Computer-Networks
Question 1 |
What is the distance of the following code 000000, 010101, 000111, 011001, 111111?
A | 2 |
B | 3 |
C | 4 |
D | 1 |
010101 ⊕ 011001 = 001100
Question 2 |
Assume that you have made a request for a web page through your web browser to a web server. Initially the browser cache is empty. Further, the browser is configured to send HTTP requests in non-persistent mode. The web page contains text and five very small images. The minimum number of TCP connections required to display the web page completely in your browser is ______.
A | 6 |
Hence, 1 Text + 5 Image = 6 Objects
Question 3 |
Consider the following statements about the functionality of an IP based router.
- I. A router does not modify the IP packets during forwarding.
II. It is not necessary for a router to implement any routing protocol.
III. A router should reassemble IP fragments if the MTU of the outgoing link is larger than the size of the incoming IP packet.
Which of the above statements is/are TRUE?
A | I and II only |
B | II only |
C | I only |
D | II and III only
|
II: Is True.
III: Reassemble is not necessary at the router.
Question 4 |
An organization requires a range of IP addresses to assign one to each of its 1500 computers. The organization has approached an Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space 202.61.0.0/17. The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP’s router using route aggregation. Which of the following address spaces are potential candidates from which the ISP can allot any one to the organization?
I. 202.61.84.0/21 II. 202.61.104.0/21 III. 202.61.64.0/21 IV. 202.61.144.0/21
A | I and II only
|
B | III and IV only
|
C | II and III only
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D | I and IV only
|
And to Assign an IP address for 1500 computer, we require 11 bit from HID part.
So NID + SID = 17 + 4 = 21 bits and HID = 11 bits
NID HID
202.61.0 0000 000.00000000
So, from the given option, possible IP Address is
I. 84 -> 0 1010 100 (Because in HID bit 1 is not possible)
II. 104 -> 0 1101 000
III. 64 -> 0 1000 000
IV. 144 -> 1 0010 000 (Because in NID bit 1 is not possible )
Question 5 |
Consider a TCP connection between a client and a server with the following specifications: the round trip time is 6 ms, the size of the receiver advertised window is 50 KB, slow start threshold at the client is 32 KB, and the maximum segment size is 2 KB. The connection is established at time t=0. Assume that there are no timeouts and errors during transmission. Then the size of the congestion window (in KB) at time t+60 ms after all acknowledgements are processed is ______.
A | 44 |
Here, t + 60 is nothing but at the 10 RTT (60/6 = 10), but here it’s asking after all acknowledgement are processed it means after the 10th RTT, i.e., at the 11RTT.
1st transmission: 2 KB
2nd transmission: 4 KB
3rd transmission: 8 KB
4th transmission: 16 KB
5th transmission: 32 KB (Threshold reached)
6th transmission: 34 KB
7th transmission: 36 KB
8th transmission: 38 KB
9th transmission: 40 KB
10th transmission: 42 KB
At the completion of 10th transmission RTT = 10*6 = 60 ms
For the 11th transmission, The congestion window size is 44 KB.
Question 6 |
Which of the following assertions is FALSE about the Internet Protocol (IP)?
A | It is possible for a computer to have multiple IP addresses
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B | IP packets from the same source to the same destination can take different routes in the network
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C | IP ensures that a packet is discarded if it is unable to reach its destination within a given number of hops
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D | The packet source cannot set the route of an outgoing packets; the route is determined only by the routing tables in the routers on the way
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Question 7 |
Which of the following functionalities must be implemented by a transport protocol over and above the network protocol?
A | Recovery from packet losses |
B | Detection of duplicate packets |
C | Packet delivery in the correct order |
D | End to end connectivity |
Question 8 |
The subnet mask for a particular network is 255.255.31.0. Which of the following pairs of IP addresses could belong to this network?
A | 172.57.88.62 and 172.56.87.233
|
B | 10.35.28.2 and 10.35.29.4 |
C | 191.203.31.87 and 191.234.31.88 |
D | 128.8.129.43 and 128.8.161.55 |
128.8.129.43 (Bitwise AND) 255.255.31.0 = 128.8.1.0
128.8.161.55 (Bitwise AND) 255.255.31.0 = 128.8.1.0
Question 9 |
A 2 km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2×108 m/s. What is the minimum packet size that can be used on this network?
A | 50 bytes |
B | 100 bytes |
C | 200 bytes |
D | None of the above |
d= 2 km = 2 x 103 m, v = 2 x 108 m/s, B= 107
Tp = d / v = 2 x 103 /(2 x 108 ) seconds = 10-5 seconds
Let L bits be minimum size of frame, then Tt = t L / B = L / 107 seconds
Now, Tt = 2Tp
L/107 = 2 x 10-5 = 200 bits = (200 / 8) bytes = 25 bytes
Question 10 |
Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 µs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 µs. What is the maximum achievable throughput in this communication?
A | 7.69 × 106 bps
|
B | 11.11 × 106 bps |
C | 12.33 × 106 bps |
D | 15.00 × 106 bps |
Transmission rate , Tt = L / B.W
Therefore, B.W. = L / Tt = 1000 bytes/ 50 μs = 8000 bits / 50 μs=160 Mbps
Efficiency = N / 1 + 2a, where a = Tp / Tt
Efficiency = 5 * 50 / (50+400) = 250/450 = 5/9
Maximum achievable throughput = Efficiency * B.W = (5/9)*160 Mbps = 88.88 Mbps = = 11.11 x 106 bytes per second
*Actual option should be in bytes per second.
Question 11 |
Choose the best matching between Group 1 and Group 2.
Group-1 Group-2 P. Data link 1. Ensures reliable transport of data over a physical point-to-point link Q. Network layer 2. Encoder/decodes data for physical transmission R. Transport layer 3. Allows end-to-end communication between two processes 4. Routes data from one network node to the next
A | P - 1, Q - 4, R - 3 |
B | P - 2, Q - 4, R - 1
|
C | P - 2, Q - 3, R - 1 |
D | P - 1, Q - 3, R - 2 |
Transport Layer :: Fourth layer of the OSI Model, Responsible for Service point addressing/Socket to socket connection or end to end connection with full reliability.
Network Layer :: Third layer of the OSI Model, Responsible for Host to Host.
Question 12 |
Which of the following is NOT true with respect to a transparent bridge and a router?
A | Both bridge and router selectively forward data packets
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B | A bridge uses IP addresses while a router uses MAC addresses
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C | A bridge builds up its routing table by inspecting incoming packets |
D | A router can connect between a LAN and a WAN |
Question 13 |
How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits, and one parity bit?
A | 600 |
B | 800 |
C | 876 |
D | 1200 |
So bit rate is 9600 bps.
To send one char we need to send (1 + 8 + 2 +1) = 12
So total char send = 9600 / 12 = 800
Question 14 |
A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is:
A | 0.5 |
B | 0.625 |
C | 0.75 |
D | 1.0 |
The probability that A wins the second back-off race = 5/8 = 0.625
More explanation in the video.
Question 15 |
The routing table of a router is shown below:
Destination Sub net mask Interface 128.75.43.0 255.255.255.0 Eth0 128.75.43.0 255.255.255.128 Eth1 192.12.17.5 255.255.255.255 Eth3 Default Eth2
On which interfaces will the router forward packets addressed to destinations 128.75.43.16 and 192.12.17.10 respectively?
A | Eth1 and Eth2 |
B | Eth0 and Eth2 |
C | Eth0 and Eth3 |
D | Eth1 and Eth3 |
If results of ANDing subnet masks and IP address are same then subnet mask with higher number of 1s is preferred.
IP address 128.75.43.16 is AND with 255.255.255.0 results 128.75.43.0 Net ID which is similar to destination of this mask, but ANDing 128.75.43.16 with 255.255.255.128 also results same destination. So, here, mask with higher number of one is considered and router will forward packet to Eth1.
ANDing 192.12.17.10 with three subnet mask in table does not result in destination Net ID so router will forward this packet to default network via Eth2.
Question 16 |
Consider three IP networks A, B and C. Host HA in network A sends messages each containing 180 bytes of application data to a host HC in network C. The TCP layer prefixes a 20 byte header to the message. This passes through an intermediate network B. The maximum packet size, including 20 byte IP header, in each network is:
A : 1000 bytes B : 100 bytes C : 1000 bytes
The network A and B are connected through a 1 Mbps link, while B and C are connected by a 512 Kbps link (bps = bits per second).
Assuming that the packets are correctly delivered, how many bytes, including headers, are delivered to the IP layer at the destination for one application message, in the best case? Consider only data packets.
A | 200 |
B | 220 |
C | 240 |
D | 260 |
Data will be divided in three packets as:
First packet: 80 bytes + 20 byte of header
Second packet: 80 bytes + 20 byte of header
Third packet: 40 bytes + 20 byte of header
Note: Defragmentation (grouping of fragments) is done only at destination.
HC will receive total 260 bytes including header.
Question 17 |
Consider three IP networks A, B and C. Host HA in network A sends messages each containing 180 bytes of application data to a host HC in network C. The TCP layer prefixes a 20 byte header to the message. This passes through an intermediate network B. The maximum packet size, including 20 byte IP header, in each network is:
A : 1000 bytes B : 100 bytes C : 1000 bytes
The network A and B are connected through a 1 Mbps link, while B and C are connected by a 512 Kbps link (bps = bits per second).
What is the rate at which application data is transferred to host HC? Ignore errors, acknowledgements, and other overheads.
A | 325.5 Kbps
|
B | 354.5 Kbps |
C | 409.6 Kbps |
D | 512.0 Kbps |
Application data is transferred at rate of (180/260) x 512 Kbps = 354.46 Kbps
Question 18 |
A simple and reliable data transfer can be accomplished by using the ‘handshake protocol’. It accomplishes reliable data transfer because for every data item sent by the transmitter __________.
A | in this case receiver has to respond that receiver can be able to receive the data item. |
Question 19 |
Start and stop bits do not contain an ‘information’ but are used in serial communication for
A | Error detection |
B | Error correction |
C | Synchronization |
D | Slowing down the communications |
Question 20 |
Match the pairs in the following questions by writing the corresponding letters only.
(A) IEEE 488 (P) specifies the interface for connecting a single device (B) IEEE 796 (Q) specifies the bus standard for connecting a computer to other devices including CPU’s (C) IEEE 696 (R) specifies the standard for an instrumentation bus (D) RS232-C (S) specifies the bus standard for the “backplane” bus called multibus.
A | Out of syllabus. |
Question 21 |
A | If the client was waiting to receive a packet, it may wait indefinitely. |
B | If the client sends a packet after the server reboot, it will receive a RST segment. |
C | The TCP server application on S can listen on P after reboot. |
D | If the client sends a packet after the server reboot, it will receive a FIN segment. |
- True
Since broken connections can only be detected by sending data, the receiving side will wait forever. This scenario is called a “half-open connection” because one side realizes the connection was lost but the other side believes it is still active. - True
The situation resolves itself when client tries to send data to server over the dead connection, and server replies with an RST packet (not FIN). - True
Yes, a TCP Server can listen to the same port number even after reboot. For example, the SMTP service application usually listens on TCP port 25 for incoming requests. So, even after reboot the port 25 is assigned to SMTP. - False
The situation resolves itself when client tries to send data to server over the dead connection, and server replies with an RST packet (not FIN), causing client to finally to close the connection forcibly.
FIN is used to close TCP connections gracefully in each direction (normal close of connection), while TCP RST is used in a scenario where TCP connections cannot recover from errors and the connection needs to reset forcibly.
Question 22 |
- The time taken for processing the data frame by the receiver is negligible.
- The time taken for processing the acknowledgement frame by the sender is negligible.
- The sender has an infinite number of frames available for transmission.
- The size of the data frame is 2,000 bits and the size of the acknowledgment frame is 10 bits.
- The link data rate in each direction is 1 Mbps (=106bits per second).
- One way propagation delay of the link is 100 milliseconds.
A | 51 |
Tt(packet) = L / B.W => 2000 bits / 10^6 bps = 2 x 10^-3 sec = 2 millisec
Tt(Ack) = L / B.W. => 10 bits / 10^6 bps = 10^-5 sec = 10^-2 millisec = 0.01 millisec
Tp = 100 millisec
Total time = Tt(packet) + 2 x Tp + Tt(Ack)
=> 2 + 2 x 100 + 0.01 = 202.01 millisec
Efficiency = 50 % = ½
Efficiency = Useful Time / Total time
½ = n x Tt / Total time
=> 2 x n x Tt = Total time
=>2 x n x 2 = 202.01
=> n = 202.01 / 4 => 50.50
For minimum, we have to take ceil, Hence size of window = 51
Question 23 |
A | If the second fragment is lost, R will resend the fragment with the IP identification value 0x1234. |
B | If the second fragment is lost, P is required to resend the whole TCP segment. |
C | Two fragments are created at R and the IP datagram size carrying the second fragment is 620 bytes. |
D | TCP destination port can be determined by analysing only the second fragment. |
Question 24 |
S1: Destination MAC address of an ARP reply is a broadcast address.
S2: Destination MAC address of an ARP request is a broadcast address.
Which of the following choices is correct?
A | Both S1and S2are true. |
B | S1is true and S2is false. |
C | S1is false and S2is true. |
D | Both S1and S2are false. |
Question 25 |
A | SYN bit = 1, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 0 |
B | SYN bit = 0, SEQ number = X+1, ACK bit = 0, ACK number = Y, FIN bit = 1 |
C | SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X+1, FIN bit = 0 |
D | SYN bit = 1, SEQ number = Y, ACK bit = 1, ACK number = X, FIN bit = 0 |
Q will send the SYN bit = 1 to the connection establishment.
Q Seq number will be Y different from X
ACK bit = 1 because sending the ACK
ACK number = X+1 (Next seq number id)
FIN bit = 0 (Because establishing the connection)
Question 26 |
A | 111 |
B | 100 |
C | 101 |
D | 110 |
Question 27 |
A | The next hop router for a packet from R to P is Y. |
B | The distance from R to Q will be stored as 7. |
C | The next hop router for a packet from R to Q is Z. |
D | The distance from R to P will be stored as 10. |
Given R gets the distance vector (3,2,5)
After the one iteration distance vector from X to P, Y to P, and Z to P is (7, 6, 5) respectively
The distance vector from R to P via X Y Z is (3+7, 2+6, 5+5) =(10, 8, 10)
So Take minimum distance from R to P which is 8 via Y
After the iteration distance vector from X to Q, Y to Q, Z to Q is ( 4, 6, 8) respectively
The distance vector from R to Q via X Y Z is (3+4, 2+6, 5+8) = (7, 8 13)
So Take minimum distance from R to Q which is 7 via X.
Question 28 |
A | 135 |
1 frames takes = Tt = L / B.w. => 1000 / 10^6 = 1 millisec
1000 frame Tt = 1000 x 1 millisec = 1 sec
In 1 sec, 1000 frames sends, which is 1 millisec per frame.
So, G = 1
Efficiency of Pure Aloha (η) = G x e-2G
where G = Number of requests per time slot willing to transmit.
e = Mathematical constant approximately equal to 2.718
So, η = 1 x 2.718(-2 x 1) = 0.1353
Therefore, In 1 sec1000 frames = 0.1353 x 1000 = 135.3(closest integer) =>135
Throughput => 135
Question 29 |
- For any two letters, the code assigned to one letter must not be a prefix of the code assigned to the other letter.
- For any two letters of the same frequency, the letter which occurs earlier in the dictionary order is assigned a code whose length is at most the length of the code assigned to the other letter.
A | 21 |
B | 30 |
C | 23 |
D | 25 |
Input String : abbccddeee
The character frequencies are
Character |
a |
b |
c |
d |
e |
Frequency |
1 |
2 |
2 |
2 |
3 |
Binary Code |
? |
? |
? |
? |
? |
Question 30 |
Traceroute reports a possible route that is taken by packets moving from some host A to some other host B. Which of the following options represents the technique used by traceroute to identify these hosts
A | By progressively querying routers about the next router on the path to B using ICMP packets, starting with the first router |
B | By requiring each router to append the address to the ICMP packet as it is forwarded to B. The list of all routers en-route to B is returned by B in an ICMP reply packet |
C | By ensuring that an ICMP reply packet is returned to A by each router en-route to B, in the ascending order of their hop distance from A |
D | By locally computing the shortest path from A to B |
So the first router forwards the packets, but the second router drops them and replies with ICMP time exceeded. Proceeding in this way, traceroute uses the returned ICMP time exceeded messages to build a list of routers that packets traverse, until the destination is reached and returns an ICMP echo reply message.
Question 31 |
Which of the following statements is TRUE about CSMA/CD
A | IEEE 802.11 wireless LAN runs CSMA/CD protocol |
B | Ethernet is not based on CSMA/CD protocol |
C | CSMA/CD is not suitable for a high propagation delay network like satellite network |
D | There is no contention in a CSMA/CD network |
For networks with high propagation delay this time becomes too long hence the minimum packet size required becomes too big to be feasible.
Question 32 |
Which of the following statements is FALSE regarding a bridge?
A | Bridge is a layer 2 device |
B | Bridge reduces collision domain |
C | Bridge is used to connect two or more LAN segments |
D | Bridge reduces broadcast domain |
The bridge acts as a interface between two networks and speed the traffic between them and there by reduces the collision domain.
So, option B also True.
Question 33 |
Count to infinity is a problem associated with
A | link state routing protocol. |
B | distance vector routing protocol. |
C | DNS while resolving host name. |
D | TCP for congestion control. |
The Bellman-Ford algorithm does not prevent routing loops from happening and suffers from the count-to-infinity problem.
Question 34 |
A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1 km cable with no repeaters. The signal speed in the cable is 2 × 108 m/sec. The minimum frame size for this network should be
A | 10000 bits |
B | 10000 bytes |
C | 5000 bits |
D | 5000 bytes |
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
10000 bits.
Question 35 |
A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimum frame size should be
A | 80 bytes |
B | 80 bits |
C | 160 bytes |
D | 160 bits |
Question 36 |
On a TCP connection, current congestion window size is Congestion Window=4 KB. The window size advertised by the receiver is Advertise Window=6 KB. The last byte sent by the sender is LastByteSent=10240 and the last byte acknowledged by the receiver is LastByteAcked=8192. The current window size at the sender is
A | 2048 bytes |
B | 4096 bytes |
C | 6144 bytes |
D | 8192 bytes |
= min (4KB, 6KB)
= 4KB
Question 37 |
In a communication network, a packet of length L bits takes link L1 with a probability of p1 or link L2 with a probability of p2. Link L1 and L2 have bit error probability of b1and b2 respectively. The probability that the packet will be received without error via either L1 or L2 is
A | (1 – b1)L p1 + (1 – b2)Lp2 |
B | [1 – (b1 + b2)L]p1p2 |
C | (1 – b1)L (1 – b2)Lp1p2 |
D | 1 – (b1 Lp1 + b2 Lp2) |
Probability for no bit error for any single bit = (1 - b1)
Similarly, for link L2,
Probability of no bit error = (1 - b2)
Packet can go either through link L1 or L2, they are mutually exclusive events.
Probability packet will be received without any error = Probability of L1 being chosen and no errors in any of L bits + Probability of L2 being chosen and no error in any of the L bits
= (1 - b1)L p1 + (1 - b2)L p2
Hence, answer is option A.
Question 38 |
In a TDM medium access control bus LAN, each station is assigned one time slot per cycle for transmission. Assume that the length of each time slot is the time to transmit 100 bits plus the end-to-end propagation delay. Assume a propagation speed of 2 × 108 m/sec. The length of the LAN is 1 km with a bandwidth of 10 Mbps. The maximum number of stations that can be allowed in the LAN so that the throughput of each station can be 2/3 Mbps is
A | 3 |
B | 5 |
C | 10 |
D | 20 |
Question 39 |
A company has a class C network address of 204.204.204.0. It wishes to have three subnets, one with 100 hosts and two with 50 hosts each. Which one of the following options represents a feasible set of subnet address/subnet mask pairs?
A | 204.204.204.128/255.255.255.192 204.204.204.0/255.255.255.128 204.204.204.64/255.255.255.128 |
B | 204.204.204.0/255.255.255.192 204.204.204.192/255.255.255.128 204.204.204.64/255.255.255.128 |
C | 204.204.204.128/255.255.255.128 204.204.204.192/255.255.255.192 204.204.204.224/255.255.255.192 |
D | 204.204.204.128/255.255.255.128 204.204.204.64/255.255.255.192 204.204.204.0/255.255.255.192 |
10000000/128 (mask) - subnet id bit (1) (subnet 1)
01000000/192 (mask) - subnet id bit (01) (subnet 2)
0000000/192 (mask) - subnet id bit (00) (subnet 3)
Question 40 |
Assume that “host1.mydomain.dom” has an IP address of 145.128.16.8. Which of the following options would be most appropriate as a subsequence of steps in performing the reverse lookup of 145.128.16.8? In the following options “NS” is an abbreviation of “nameserver”.
A | Query a NS for the root domain and then NS for the “dom” domains |
B | Directly query a NS for “dom” and then a NS for “mydomain.dom” domains |
C | Query a NS for in-addr.arpa and then a NS for 128.145.in-addr.arpa domains |
D | Directly query a NS for 145.in-addr.arpa and then a NS for 128.145.in-addr.arpa domains |
First we need to locate in-addr.apra, then perform reverse lookup of 8.16.128.145.in-addr.arpa which will point to host1.mydomain.com.
Question 41 |
Consider the following message M = 1010001101. The cyclic redundancy check (CRC) for this message using the divisor polynomial x5 + x4 + x2 + 1 is:
A | 01110 |
B | 01011 |
C | 10101 |
D | 10110 |
M = 1010001101
append 5 zeroes = M = 101000110100000
∴ CRC = 01110
Question 42 |
Suppose that two parties A and B wish to setup a common secret key (D-H key) between themselves using the Diffie-Hellman key exchange technique. They agree on 7 as the modulus and 3 as the primitive root. Party A chooses 2 and party B chooses 5 as their respective secrets. Their D-H key is
A | 3 |
B | 4 |
C | 5 |
D | 6 |
where p is the primitive root and n is the modulus and 'a' and 'b' are the secret values selected by parity A & B.
So answer is,
32×5 mod 7 = 310 mod 7 = 4
Question 43 |
Which one of the following statements is FALSE?
A | Packet switching leads to better utilization of bandwidth resources than circuit switching. |
B | Packet switching results in less variation in delay than circuit switching. |
C | Packet switching requires more per packet processing than circuit switching. |
D | Packet switching can lead to reordering unlike in circuit switching. |
Question 44 |
Which one of the following statements is FALSE?
A | TCP guarantees a minimum communication rate |
B | TCP ensures in-order delivery |
C | TCP reacts to congestion by reducing sender window size |
D | TCP employs retransmission to compensate for packet loss
|
Sequence numbers can allow receivers to discard duplicate packets and properly sequence reordered packets.
Option C:
If the congestion is deleted, the transmitter decreases the transmission rate by a multiplicative factor.
Option D:
Acknowledgement allows the sender to determine when to retransmit lost packets.
Question 45 |
Which one of the following statements is FALSE?
A | HTTP runs over TCP |
B | HTTP describes the structure of web pages |
C | HTTP allows information to be stored in a URL |
D | HTTP can be used to test the validity of a hypertext link |
Question 46 |
A sender is employing public key cryptography to send a secret message to a receiver. Which one of the following statements is TRUE?
A | Sender encrypts using receiver’s public key |
B | Sender encrypts using his own public key |
C | Receiver decrypts using sender’s public key |
D | Receiver decrypts using his own public key |
Question 47 |
A subnet has been assigned a subnet mask of 255.255.255.192. What is the maximum number of hosts that can belong to this subnet?
A | 14 |
B | 30 |
C | 62 |
D | 126 |
= 26- 2
= 64 - 2
= 62
Question 48 |
A host is connected to a Department network which is part of a University network. The University network, in turn, is part of the Internet. The largest network in which the Ethernet address of the host is unique is:
A | the subnet to which the host belongs |
B | the Department network |
C | the University network |
D | the Internet |
Question 49 |
In TCP, a unique sequence number is assigned to each
A | byte |
B | word |
C | segment |
D | message |
Question 50 |
Which of the following objects can be used in expressions and scriplets in JSP (Java Server Pages) without explicitly declaring them?
A | session and request only |
B | request and response only |
C | response and session only |
D | session, request and response |