wireless network please answer correct number 7a b c . Calculate the total time

Question

wireless network
please answer correct number 7a b c

. Calculate the total time required to transfer a 200-B (KB: Kilo-Byte) fluk m, following cases, assuming a Round Trip Time (RTT) of 100ms, a packet size of 1KB, and otal time required to transfer a 2000-KB (KB: Kilo-Byte) file in the an initial 2*RTT of han a) the data rate bandwidth is 2.5Mbps (million bits per seco continuously. Calculate the total time required to transfer this file, in seconds. C6 dshaking to establish the connection before data is sent (10 points) e nd), and data packets can be sent b) the data rate bandwidth is still 2.5Mbps, but we use a protocol wi reliability. After we finish sending cach data packet we must wait one RTT to receive ACK before sending the next. Calculate the total time required to transfer this file, in minutes. (3 points) th ACK to improve c) the data rate bandwidth is 2.5kbps, and we still use the protocol with ACK to improve reliability. After we finish sending each data packet we must wait one RTT to receive ACK before sending the next. Calculate the total time required to transfer this file, in hours. (4 points)

Explanation / Answer

7.

Data Size 2000 KB = 2000*1024*8 bits = 16384000 bits

RTT = 100 ms

Handshake = 2*RTT = 200 ms

Number of packets = 2000

a . Data rate = 2500000 bits /sec

Propagation Time = 100 ms

Transmit Time = Data Size / Data Rate = 16384000 / 2500000 = 6.5536 sec

Total time = Handshake Time + Transmit Time + Propagation

= 0.200 sec+ 6.5536 sec + 0.100 sec = 6.8536 sec

b.

Number of packets = 2000 . For the transfer to complete 1999 ACKs are required there is no ACK for last packet or even if there is there is no delay because of that .

Each Packet ( First 1999 ) requires Trasmit + Propagation Time + ACK = 1024*8 *1000/ 2500000 ms + 100ms + 100 ms = 203.2768 ms

Once the last last packet reaches the transmission is complete and ACK is irrelevant so tie required for last packet is 103.2768 ms

Total time required = Handshake + Time for 1999 packets + 1 last Packet = 0.200 sec + 1999 * 0.2032768 sec + 0.1032768 sec = 406.6536 sec = 6.77756 mins

c. It is same as b. the only difference being data rate is reduced a 1000 times and is 2.5 kbps

Each Packet ( First 1999 ) requires Trasmit + Propagation Time + ACK = 1024*8*1000 / 2500 + 100ms + 100 ms = 3476.8 ms

Once the last last packet reaches the transmission is complete and ACK is irrelevant so tie required for last packet is 3376.8 ms

Total time required = Handshake + Time for 1999 packets + 1 last Packet = 0.200 sec + 1999 * 3.4768 sec + 3.3768 sec = 6653.7 sec = 1.9315833 hours

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