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Help on Network Calculations Problems
19 years 5 months ago #9230
by alongkuku
Help on Network Calculations Problems was created by alongkuku
Hi Gurus out there, I'm a newbie here have 2 questions here dat needs ur help and will greatly appreciate if anyone can lend a helping hand
here it goes...
Q1. A bus topology LAN consisting two stations, PC A and PC B 500m apart. The LAN operates at a data rate of 10Mbps. In the scenario below, PC A is transmitting a file containing 1 million characters (bytes) to PC B. During the transmission, the average data packet size is 384 bits with an additional 64 bits of overhead. Each received packet is acknowledged with an 88-bit packet before the next is sent. Assuming the propagation speed on the bus is 200m/us calculate the following.
1) the total elasped time to complete the file transmission
2) the effective throughput.
Q2.Consider the transfer of a file containing one million characters from one station to another. What is the total elasped time and effective throughput for the follow cases.
a) a circuit siwthced, star topology local network. call setuptime is negligible and he data rate on the medium is 64kbps.
b) a bus topology local network with 2 stations a Distance D apart, a data rate of B bps, and a packet size P with 80 bits of overhead. Each packet is acknowledged with an 88bit packet before the next is sent. The propagation speed on the bus is 200m/us. Solve for
D=1km, B=1Mbps, P=256bits
D=1km, B=10Mbps, P=256bits
D=10km, B=1Mbps, P=256bits
D=1km, B=50Mbps, P=10000bits
c) a ring topology with a total circular length of 2D, with the stations at a distance D apart. Acknowledgement is achieved by allowing a packet to circulate past the destination station, back to the source station. There are N repeaters on the ring, each of which introduces a delay of one bit time. Repeat the calculation for each of b(1) through b(4) for N= 10; 100; 1000.
here it goes...
Q1. A bus topology LAN consisting two stations, PC A and PC B 500m apart. The LAN operates at a data rate of 10Mbps. In the scenario below, PC A is transmitting a file containing 1 million characters (bytes) to PC B. During the transmission, the average data packet size is 384 bits with an additional 64 bits of overhead. Each received packet is acknowledged with an 88-bit packet before the next is sent. Assuming the propagation speed on the bus is 200m/us calculate the following.
1) the total elasped time to complete the file transmission
2) the effective throughput.
Q2.Consider the transfer of a file containing one million characters from one station to another. What is the total elasped time and effective throughput for the follow cases.
a) a circuit siwthced, star topology local network. call setuptime is negligible and he data rate on the medium is 64kbps.
b) a bus topology local network with 2 stations a Distance D apart, a data rate of B bps, and a packet size P with 80 bits of overhead. Each packet is acknowledged with an 88bit packet before the next is sent. The propagation speed on the bus is 200m/us. Solve for
D=1km, B=1Mbps, P=256bits
D=1km, B=10Mbps, P=256bits
D=10km, B=1Mbps, P=256bits
D=1km, B=50Mbps, P=10000bits
c) a ring topology with a total circular length of 2D, with the stations at a distance D apart. Acknowledgement is achieved by allowing a packet to circulate past the destination station, back to the source station. There are N repeaters on the ring, each of which introduces a delay of one bit time. Repeat the calculation for each of b(1) through b(4) for N= 10; 100; 1000.
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19 years 5 months ago #9234
by cybersorcerer
"He who breaks something to find out what it is, has left the path of wisdom."
Gandalf the Grey
Replied by cybersorcerer on topic Re: Help on Network Calculations Problems
This equation should be able to solve them, although I'd rather you complete them yourself before I just hand out the answers.
T = S/BW
Where T = time, S = the size of the file and BW = the maximum bandwidth capacity. Also, since we all know in a real scenario, the BW variable is worthless since the theoretical bandwidth is hard to achieve in a production environment, there is an alternative equation.
T = S/P
Where P = the present bandwidth capacity. Both these should aid you in throughput calculation. Hope this helps
T = S/BW
Where T = time, S = the size of the file and BW = the maximum bandwidth capacity. Also, since we all know in a real scenario, the BW variable is worthless since the theoretical bandwidth is hard to achieve in a production environment, there is an alternative equation.
T = S/P
Where P = the present bandwidth capacity. Both these should aid you in throughput calculation. Hope this helps
"He who breaks something to find out what it is, has left the path of wisdom."
Gandalf the Grey
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