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Help in computer networking! SOS!
bellewinnie
Registered Users Posts: 1 ■□□□□□□□□□
in Network+
Hi guys,
Im stuck on this question, i tried..but im not sure if i am at the right path as well.....
I went to seek help from an online tutor jus now but he is making me even more confuse as what he said and what my lecturer said is pretty conflicting...
So here i am,
seeking help in forum, looking for some guidance. T_T
my ans :
one packet = 3000bits - 100bits = 2900 bits
Total datafile = 2 x 10^6 bytes = 16 x 10^6 bits
16x10^6bits = 5518 packets
online tutor ans:
one packet = 3000bits (including header)
Total datafile = 2 x 10^6 bytes = 16 x 10^6 bits
16x10^6bits = 5334 packets
(ii) Determine the time it takes each host along the path to transmit the entire datafile.
my ans:
propagation time = propagation delay = 0.005 sec
transmission time = 16 x 10^6 / 1x10^9 = 0.016 seconds
total time = 0.005 + 0.016 seconds = 0.021 seconds.
online tutor ans:
total time = ( 5334 x 5(propagation delay) ) x 5 (each packet transmitted 5 times) = 133350ms
(iii) Compute the end to end delay using datagram packet switching.
I dont even know how to start this..
Im stuck on this question, i tried..but im not sure if i am at the right path as well.....
I went to seek help from an online tutor jus now but he is making me even more confuse as what he said and what my lecturer said is pretty conflicting...
So here i am,
seeking help in forum, looking for some guidance. T_T
- Consider the transfer of a data file of 2000000 bytes from a source to a destination forthe following switching network:
- The data rate of the line is 1 Gbps.
- The file is divided into fixed packet size, each of which has a size of 3000 bits
(including data and header), before transmitting. There are 100 bits in the header.If there is not enough data to fill a packet, it is filled with zeros before transmission.All frames sent have the same size. - The propagation delay per transmission is 0.005 sec.
- Each packet must be transmitted 5 times before reaching destination.
Assume that there are 8 bits in a byte.
- The data rate of the line is 1 Gbps.
my ans :
one packet = 3000bits - 100bits = 2900 bits
Total datafile = 2 x 10^6 bytes = 16 x 10^6 bits
16x10^6bits = 5518 packets
online tutor ans:
one packet = 3000bits (including header)
Total datafile = 2 x 10^6 bytes = 16 x 10^6 bits
16x10^6bits = 5334 packets
(ii) Determine the time it takes each host along the path to transmit the entire datafile.
my ans:
propagation time = propagation delay = 0.005 sec
transmission time = 16 x 10^6 / 1x10^9 = 0.016 seconds
total time = 0.005 + 0.016 seconds = 0.021 seconds.
online tutor ans:
total time = ( 5334 x 5(propagation delay) ) x 5 (each packet transmitted 5 times) = 133350ms
(iii) Compute the end to end delay using datagram packet switching.
I dont even know how to start this..