[SOLVED] 代写 html EE450: Midterm Solutions

EE450: Midterm Solutions
1. T, T, F, F, F, F, F, T, T, F, T, F, T, T, F, T, F, F, T, T, F, F, F, (a, b, c, d), (b, d)
2. 80Km, 11.1%, 1440 Hz, 0110110011010, 50,000 sec, 15,000 sec, 38 Bits, 1425 bps, 37.5 frames/sec, 1.4 msec, 56.6%, 37.6 dB, 4RTTs, 0.568 sec, 0.448 sec, 0.368 sec,
0.488 sec, 0.448 sec, 4.5 sec, 3.501 sec, 3, 8
3.
a. Transmitted pattern: 101011010. The red bits are the FCS bits
b. Received Pattern 010010111. Errors did occur (Since the received Pattern is NOT the
same as the Transmitted Pattern). Receiver will divide the received sequence by the generator sequence and observe a zero remainder. The receiver failed to detect the error.
c. Received pattern: 101011010 100010111 = 001001101. When we divide this sequence by the generator sequence, the reminder is 011 and hence the error is detected. Note the receiver does NOT know the error pattern. He only observes the received pattern.

4. The color codes areBlack (Frames received with no errors detected), Red (Frame retransmitted due to expiration of time out), Green (Acks, sent every other frame), Star (frame received in error), Blue (Frame lost). Each time division is 0.5 sec (one-way propagation delay). It will take 11 seconds for “A” to finish transmitting all 5 frames and receive ACKs for all of them. Hence, the throughput is 5frames/11 sec. If each frame is 1000 bits long, then the throughput is 454.5 bps bits/sec. The link utilization is therefore ~ 45.5%. Timing diagram is shown below. Note, I am assuming that that both sides know that there are ONLY 5 frames to transmit that is why when the receiver did not wait 1 second to acknowledge F4. Otherwise, it will take 12 seconds for a throughput of 416.6 bps and utilization of 41.6%. Both assumptions are acceptable.
Fo
F2
F3
Fo
F1
F3
F4
F3
F2
F4
Both F0 and F1 are accepted Fo is dropped (Out of order)
ACK2 Both F0 and F1
are Acked
ACK2
Is repeated
Frame F2is received in error
F3 is dropped (Out of order)
F2 is accepted & acked
ACK3
F2 is Acked
F3 is accepted
F3 is acked
ACK4
F3 is Acked
F3is dropped(Out of order)
F4 is accepted & ack’ed
ACK5
F4 is Acked

5.
a. Transmission Time of the HTML file over LAN = 1Gbit/1Gbps = 1 sec
Transmission Time of file over WAN Link (R1 —- R2) = 1Gbit/1Mbps = 1000 sec
Step
Action
Delay (sec)
1
Host A contact the Local DNS server about IP address of X
0.1
2
Local DNS server contact the RNS and get IP address of TLD
0.8
3
Local DNS server contact the TLD and get the IP address of X
0.6
4
Local DNS server cache the address and return it to the Client
0.1
5
Client establish a TCP connection with server X (Handshaking)
0.4
6
Client request downloading the HTML file
0.2
7
HTML file downloaded over LAN & WAN Links (Transmission + Propagation) = 1+1000+0.2+1 = 1002.2
1002.2
8
Total Delay
1004.4
b. The additional time it takes for the client to download the embedded image is = 0.2 + 10M/1G + 0.2 +10M/1M + 10M/1G = 10.42 sec
c. It will take Client B just 2(0.10) = 0.2 sec to fetch the IP address from the Local DNS
d. Average Rate = (0.5)(1Mbps)+(0.5)(1Gbps) = 500.5 Mbps