Question

Networks.

1. Consider the two 16-bit words below. Compute the Internet checksum of these 16-bit words: 00111111 11101001      and 11010010 10001111.

2. Suppose that TCP's current estimated values for the round trip time (estimatedRTT) and deviation in the RTT (DevRTT) are 390 msec and 38 msec, respectively. Suppose that the next measured value of the RTT is 380. Compute TCP's new value of estimatedRTT, DevRTT, and the TCP timeout value for the RTT value obtained. Use the values of ? = 0.125 and ? = 0.25. Show your answer in millisecs to 3 decimal places.

3. The following 8-bit bytes exist in a UDP segment: 01010011, 01100110, 01110100. What will the checksum value in the segment.
a.   00101101      
b.   00101110      
c.   11010001      
d.   11010010      
e.   None of the above

4. What will the TCP sender do if 3 duplicate ACKS s are received?
a.   Wait for a timeout to occur so that the TCP segment can be retransmitted  
b.   Retransmit the last transmitted segment      
c.   Wait for a further ACK from the receiver before retransmission      
d.   Send a SYN segment      
e.   None of the above

5. Which reliable data transfer protocol handles packet loss?
a.   RDT2.0      
b.   RDT2.2      
c.   RDT3.0      
d.   A, B & C      
e.   None of the above

6. With reference to TCP congestion control, when a timeout occurs, the value of cwnd is set to:
a.   Half of its prior value      
b.   Half of the value of ssthresh      
c.   1      
d.   1 x Maximum segment size      
e.   None of the above

Kindly assist guys. Thanks

Answer 1

1
Checksum of
0011111111101001 = 11011000
1101001010001111 = 10011111

2.
The formulas are,

EstimatedRTT = (1-α)*EstimatedRTT_prev + α*SampleRTT = (1 - 0.125) * 390 + (0.125 * 380) = 388.85 milli sec.
DevRTT = (1-β)*DevRTT_prev + β*|SampleRTT - EstimatedRTT| = (1 - 0.25) * 38 + (0.25 * | 380 - 388.85| ) = 30.712 milli sec.
timeout = EstimatedRTT + 4*DevRTT = 388.85 + (4 * 30.712) = 511.7 milli sec,

3. d) 11010010
4. b) Retransmit the last transmitted segment