4) In a token ring network the transmission speed is 10^7 bps and the propagation speed is 200 metres/micro second. The 1-bit delay in this network is equivalent to:
(A) 500 metres of cable.
(B) 200 metres of cable.
(C) 20 metres of cable.
(D) 50 metres of cable.
Answer (C)
Transmission delay for 1 bit t = 1/(10^7) = 0.1 micro seconds.
200 meters can be traveled in 1 micro second. Therefore, in 0.1 micro seconds, 20 meters can be traveled.
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Series: 1 + x + x2 + x3 + x4 + x5
Sum of Series is in GP = (1-x6)/(1-x)
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2) The message 11001001 is to be transmitted using the CRC polynomial x^3 + 1 to protect it from errors. The message that should be transmitted is:
(A) 11001001000
(B) 11001001011
(C) 11001010
(D) 110010010011
Answer (B)
The polynomial x^3+1 corresponds to divisor is 1001.
11001001 000 <--- input right padded by 3 bits
1001 <--- divisor
01011001 000 <---- XOR of the above 2
1001 <--- divisor
00010001 000
1001
00000011 000
10 01
00000001 010
1 001
00000000 011 <------- remainder (3 bits)
See this for division process.
After dividing the given message 11001001 by 1001, we get the remainder as 011 which is the CRC. The transmitted data is, message + CRC which is 11001001 011.
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3) The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:
Answer (C)
Distance between stations = L KM
Propogation delay per KM = t seconds
Total propagation delay = Lt seconds
Frame size = k bits
Channel capacity = R bits/second
Transmission Time = k/R
Let n be the window size.
UtiliZation = n/(1+2a) where a = Propagation time / transmission time
= n/[1 + 2LtR/k]
= nk/(2LtR+k)
For maximum utilization: nk = 2LtR + k
Therefore, n = (2LtR+k)/k
Number of bits needed for n frames is Log(n).
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2) Consider an instance of TCP’s Additive Increase Multiplicative Decrease(AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a time out occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.
(A) 8 MSS
(B) 14 MSS
(C) 7 MSS
(D) 12 MSS
Answer (C)
Since Slow Start is used, window size is increased by the number of segments successfuly sent. This happens until either threshold value is reached or time out occurs.
In both of the above situations AIMD is used to avoid congestion. If threshold is reached, window size will be increased linearly. If there is timeout, window size will be reduced to half.
Window size for 1st transmission = 2 MSS
Window size for 2nd transmission = 4 MSS
Window size for 3rd transmission = 8 MSS
threshold reached, increase linearly (according to AIMD)
Window size for 4th transmission = 9 MSS
Window size for 5th transmission = 10 MSS
time out occurs, resend 5th with window size starts with as slow start.
Window size for 6th transmission = 2 MSS
Window size for 7th transmission = 4 MSS
threshold reached, now increase linearly (according to AIMD)
Additive Increase: 5 MSS (since 8 MSS isn’t permissible anymore)
Window size for 8th transmission = 5 MSS
Window size for 9th transmission = 6 MSS
Window size for 10th transmission = 7 MSS
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SMTP is an application layer protocol used for e-mail transmission.
TCP is a core transport layer protocol.
BGP is a network layer protocol backing the core routing decisions on the Internet
PPP is a data link layer protocol commonly used in establishing a direct connection between two networking nodes.
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1) The address of a class B host is to be split into subnets with a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts in each subnet?
(A) 62 subnets and 262142 hosts.
(B) 64 subnets and 262142 hosts.
(C) 62 subnets and 1022 hosts.
(D) 64 subnets and 1024 hosts.
Answer (C)
Maximum number of subnets = 2^6-2 =62.
Note that 2 is subtracted from 2^6. The RFC 950 specification reserves the subnet values consisting of all zeros (see above) and all ones (broadcast), reducing the number of available subnets by two.
Maximum number of hosts is 2^10-2 = 1022.
2 is subtracted for Number of hosts is also. The address with all bits as 1 is reserved as broadcast address and address with all host id bits as 0 is used as network address of subnet.
In general, the number of addresses usable for addressing specific hosts in each network is always 2^N – 2 where N is the number of bits for host id.
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2) The message 11001001 is to be transmitted using the CRC polynomial x^3 + 1 to protect it from errors. The message that should be transmitted is:
(A) 11001001000
(B) 11001001011
(C) 11001010
(D) 110010010011
Answer (B)
The polynomial x^3+1 corresponds to divisor is 1001.
11001001 000 <--- input right padded by 3 bits
1001 <--- divisor
01011001 000 <---- XOR of the above 2
1001 <--- divisor
00010001 000
1001
00000011 000
10 01
00000001 010
1 001
00000000 011 <------- remainder (3 bits)
See this for division process.
After dividing the given message 11001001 by 1001, we get the remainder as 011 which is the CRC. The transmitted data is, message + CRC which is 11001001 011.
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3) The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:
Answer (C)
Distance between stations = L KM
Propogation delay per KM = t seconds
Total propagation delay = Lt seconds
Frame size = k bits
Channel capacity = R bits/second
Transmission Time = k/R
Let n be the window size.
UtiliZation = n/(1+2a) where a = Propagation time / transmission time
= n/[1 + 2LtR/k]
= nk/(2LtR+k)
For maximum utilization: nk = 2LtR + k
Therefore, n = (2LtR+k)/k
Number of bits needed for n frames is Logn.
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1) Station A uses 32 byte packets to transmit messages to
Station B using a sliding window protocol. The round trip delay between A
and B is 80 milliseconds and the bottleneck bandwidth on the path
between A and B is 128 kbps. What is the optimal window size that A
should use?
(A) 20
(B) 40
(C) 160
(D) 320
Answer (B)
Round Trip propagation delay = 80ms
Frame size = 32*8 bits
Bandwidth = 128kbps
Transmission Time = 32*8/(128) ms = 2 ms
Let n be the window size.
UtiliZation = n/(1+2a) where a = Propagation time / transmission time
= n/(1+80/2)
For maximum utilization: n = 41 which is close to option (B)
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2) Frames of 1000 bits are sent over a 10^6 bps duplex link
between two hosts. The propagation time is 25ms. Frames are to be
transmitted into this link to maximally pack them in transit (within the
link).
What is the minimum number of bits (i) that will be required to
represent the sequence numbers distinctly? Assume that no time gap needs
to be given between transmission of two frames.
(A) i=2
(B) i=3
(C) i=4
(D) i=5
Answer (D)
Transmission delay for 1 frame = 1000/(10^6) = 1 ms
Propagation time = 25 ms
The sender can atmost transfer 25 frames before the first frame reaches the destination.
The number of bits needed for representing 25 different frames = 5
3) Consider the data of previous question. Suppose that the
sliding window protocol is used with the sender window size of 2^i where
is the number of bits identified in the previous question and
acknowledgments are always piggybacked. After sending 2^i frames, what
is the minimum time the sender will have to wait before starting
transmission of the next frame? (Identify the closest choice ignoring
the frame processing time.)
(A) 16ms
(B) 18ms
(C) 20ms
(D) 22ms
Answer (B)
Size of sliding window = 2^5 = 32
Transmission time for a frame = 1ms
Total time taken for 32 frames = 32ms
The sender cannot receive acknoledgement before round trip time which is 50ms
After sending 32 frames, the minimum time the sender will have to wait
before starting transmission of the next frame = 50 – 32 = 18
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