An IP router with a Maximum Transmission Unit (MTU) of 1500 bytes has received an IP packet of size 4404 bytes with an IP header of length 20 bytes. The values of the relevant fields in the header of the third IP fragment generated by the router for this packet are
Prefix Output Interface Identifier 18.104.22.168/ 12 3 22.214.171.124/ 14 5 126.96.36.199/ 16 2 188.8.131.52/ 15 1
The identifier of the output interface on which this packet will be forwarded is ______.
If we do AND operation between 255.254.0.0 and given IP 184.108.40.206, gives 220.127.116.11 which is matching with interface 1.
Assume that source S and destination D are connected through two intermediate routers labeled R. Determine how many times each packet has to visit the network layer and the data link layer during a transmission from S to D.
Network layer – 4 times and Data link layer – 4 times
Network layer – 4 times and Data link layer – 3 times
Network layer – 4 times and Data link layer – 6 times
Network layer – 2 times and Data link layer – 6 times
From above given diagram, its early visible that packet will visit network layer 4 times, once at each node [S, R, R, D] and packet will visit Data Link layer 6 times. One time at S and one time at D, then two times for each intermediate router R as data link layer is used for link to link communication.
One at packet reaches R and goes up from physical –DL-Network and second time when packet coming out of router in order Network –DL-Physical.
In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are
Last fragment, 2400 and 2789
First fragment, 2400 and 2759
Last fragment, 2400 and 2759
Middle fragment, 300 and 689
HLEN = 10 - So header length is 4×10 = 40, as 4 is constant scale factor.
Total Length = 400 (40 Byte Header + 360 Byte Payload)
Fragment Offset = 300, that means 300×8 Byte = 2400 bytes are before this last fragment.
So the position of datagram is last fragment.
Sequence number of First Byte of Payload = 2400 (as 0 to 2399 Sequence no are used)
Sequence number of Last Byte of Payload = 2400+360-1 = 2759
A link of capacity 100 Mbps is carrying traffic from a number of sources. Each source generates an on-off traffic stream; when the source is on, the rate of traffic is 10 Mbps, and when the source is off, the rate of traffic is zero. The duty cycle, which is the ratio of on-time to off-time, is 1:2. When there is no buffer at the link, the minimum number of sources that can be multiplexed on the link so that link capacity is not wasted and no data loss occurs is S1. Assuming that all sources are synchronized and that the link is provided with a large buffer, the maximum number of sources that can be multiplexed so that no data loss occurs is S2. The values of S1 and S2 are, respectively,
10 and 30
12 and 25
5 and 33
15 and 22
Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.
Since data should not be lost, we calculate for the extreme case when all sources are ontime (that is transmitting).
10Mbps × n-station ≤ 100Mbps
n = 10 = S1
In the next part of question, it is given that the link is provided with large buffer and we are asked to find out large no. of stations.
For that we calculate expected value of bandwidth usage,
E = 1/3 × 10 + 1/3 × 10 + .......+ ....... n-station times ≤ 100Mbps
⇒ 1/3 × 10 × n-station ≤ 100Mbps ⇒ n-station = 30 = S2
So option (A) is answer.
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
By progressively querying routers about the next router on the path to B using ICMP packets, starting with the first router
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
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
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.
Which of the following statements is FALSE regarding a bridge?
Bridge is a layer 2 device
Bridge reduces collision domain
Bridge is used to connect two or more LAN segments
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.
Count to infinity is a problem associated with
link state routing protocol.
distance vector routing protocol.
DNS while resolving host name.
TCP for congestion control.
The Bellman-Ford algorithm does not prevent routing loops from happening and suffers from the count-to-infinity problem.
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
(1 – b1)L p1 + (1 – b2)Lp2
[1 – (b1 + b2)L]p1p2
(1 – b1)L (1 – b2)Lp1p2
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.
A company has a class C network address of 18.104.22.168. 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?
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)
Assume that “host1.mydomain.dom” has an IP address of 22.214.171.124. Which of the following options would be most appropriate as a subsequence of steps in performing the reverse lookup of 126.96.36.199? In the following options “NS” is an abbreviation of “nameserver”.
Query a NS for the root domain and then NS for the “dom” domains
Directly query a NS for “dom” and then a NS for “mydomain.dom” domains
Query a NS for in-addr.arpa and then a NS for 128.145.in-addr.arpa domains
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 188.8.131.52.in-addr.arpa which will point to host1.mydomain.com.
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?
= 26- 2
= 64 - 2
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:
the subnet to which the host belongs
the Department network
the University network
In TCP, a unique sequence number is assigned to each
Which of the following objects can be used in expressions and scriplets in JSP (Java Server Pages) without explicitly declaring them?
session and request only
request and response only
response and session only
session, request and response
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 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?
In the TCP/IP protocol suite, which one of the following is NOT part of the IP header?
Source IP address
Destination IP address
Destination port number
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?
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.
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
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?
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
TP1, TP3, TP4
TP0, TP1, TP2, TP3, TP4
Transport Protocol Class 3 (TP3) offers error recovery, segmentation and reassembly, and multiplexing and demultiplexing of data streams over a single virtual circuit. TP3 also sequences PDUs and retransmits them or re-initiates the connection if an excessive number are unacknowledged.