A 10-story office building has the floor plan of the following figure for each f
ID: 640423 • Letter: A
Question
- A 10-story office building has the floor plan of the following figure for each floor. A local network is to be installed that will allow attachment of a device from each office on each floor. Attachment is to take place along the outside wall at the baseboard. Cable or wire can be run vertically through the indicated closet and horizontally along the baseboards. The height of each story is 10 feet. What is the minimum total length of cable or wire required for bus, tree, ring, and star topology? Which floor
Explanation / Answer
2)
a)
Total elapsed time is,
Therefore, the total elapsed time is 125 seconds.
b)
i)
D = 1 km, B = 1 Mbps, P = 256 bits
Overhead = 80 bits
Ack = 88 bits
The file transfer rate is,
Therefore, the number of packets (rounding up) is 45455.
Td = total transmission time of a frame
Td = 2Tprop + Tframe + Tack Where, Tprop = propagation time
Tframe=time taken by frame
Tack=time taken by the acknowledgement
a =
Utilization =
Data packet = 256-80 => 176
Therefore, seconds
TE(Elapsed time) = 45455 * (354*10-6)
=> 16.09107 seconds
TE=16.09107 seconds
Efficiency = => 0.92624
Throughput = 1 Mbps * 0.92624 => 0.92624 Mbps
ii)
D = 1Km, B = 10Mbps, P = 256 bits
The data packet is P = 256-80
=> 176
Size of the file,
Therefore, the number of packets (rounding up) is 45455.
The total transmission of a frame is,
Td = 2Tprop + Tframe + Tack
Where, Tprop = time taken for proper transmission.
Tframe=time taken by frame
Tack=time taken by the acknowledgement
The total time taken by the frame is, 44.4*10-6 seconds.
The value of the elapsed time will be, TE = number of packets transferred * Td
=45455*44.4*10-6
=2.0182
Therefore the elapsed time is TE = 2.0182 seconds
Elapsed time = 45455 * (44.4*10-6) => 2.018202 seconds
The acknowledgment is,
The efficiency is, E=
The throughput is = 10 Mbps * 0.71910 => 7.1910 Mbps
3)
i)
The delay of the total repeater is Trep
Trep = N/B
Apply these equations for N = 10 ;100 ;1000 for each part above.
D = 1 km, B = 1 Mbps, P = 256 bits
Overhead = 80 bits
Ack = 88 bits
The file transfer rate is,
Therefore, the number of packets (rounding up) is 45455.
Td = total transmission time of a frame
Td = 2Tprop + Tframe + Trep Where, Tprop = propagation time
Tframe=time taken by frame
Tack=time taken by the acknowledgement
a =
Utilization =
Data packet = 256-80 => 176
Therefore the total elapsed time is, 276*10-6*45455=12 seconds.
For N=100
Therefore, the total elapsed time is 276*10-6*45455=16seconds
When N=1000;
Therefore, the total elapsed time is 1266*10-6*45455=57seconds
ii)
D = 1Km, B = 10Mbps, P = 256 bits
The data packet is P = 256-80
=> 176
Size of the file,
Therefore, the number of packets (rounding up) is 45455.
The total transmission of a frame is,
Td = 2Tprop + Tframe + Trep
Where, Tprop = time taken for proper transmission.
Tframe=time taken by frame
Trep=time taken by repeater
When N=10
The value of the elapsed time will be, TE = number of packets transferred * Td
=45455*36.6*10-6
=1second
Therefore the elapsed time is TE = 1 seconds
When N=100
The value of the elapsed time will be, TE = number of packets transferred * Td
=45455*45.6*10-6
=2.07seconds
Therefore the elapsed time is TE = 2.07seconds
When N=1000
The value of the elapsed time will be, TE = number of packets transferred * Td
=45455*135.6*10-6
=6.16seconds
Therefore the elapsed time is TE = 6.16seconds
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