | 7 Yes | 5 No |
Answer / john
I agree this is the correct one.
10kb=10*1024bytes=10240 bytes
28Kilobitsps=28/8KiloBytesps=3.5KBps=3.5*1024Bytesps=3584Bps
time=10240/3584=2.85 s
do same with 28.8Mbps=28.8*1024*1024.
Is This Answer Correct ? | 2 Yes | 3 No |
Is youtube a tcp or udp?
What is ipv6?
what is the difference between server and database?
How do you check if a port is blocked?
is the private address 192.168.1.1 assigned by the dhcp?
Do I need a dedicated ip?
From the perspective of a computer's network configuration, the name given to the IP address of the router to which the hosts in a particular subnet must go through to get to other networks/subnets is the
Does DNS use TCP or UDP or both?
What is a subnet mask?
How to instal two operarting system in one pc?
How do I find what ports are being used on my mac?
What is the role of tcp/ip in data transmission from source to destination?