Computer-Networks
Question 1 |
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 2 |
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 3 |
- 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 4 |
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 5 |
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 6 |
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 7 |
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
|
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 8 |
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 9 |
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 10 |
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 timeout 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 |
Time = 1 during 1st trans. , window size = 2 (Slow start),
Time = 2 congestion window size = 4 (double the no. of ack.)
Time = 3 congestion window = 8
Time = 4 congestion window size = 9, after threshold, increase by one additive increase.
Time = 5 transmit 10 MSS, but time out occur congestion window size = 10
Hence threshold = (congestion window size)/2 = 10/2 = 5
Time = 6 transmit 2(since in the question, they are saying ss is starting from 2)
Time = 7 transmit 4
Time = 8 transmit 5
Time = 9 transmit 6
Time = 10 transmit 7
Question 11 |
Consider a source computer (S) transmitting a file of size 106 bits to a destination computer (D) over a network of two routers (R1 and R2) and three links (L1, L2 and L3). L1 connects S to R1; L2 connects R1 to R2; and L3 connects R2 to D. Let each link be of length 100 km. Assume signals travel over each link at a speed of 108 meters per second. Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D?
A | 1005 ms |
B | 1010 ms |
C | 3000 ms |
D | 3003 ms |
Propagation delay = (Distance) / (Velocity) = 3*105/108 = 3ms
Total transmission delay for 1 packet = 3 * L / B = 3*(1000/106) = 3ms. Because at source and 2 routers, we need to transmit the bits.
The first packet will reach destination = Tt + Tp = 6ms.
While the first packet was reaching to D, other packets must have been processing in parallel. So D will receive remaining packets 1 packet per 1 ms from R2. So remaining 999 packets will take 999 ms.
And total time will be 999 + 6 = 1005 ms
Question 12 |
An Internet Service Provider (ISP) has the following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20. The ISP wants to give half of this chunk of addresses to Organization A, and a quarter to Organization B, while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?
A | 245.248.136.0/21 and 245.248.128.0/22 |
B | 245.248.128.0/21 and 245.248.128.0/22 |
C | 245.248.132.0/22 and 245.248.132.0/21 |
D | 245.248.136.0/24 and 245.248.132.0/21 |
Question 13 |
In the IPv4 addressing format, the number of networks allowed under Class C addresses is
A | 214 |
B | 27 |
C | 221 |
D | 224 |
Question 14 |
Which of the following transport layer protocols is used to support electronic mail?
A | SMTP |
B | IP |
C | TCP |
D | UDP |
Question 15 |
The protocol data unit (PDU) for the application layer in the Internet stack is
A | Segment |
B | Datagram |
C | Message |
D | Frame |
Question 16 |
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 17 |
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 18 |
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 19 |
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 20 |
Which of the following assertions is FALSE about the Internet Protocol (IP)?
A | It is possible for a computer to have multiple IP addresses
|
B | IP packets from the same source to the same destination can take different routes in the network
|
C | IP ensures that a packet is discarded if it is unable to reach its destination within a given number of hops
|
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
|
Question 21 |
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 22 |
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 23 |
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 24 |
- 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 25 |
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 26 |
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 27 |
A | 111 |
B | 100 |
C | 101 |
D | 110 |
Question 28 |
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 29 |
Suppose that the maximum transmit window size for a TCP connection is 12000 bytes. Each packet consists of 2000 bytes. At some point of time, the connection is in slow-start phase with a current transmit window of 4000 bytes. Subsequently, the transmitter receives two acknowledgements. Assume that no packets are lost and there are no time-outs. What is the maximum possible value of the current transmit window?
A | 4000 bytes |
B | 8000 bytes |
C | 10000 bytes |
D | 12000 bytes |
and packet size =2000 B (or MSS)
Receiver window size = 6 MSS and
Current sender window size = 2 MSS
Slow start threshold = receiver window/2 = 3 MSS
Now current sender window size = 2 MSS <3 MSS,
which implies transmission is in slow start phase.
After receiving first Ack: Current sender window should increase exponentially to 4 MSS but since threshold = 3 MSS, current sender window size goes to threshold which is 3 MSS, then after receiving second Ack: Since now it is in congestion avoidance phase, sender window size increases linearly which makes current sender window
= 4 MSS
= 4 × 2000 B
= 8000 B
Question 30 |
A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the second network for this transmission?
A | 40 bytes |
B | 80 bytes |
C | 120 bytes |
D | 160 bytes |
2120B reach R1's network layer. It removes original IP header, fragments data part at IP and then appends IP header to all fragments and forwards . So, it divides 2100 Bytes into two fragments of size 1200 and 900. And both fragments are sent to R2.
At Router-2:
Both fragments that reach R2 exceed MTU at R2. So, both are fragmented. First packet of 1200B is fragmented into 3 packets of 400 Bytes each. And second packet of 900B is fragmented into 3 fragments of 400, 400 and 100 Bytes respectively.
So, totally 6 packets reach destinations.
So, total IP overhead = 6 × 20 = 120 Bytes
Question 31 |
Consider a simplified time slotted MAC protocol, where each host always has data to send and transmits with probability p = 0.2 in every slot. There is no backoff and one frame can be transmitted in one slot. If more than one host transmits in the same slot, then the transmissions are unsuccessful due to collision. What is the maximum number of hosts which this protocol can support, if each host has to be provided a minimum through put of 0.16 frames per time slot?
A | 1 |
B | 2 |
C | 3 |
D | 4 |
0.16 = 0.2 × 0.8N-1
⇒ 0.8 = 0.8N-1
⇒ N = 2
Question 32 |
In the TCP/IP protocol suite, which one of the following is NOT part of the IP header?
A | Fragment Offset |
B | Source IP address |
C | Destination IP address |
D | Destination port number |
Question 33 |
A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?
A | 5 Kbps |
B | 10 Kbps |
C | 15 Kbps |
D | 20 Kbps |
Question 34 |
In a data link protocol, the frame delimiter flag is given by 0111. Assuming that bit stuffing is employed, the transmitter sends the data sequence 01110110 as
A | 01101011 |
B | 011010110 |
C | 011101100 |
D | 0110101100 |
Thus using the above logic,
Delimiter flag: 0111
Data sequence: 01110110
So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.
0 1 1 0 1 0 1 1 0 0
In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).
Question 35 |
In a sliding window ARQ scheme, the transmitter's window size is N and the receiver's window size is M. The minimum number of distinct sequence numbers required to ensure correct operation of the ARQ scheme is
A | min (M,N) |
B | max (M,N) |
C | M + N |
D | MN |
Question 36 |
Consider a 10 Mbps token ring LAN with a ring latency of 400 µs. A host that needs to transmit seizes the token. Then it sends a frame of 1000 bytes, removes the frame after it has circulated all around the ring, and finally releases the token. This process is repeated for every frame. Assuming that only a single host wishes to transmit, the effective data rate is
A | 1Mbps |
B | 2Mbps |
C | 5Mbps |
D | 6Mbps |
Question 37 |
A serial transmission T1 uses 8 information bits, 2 start bits, 1 stop bit and 1 parity bit for each character. A synchronous transmission T2 uses 3 eight bit sync characters followed by 30 eight bit information characters. If the bit rate is 1200 bits/second in both cases, what are the transfer rates of T1 and T2?
A | 100 characters/sec, 153 characters/sec |
B | 80 characters/sec, 136 characters/sec |
C | 100 characters/sec, 136 characters/sec |
D | 80 characters/sec, 153 characters/sec |
Transfer rate = 1200/12 = 100 char/sec
T2: Transfer character in bits = 24 + 240 = 264 bits
In 264 = 30 characters
Then in 1200 = ? 264/30 = 1200/x
x = 136.3 char/sec
So, correct option is (C).
Question 38 |
Consider the following statements:
I. telnet, ftp and http are application layer protocols.
II.l EJB (Enterprise Java Beans) components can be deployed in a J2EE (Java2 Enterprise Edition) application server.
III. If two languages conform to the Common Language Specification (CLS) of the Microsoft.NET framework, then a class defined in any one of them may be inherited in the other.
Which statements are true?
A | l and II only |
B | II and III only |
C | l and III only |
D | I, II and III |
Then there are certain compliance rules which may be used for inheritance. So other statement (I) and (II) are True.
Question 39 |
In TCP, a unique sequence number is assigned to each
A | byte |
B | word |
C | segment |
D | message |
Question 40 |
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 |
Question 41 |
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 42 |
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 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 |
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 48 |
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 49 |
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 50 |
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