Which one of the following is not a client server application?
Match the following:
(P) SMTP (1) Application layer (Q) BGP (2) Transport layer (R) TCP (3) Data link layer (S) PPP (4) Network layer (5) Physical layer
P – 2 Q – 1 R – 3 S – 5
P – 1 Q – 4 R – 2 S – 3
P – 1 Q – 4 R – 2 S – 5
P – 2 Q – 4 R – 1 S – 3
Question 2 Explanation:
See question 4 of http://www.geeksforgeeks.org/computer-networks-set-10/
In the following pairs of OSI protocol layer/sub-layer and its functionality, the INCORRECT pair is
Network layer and Routing
Data Link Layer and Bit synchronization
Transport layer and End-to-end process communication
Medium Access Control sub-layer and Channel sharing
Question 3 Explanation:
1) Yes, Network layer does Rotuing 2) No, Bit synchronization is provided by Physical Layer 3) Yes, Transport layer provides End-to-end process communication 4) Yes, Medium Access Control sub-layer of Data Link Layer provides Channel sharing.
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
P-1, Q-4, R-3
P-2, Q-4, R-1
P-2, Q-3, R-1
P-1, Q-3, R-2
Question 4 Explanation:
Data link layer is the second layer of the OSI Model. This layer is responsible for data transfer between nodes on the network and providing a point to point local delivery framework. So, P matches with 1. Network layer is the third layer of the OSI Model. This layer is responsible for forwarding of data packets and routing through intermediate routers. So, Q matches with 4. Transport layer is the fourth layer of the OSI Model. This layer is responsible for delivering data from process to process. So, R matches with 3. Thus, A is the correct option. Please comment below if you find anything wrong in the above post.
Which of the following is NOT true with respect to a transparent bridge and a router?
Both bridge and router selectively forward data packets
A bridge uses IP addresses while a router uses MAC addresses
A bridge builds up its routing table by inspecting incoming packets
A router can connect between a LAN and a WAN
Question 5 Explanation:
Network Devices Bridges are used to connect two or more networks and create a bigger network. A bridge works up upto the second layer only, i.e. it only implements the Physical and Data Link layers. Therefore it does not use IP addresses for its routing. Therefore option (B) is correct. This explanation is provided by Chirag Manwani.
Host A sends a UDP datagram containing 8880 bytes of user data to host B over an Ethernet LAN. Ethernet frames may carry data up to 1500 bytes (i.e. MTU = 1500 bytes). Size of UDP header is 8 bytes and size of IP header is 20 bytes. There is no option field in IP header. How may total number of IP fragments will be transmitted and what will be the contents of offset field in the last fragment?
6 and 925
6 and 7400
7 and 1110
7 and 8880
Question 6 Explanation:
UDP data = 8880 bytes UDP header = 8 bytes IP Header = 20 bytes Total Size excluding IP Header = 8888 bytes. Number of fragments = ⌈ 8888 / 1480 ⌉ = 7 Refer the Kurose book slides on IP (Offset is always scaled by 8) Offset of last segment = (1480 * 6) / 8 = 1110
Since it is a network that uses switch, every packet goes through two links, one from source to switch and other from switch to destination. Since there are 10000 bits and packet size is 5000, two packets are sent. Transmission time for each packet is 5000 / 1077 bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 10000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is _________.
Question 7 Explanation:
Sender host transmits first packet to switch, the transmission time is 5000/107 which is 500 microseconds. After 500 microseconds, the second packet is transmitted. The first packet reaches destination in 500 + 35 + 20 + 20 + 500 = 1075 microseconds. While the first packet is traveling to destination, the second packet starts its journey after 500 microseconds and rest of the time taken by second packet overlaps with first packet. So overall time is 1075 + 500 = 1575.
Which one of the following statements is FALSE?
TCP guarantees a minimum communication rate
TCP ensures in-order delivery
TCP reacts to congestion by reducing sender window size
TCP employs retransmission to compensate for packet loss
Question 8 Explanation:
Transport Layer protocols offer certain services to the applications using them. These services include-
- Reliable data transfer.
- It ensures in-order delivery of packets by numbering them.
- It avoids congestion by limiting the sending window size.
- It retransmits packets which are lost in the path.
Which one of the following statements is FALSE?
HTTP runs over TCP
HTTP describes the structure of web pages
HTTP allows information to be stored in a URL
HTTP can be used to test the validity of a hypertext link
Question 9 Explanation:
HTML describes structure of page not HTTP. HTTP is the set of rules for transferring files (text, graphic images, sound, video, and other multimedia files) on the World Wide Web.
A serial transmission Ti 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 Ti and T2?
100 characters/sec, 153 characters/sec
80 characters/sec, 136 characters/sec
100 characters/sec, 136 characters/sec
80 characters/sec, 153 characters/sec
Question 10 Explanation:
Serial communication :
Total number of bits transmitted = 8 + 2 + 1 + 1 = 12 bits Bit rate = 1200 / second Transfer Rate = 1200 * (8/12) = 800 bits/sec = 100 bytes/sec = 100 characters/sec
Synchronous transmission :
Total number of bits transmitted = 3 + 30 = 33 bits Transfer Rate = 1200 * (30/33) = 1090.90 bits/sec Given size of one character is 8 bits. So, Transfer Rate = 1090.90 bits/sec = 1090/8 = 136 characters/sec. So,
Thus, option (C) is correct.
Please comment below if you find anything wrong in the above post.
There are 44 questions to complete.