Computer Network Questions and Answers

Q-1 What are the goals in mind of IEEE 802 committee?
Ans: IEEE 802 committee has few goals in mind, namely
–To promote compatibility
–Implementation with minimum efforts
–Accommodate diverse applications

Q-2. List the functions performed by the physical layer of 802.3 standard?
Ans. Functions of physical layer are:
i) Data encoding/decoding (To facilitate synchronization and efficient transfer of signal through the medium).
ii) Collision detection (It detects at the transmit side)
iii) Carrier sensing (Channel access senses a carrier on the channel at both the transmit and receive sides)
iv) Transmit/receive the packets (Frame transmitted to all stations connected to the channel)
v) Topology and medium used (Mediums are co-axial cable, twisted pair and fiber optic cable)

Q-3. Why do you require a limit on the minimum size of Ethernet frame?
Ans. To detect collision, it is essential that a sender continue sending a frame and at the same time receives another frame sent by another station. Considering maximum delay with five Ethernet segments in cascade, the size of frame has been found to be 64 bytes such that the above condition is satisfied.

Q-4. What are the different types of cabling supported by Ethernet standard?
Ans. Types of cabling are:
i) 10 BASE 5 - Maximum cable length is 500 meters using 4” diameter coaxial cable.
ii) 10 BASE 2 - Maximum cable length is 185 meters using 0.25” diameter CATV cable.
iii) 10 BASE T - Maximum cable length is 100 meters using twisted-pair cable (CAT-3 UTP).
iv) 10 BASE FL - Maximum cable length is 2 Km using multimode fiber optic cable (125/62.5 micrometer).

Q-5. Explain the basic difference between IEEE 802.3 and switched Ethernet, as far as implementation is concerned.
Ans: In Ethernet (IEEE 802.3) the topology, though physically is start but logically is BUS. i.e. the collision domain of all the nodes in a LAN is common. In this situation only one frame can send the frame, if more than one station sends the frame, there is a collision.
In Switched Ethernet, this collision domain is separated. Hub is replaced by a switch, a device that can recognize the destination address and can route the frame to the port to which the destination station is connected, the rest of the media is not involved in the transmission process. The switch can receive another frame from another station at the same time and can route this frame to its own final destination.

Q-6. Explain the two techniques for implementing Ethernet switches.
Ans: There are two techniques used in the implementation of Ethernet switches: store-and-forward and cut-through. In the first case, the entire frame is captured at the incoming port, stored in the switch’s memory, and after an address lookup to determine the LAN destination port, forwarded to the appropriate port. The lookup table is automatically built up. On the other hand, a cut-through switch begins to transmit the frame to the destination port as soon as it decodes the destination address from the frame header.
Store-and-forward approach provides a greater level of error detection because damaged frames are not forwarded to the destination port. But, it introduces longer delay of about 1.2 msec for forwarding a frame and suffers from the chance of loosing data due to reliance on buffer memory. The cut-through switches, on the other hand, has reduced latency but has higher switch cost.

Q-7. What are the different categories of Fast Ethernet?
Ans: IEEE has designed two categories of Fast Ethernet: 100Base-X and 100Base-T4. 100Base-X uses two cables between hub and the station while 100Base-T4 uses four. 100-Base-X itself is divided into two: 100Base-TX and 100base-FX.
* 100 BASE-T4: This option is designed to avoid overwriting. It is used for half-duplex communication using four wire-pairs of the existing category 3 UTP cable, which is already available for telephone services in homes/offices. Two of four pairs are bi-directional; other two are unidirectional. This means that there are 3 pairs to be used for carrying data, in each direction (2 bi-directional and 1 uni-directional). Because 100Mbps data cannot be handled by voice-grade UTP, this specification splits the 100 Mbps flow into three 33.66Mbps flow.
* 100 BASE TX: This option uses two category 5 UTP or two shielded (STP) cable to connect a station to hub. One pair is used to carry frames from the hub to the station and other to carry frames from station to hub. Encoding is 4B/5B to handle 100 Mbps; signaling is NRZ-I. The distance between station and hub should be less than 100 meters.
* 100 BASE FX: This option uses two Fiber optic cables, one carry frames from station to hub and other from hub to station. The encoding is 4B/5B and signaling in NRZ-I. the distance between station and hub should be less than 2000 meters.

Q-8. What are the Objectives of The Gigabit Ethernet Alliance?
Ans: The objectives of the alliance are:
• supporting extension of existing Ethernet and Fast Ethernet technology in response to demand for higher network bandwidth.
• developing technical proposals for the inclusion in the standard
• establishment of inter-operability test procedures and processes

Q-9. Explain GMII (Gigabit Media Independent Interface) in brief.
Ans: The GMII is the interface between the MAC layer and the Physical layer. It allows any physical layer to be used with the MAC layer. It is an extension of the MII (Media Independent Interface) used in Fast Ethernet. It uses the same management interface as MII. It supports 10, 100 and 1000 Mbps data rates. It provides separate 8-bit wide receive and transmit data paths, so it can support both full-duplex as well as half-duplex operation.
The GMII provides 2 media status signals: one indicates presence of the carrier, and the other indicates absence of collision. With the GMII, it is possible to connect various
media types such as shielded and unshielded twisted pair, and single-mode and multi mode optical fiber, while using the same MAC controller. It has three sub-layers namely: PCS (Physical Coding Sublayer), PMA (Physical Medium Attachment) and PMD (Physical Medium Dependent)

Computer Network Questions and Answers

1. The IP datagram for a TCP ACK message is 40 bytes long: It contains 20 bytes of TCP header and 20 bytes of IP header. Assume that this ACK is traversing an ATM network that uses AAL5 to encapsulate IP packets. How many ATM layer packets will it take to carry the ACK for each of AAL layer?
a. AAL5
b. AAL3/4
ANS
a)) The length of the AAL5 CS-PDU into which the ACK is encapsulated is exactly 48 bytes, and this fits into a single ATM cell.
b)) When AAL3/4 is used the CS-PDU is again 48 bytes, but now the per-cell payload is only 44 bytes and two cells are necessary.

2. Suppose there are N stations on a LAN that has capacity (transmission rate) C. All packets have a fixed length L and the end-to-end propagation delay of the channel is P. If only one station ever has a message to send (i.e., the other N-1 stations generate no traffic). What is the maximum possible throughput seen by this single node under each of the protocols:
a. CSMA
Assuming that the station senses the channel immediately after finishing its transmission, and finding it idle, sends another packet, the throughput is C. Note that this ignores the overhead time needed to do channel sensing.

b. Slotted Aloha
Assuming the station always has data to send, it will always to do under slotted Aloha, so the throughput would be C.

c. Token Passing
After sending a message, a station gives up the token. The token circulates around the ring (taking time P). Assuming that L is the amount of time needed to transmit a message, the throughput is thus (L/(L+PC))*C.


3. Consider sending a 1500-byte datagram into a link that has an MTU of 500 bytes. Suppose the original datagram is stamped with the identification number 1. Assume that IPv4 is used.
a. Where does fragmentation happen? Where are the fragments reassembled?
Fragmentation happens in the router preceding the link with the small MTU. The fragments are reassembled in the end system.

b. How many fragments are generated?
The maximum size of the data field in each fragment = 480 (because there are 20 bytes IP header). Thus the number of required fragments= 1500-20/480 = 4

c. What are the values of the fragmentation-related fields in the generated IP datagram(s)?
Each fragment will have an identical identification number. Each fragment except the last one will be of size 500 bytes (including the IP header). The last datagram will be of size 60 bytes (including the IP header). The offsets of the 4 fragments will be 0, 60, 120, 180. Each of the first 3 fragments will have flag=1; the last fragment will have flag=0.

d. What changes if IPv6 were used?
The router preceding the link with the small MTU will drop the packet and send an ICMP error message “Packet Too Big” back to the source. The source is responsible for adjusting the packet size.

4. Consider building a CSMA/CD network running at 1 Gbps over a 1-km cable with no repeaters. The signal speed in the cable is 200,000 km/sec. What is the minimum frame size?
ANS
For a 1-km cable, the one-way propagation time is 5 μsec, so 2τ = 10 μsec. To make CSMA/CD work, it must be impossible to transmit an entire frame in this interval. At 1 Gbps, all frames shorter than 10,000 bits can be completely transmitted in under 10 μsec, so the minimum frame is 10,000 bits or 1250 bytes

5. In the CSMA/CD protocol, what condition on the transmission delay Ttrans and the propagation delay Tprop has to be satisfied to guarantee that a node always detects a collision?
ANS Ttrans>2Tprop


6. At low load, consider the delay of pure ALOHA versus slotted ALOHA. Which one is less? Explain your answer.
ANSWith pure ALOHA, transmission can start instantly. At low load, no collisions are expected so the transmission is likely to be successful. With slotted ALOHA, it has to wait for the next slot. This introduces half a slot time of delay.

Computer Network Questions and Answers

1. Why is it important for protocols configured on top of the Ethernet to have a length field in their header, indicating how long the message is?
Ans: It is important to indicate the message size because of the following reasons:
 There can be padded data inside the message,
 In case of collision only part of data is received, so it can checked by the message length field, otherwise, it won't be known.
 The Ethernet frame size isn't fixed. It can hold up to 1500 bytes but it can vary from network to network.

2. What kinds of problems can arise when two hosts on the same Ethernet share the same hardware address? Describe what happens and why that behavior is a problem.
Ans : 1. Hosts are using hardware address to filter whether the packet is for them or not. So, they can discard the packets which do not belong to them. However, if the same hardware address is shared with other host, it is likely to receive wrong packets which can harm the host.
2. On the same Ethernet line, hosts keep dynamic ARP table to match Internet address and physical address (MAC address), but if they see two different IP addresses having same MAC address. It is hard to tell what happens for this.
3. If hosts share the switch on the Ethernet line, the switch will get confused to update the address table. It will continuously change the address table, so it can lead to unstable state.

3. The 1982 Ethernet specification allowed between any two stations up to 1500 m of coaxial cable, 1000 m of other point-to-point link cable, and two repeaters. Each station or repeater connects to the coaxial cable via up to 50 m of “drop cable.” Typical delays associated with each device are given in Table below (where c = speed of light in a vacuum = 3×108 m/s). What is the worst-case round-trip propagation delay, measured in bits, due to the sources listed? (This list is not complete; other sources of delay include sense time and signal rise time.)

Item Delay
Coaxial cable propagation speed .77c
Link/drop cable propagation speed .65c
Repeaters approximately 0.6 μs each
Transceivers approximately 0.2 μs each
Typical delays associated with various devices
One-way delays:
Coax: 1500m 6.49 3 s
link: 1000m 5.13 3 s
repeaters two 1.20 3 s
transceivers six (two for 1.20 3 s
each repeater,
one for each
station)

drop cable 6 , 50m 1.54 3 s
Total: 15.56 3 s
The roundtrip delay is thus about 31.1 3 s, or 311 bits. The “official” total is 464 bits, which when extended by 48 bits of jam signal exactly accounts for the 512-bit minimum packet size.
The 1982 Digital-Intel-Xerox specification presents a delay budget (page 62 of that document) that totals 463.8 bit-times, leaving 20 nanoseconds for unforeseen contingencies.

4. Coaxial cable Ethernet was limited to a maximum of 500 m between repeaters, which regenerate the signal to 100% of its original amplitude. Along one 500-m segment, the signal could decay to no less than 14% of its original value (8.5 dB). Along 1500 m, then, the decay might be (0.14)3 = 0.3%. Such a signal, even along 2500 m, is still strong enough to be read; why then are repeaters required every 500 m?
Ans>>A station must not only detect a remote signal, but for collision detection it must detect a remote signal while it itself is transmitting. This requires much higher remote-signal intensity.

5. Suppose the round-trip propagation delay for Ethernet is 46.4 μs. This yields a minimum packet size of 512 bits (464 bits corresponding to propagation delay + 48 bits of jam signal).
(a) What happens to the minimum packet size if the delay time is held constant, and the signaling rate rises to 100 Mbps?
(b) What are the drawbacks to so large a minimum packet size?
(c) If compatibility were not an issue, how might the specifications be written so as to permit a smaller minimum packet size?
Ans>>(a) Assuming 48 bits of jam signal was still used, the minimum packet size would be 4640+48 bits = 586 bytes.
(b) This packet size is considerably larger than many higher-level packet sizes, resulting in considerable wasted bandwidth.
(c) The minimum packet size could be smaller if maximum collision domain diameter were reduced, and if sundry other tolerances were tightened up.

6. A group of N stations share a 56-kbps pure ALOHA channel. Each station outputs a 1000-bit frame on an average of once every 100 sec, even if the previous one has not yet been sent (e.g., the stations can buffer outgoing frames). What is the maximum value of N?
ANS>>with pure ALOHA the usable bandwidth is 0.184 × 56 kbps = 10.3 kbps.Each station requires 10 bps, so N = 10300/10 = 1030 stations

7. Consider the delay of pure ALOHA versus slotted ALOHA at low load. Which one is less? Explain your answer.
ANS>>With pure ALOHA, transmission can start instantly. At low load, no collisions are expected so the transmission is likely to be successful. With slotted ALOHA, it has to wait for the next slot. This introduces half a slot time of delay.

8. Ten thousand airline reservation stations are competing for the use of a single slotted ALOHA channel. The average station makes 18 requests/hour. A slot is 125 μsec. What is the approximate total channel load?
ANS>>Each terminal makes one request every 200 sec, for a total load of 50 requests/sec. Hence G = 50/8000 = 1/160.

9. A large population of ALOHA users manages to generate 50 requests/sec, including both originals and retransmissions. Time is slotted in units of 40 msec.
a. What is the chance of success on the first attempt?
b. What is the probability of exactly k collisions and then a success?
c. What is the expected number of transmission attempts needed?
ANS>> (a) With G = 2 the Poisson law gives a probability of e −2.
(b) (1 − e −G)ke −G = 0.135 × 0.865k .
(c) The expected number of transmissions is eG = 7.4.

10. Measurements of a slotted ALOHA channel with an infinite number of users show that 10 percent of the slots are idle.
a. What is the channel load, G?
b. What is the throughput?
c. Is the channel underloaded or overloaded?
ANS>> (a) From the Poisson law again, P0 = e −G, so G = −lnP0 = −ln 0.1 = 2.3.
(b) Using S = Ge−G with G = 2.3 and e −G = 0.1, S = 0.23.
(c) Whenever G > 1 the channel is overloaded, so it is overloaded.

11. In an infinite-population slotted ALOHA system, the mean number of slots a station waits between a collision and its retransmission is 4. Plot the delay versus throughput curve for this system.
ANS>>The number of transmissions is E = eG. The E events are separated by E – 1 intervals of four slots each, so the delay is 4(eG − 1). The throughput is given by S = Ge−G. Thus, we have two parametric equations, one for delay and one for throughput, both in terms of G. For each G value it is possible to find the corresponding delay and throughput, yielding one point on the curve.

12. 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed of 200 m/μsec. Repeaters are not allowed in this system. Data frames are 256 bits long, including 32 bits of header, checksum, and other overhead. The first bit slot after a successful transmission is reserved for the receiver to capture the channel in order to send a 32-bit acknowledgement frame. What is the effective data rate, excluding overhead, assuming that there are no collisions?
ANS>> The round-trip propagation time of the cable is 10µsec. A complete transmission has six phases: transmitter seizes cable (10µsec) transmit data (25.6µsec) Delay for last bit to get to the end (5.0 µsec) receiver seizes cable (10µsec) acknowledgement sent (3.2 µsec) Delay for last bit to get to the end (5.0 µsec) the sum of these is 58.8 µsec. In this period, 224 data bits are sent, for a rate of about 3.8 Mbps.

13. Consider building a CSMA/CD network running at 1 Gbps over a 1-km cable with no repeaters. The signal speed in the cable is 200,000 km/sec. What is the minimum frame size?
ANS>> For a 1-km cable, the one-way propagation time is 5 μsec, so 2τ = 10 μsec. To make CSMA/CD work, it must be impossible to transmit an entire frame in this interval. At 1 Gbps, all frames shorter than 10,000 bits can be completely transmitted in under 10 μsec, so the minimum frame is 10,000 bits or 1250 bytes.

14. An IP packet to be transmitted by Ethernet is 60 bytes long, including all its headers. If LLC is not in use, is padding needed in the Ethernet frame, and if so, how many bytes?
ANS>>The minimum Ethernet frame is 64 bytes, including both addresses in the Ethernet frame header, the type/length field, and the checksum. Since the header fields occupy 18 bytes and the packet is 60 bytes, the total frame size is 78 bytes, which exceeds the 64-byte minimum. Therefore, no padding is used.

15. Ethernet frames must be at least 64 bytes long to ensure that the transmitter is still going in the event of a collision at the far end of the cable. Fast Ethernet has the same 64-byte minimum frame size but can get the bits out ten times faster. How is it possible to maintain the same minimum frame size?
ANS>>The maximum wire length in fast Ethernet is 1/10 as long as in Ethernet.

16. Some books quote the maximum size of an Ethernet frame as 1518 bytes instead of 1500 bytes. Are they wrong? Explain your answer.
ANS>>The payload is 1500 bytes, but when the destination address, source address, type/length, and checksum fields are counted too, the total is indeed 1518.

17. Recall that with the CSMA/CD protocol, the adapter waits K*512 bit times after a collision, where K is drawn randomly. For K=100, how long does the adapter wait until returning to Step 2 for a 10 Mbps Ethernet? For a 100 Mbps Ethernet?
ANS>> Wait for 51,200 bit times. For 10 Mbps, this wait is

For 100 Mbps, the wait is 512 μ sec.

18. Suppose nodes A and B are on the same 10 Mbps Ethernet segment, and the propagation delay between the two nodes is 225 bit times. Suppose node A begins transmitting a frame, and before it finishes station B begins transmitting a frame. Can A finish transmitting before it detects that B has transmitted? Why or why not? If the answer is yes, then A incorrectly believes that its frame was successfully transmitted without a collision.
Hint: Suppose at time t=0 bit times, A begins transmitting a frame. In the worst case, A transmits a minimum size frame of 512+64(prembale) bit times. So A would finish transmitting the frame at t=512+64 bit times. Thus the answer is no if B's signal reaches A before bit time t=512+64 bits. In the worst case, when does B's signal reach A?

ANS>> At t = 0 A transmits. At t = 576 , A would finish transmitting. In the worst case, B begins transmitting at time t = 224 . At time t = 224 + 225 = 449 B 's first bit arrives at A . Because 449 < 576 , A aborts before completing the transmission of the packet, as it is supposed to do. Thus A cannot finish transmitting before it detects that B transmitted. This implies that if A does not detect the presence of a host, then no other host begins transmitting while A is transmitting. 19. Suppose nodes A and B are on the same 10 Mbps Ethernet segment, and the propagation delay between the two nodes is 225 bit times. Suppose A and B send frames at the same time, the frames collide, and then A and B choose different values of K in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from A and B collide? For our purposes, it suffices to work out the following example. Suppose A and B begin transmission at t=0 bit times. They both detect collisions at t=225 bit times. They finish transmitting jam signal at t= 225+48= 273 bit times. Suppose KA=0 and KB=1. At what time does B schedule its retransmission? At what time does A begin transmission? (Note, the nodes must wait for an idle channel after returning to Step 2-- see protocol.) At what time does A's signal reach B? Does B refrain from transmitting at its scheduled time?

Time Event
0 A and B begin transmission
225 A and B detect collision
273 A and B finish transmitting jam signal
273+225=498 B 's last bit arrives at A ;A detects an idle channel
498+96=594 A starts transmitting
273+512=785 B returns to Step2
B must sense idle channel for 96 bit times before it transmits
785+96=881 B starts transmitting
594+225=819 A’s transmission reaches B
Because A 's retransmission reaches B before B 's scheduled retransmission time, B refrains from transmitting while A retransmits. Thus A and B do not collide. Thus the factor 512 appearing in the exponential backoff algorithm is sufficiently large.

20. Consider a 100Mbps 100-BT Ethernet. In order to have an efficiency of .50, what should be the maximum distance between a node and the hub? Assume a frame length of 64 bytes and that there are no repeaters. Does this maximum distance also ensure that a transmitting node A will be able to detect whether any other node transmitted while A was transmitting? Why or why not? How does your maximum distance compare to the actual 100 Mbps standard?
Ans>>We want 1/(1+ 5a) = .5 or, equivalently, prop trans a = .2 = t / t . t = d /(1.8 ×108 ) prop m/sec and = (576 trans t bits ) /(108 bits/sec ) = 5.76μ sec. Solving for d we obtain d = 265 meters. For the 100 Mbps Ethernet standard, the maximum distance between two hosts is 200 m.
For transmitting station A to detect whether any other station transmitted during A 's interval, trans t must be greater than 2 = 2 ⋅ 265 prop t m/1.8 ×108 m/sec = 2.94μ sec. Because 2.94 < 5.76 , A will detect B 's signal before the end of its transmission. 21. Suppose two nodes, A and B, are attached to opposite ends of a 900 m cable, and that they each have one frame of 1000 bits (including all headers and preambles) to send to each other. Both nodes attempt to transmit at time t=0. Suppose there are four repeaters between A and B, each inserting a 20 bit delay. Assume the transmission rate is 10 Mbps, and CSMA/CD with backoff intervals of multiples of 512 bits is used. After the first collision, A draws K=0 and B draws K=1 in the exponential backoff protocol. Ignore the jam signal.
a. What is the one-way propagation delay (including repeater delays) between A and B in seconds. Assume that the signal propagation speed is 2 * 108m/sec.
b. At what time (in seconds) is A's packet completely delivered at B.
c. Now suppose that only A has a packet to send and that the repeaters are replaced with bridges. Suppose that each bridge has a 20 bit processing delay in addition to a store-and-forward delay. At what time in seconds is A's packet delivered at B?
ANS>> a)
= (4.5 10-6+8 10-6)
=12.5µsec
b) frame transmission time is (1000bits/10*106bps)=100 μsec
At time t = 0 , both A and B transmit.
At time t = 12.5μ sec , A detects a collision.
At time t = 25μ sec last bit of B 's aborted transmission arrives at A .
At time t = 37.5μ sec first bit of A 's retransmission arrives at B .
At time t= 37.5 μsec + 100 μsec =137.5 μsecA 's packet is completely delivered at B .
c) 12.5μ sec+ 5⋅100μ sec = 512.5μ sec

Overview of Data Communication

1. The ____________ is the physical path over which a message travels.
a. Protocol
b. Medium
c. Signal
d. All the above

2. The information to be communicated in a data communications system is the _________________
a. Medium
b. Protocol
c. Message
d. Transmission

3. Frequency of failure and network recovery time after a failure is measures of the _______________ of a network.
a. Performance
b. Reliability
c. Security
d. Feasibility

4. An unauthorized user is a network ____________ issue.
a. Performance
b. Reliability
c. Security
d. All the above

5. Which topology requires a central controller or hub?
a. Mesh
b. Star
c. Bus
d. Ring

6. Which topology requires a multipoint connection?
a. Mesh
b. Star
c. Bus
d. Ring

7. Communication between a computer and a keyboard involves _______ transmission.
a. Simplex
b. Half-duplex
c. Full-duplex
d. Automatic

8. In a network with 25 computers, which topology would require the most extensive cabling?
a. Mesh
b. Star
c. Bus
d. Ring

9. A television broadcast is an example of ____________ transmission.
a. Simplex
b. Half-duplex
c. Full-duplex
d. Automatic

10. A _____________connection provides a dedicated link between two devices.
a. Point-to-point
b. Multipoint
c. Primary
d. Secondary

11. In a ____________ connection, more than two devices can share a single link.
a. Point-to-point
b. Multipoint
c. Primary
d. Secondary

12. In a transmission, the channel capacity is shared by both communicating devices at all times.
a. Simplex
b. Half-duplex
c. Full-duplex
d. Half-simplex

13. A cable break in a ___________ topology stops all transmission.
a. Mesh
b. Bus
c. Star
d. Primary

14. Which organisation has authority over interstate and international commerce in the communications filed?
a. ITU-T
b. IEEE
c. FCC
d. ISO

15. The Internet model consists of ____________ layers.
a. Three
b. Five
c. Seven
d. Eight

16. The process-to-process delivery of the entire message is the responsibility of the ___________ layer.
a. Network
b. Transport
c. Application
d. Physical

17. The _____________ layer is the layer closest to the transmission medium.
a. Physical
b. Data link
c. Network
d. Transport

18. Mail services are available to network users through the ___________ layer.
a. Data link
b. Physical
c. Transport
d. Application

19. As the data packet moves from the lower to the upper layers, headers are_________
a. Added
b. Subtracted
c. Rearranged
d. Modified

20. As the data packet moves from upper to lower layers, headers are____________
a. Added
b. Removed
c. Rearranged
d. Modified

21. The __________ layer lies between the network layer and the application layer.
a. Physical
b. Data link
c. Transport
d. None of the above

22. Layer 2 lies between the physical and the _______________ layer.
a. Network
b. Data link
c. Transport
d. None of the above

23. When data are transmitted from device A to device B, the header from A’s layer 4 is read by B’s _____________ layer.
a. Physical
b. Transport
c. Application
d. None of the above

24. The _________________ layer changes bits into electromagnetic signals.
a. Physical
b. Data link
c. Transport
d. None of the above

25. The physical layer is concerned with the transmission of ___________ over the physical medium.
a. Programs
b. Dialogs
c. Protocols
d. Bits

26. Which layer functions as a liaison between user support layers and network support layers?
a. Network
b. Physical
c. Transport
d. Application

27. What is the main function of the transport layer?
a. Node-to-node delivery
b. Process-to-process delivery
c. Synchronization
d. Updating and maintenance of routing tables

28. Which of the following is an application layer service?
a. Remote log-in
b. File transfer and access
c. Mail service
d. All the above
_________________________________________________________________________________________________

Computer Network: MCQ

1. If a computer on the network shares resources for others to use, it is called ____
a. Server
b. Client
c. Mainframe
d. PC
Answer : a

2. Terminators are used in ______ topology.
a. Bus
b. Star
c. Mesh
d. Ring
Answer : a

3. In _____ topology, if a computer’s network cable is broken, whole network goes down.
a. Bus
b. Star
c. Mesh
d. Ring
Answer : a

4. For large networks, _______ topology is used.
a. Bus
b. Star
c. Ring
d. Tree
Answer : b

5. ISO stands for
a. International Standard Organization
b. International Student Organization
c. Integrated Services Organization
d. Indian Standard Organisation
Answer : a

6. ISO OSI model is used in
a. Stand alone PC
b. Network environment
c. Mini Computer
d. Mainframe
Answer : b

7. Network cable lies on _____ layer
a. Application
b. Network
c. Physical
d. Transport
Answer : c

8. ____ layer decides which physical pathway the data should take.
a. Application
b. Network
c. Physical
d. Session
Answer : c

9. ISDN is an example of ______ network
a. Circuit switched
b. Packet switched
c. Message Switched
d. Virtual Circuit Switched
Answer : a

10. X.25 is an example of ______ network
a. Circuit switched
b. Packet switched
c. Message Switched
d. Virtual Circuit Switched
Answer : b

11. _____________ allows LAN users to share computer programs and data.
a. Communication server
b. Print server
c. File server
d. Client-Server
Answer : c

12. Print server uses ________ which is a buffer that holds data before it is send to the printer.
a. Queue
b. Spool
c. Node
d. Host
Answer : b

13. A standalone program that has been modified to work on a LAN by including concurrency controls such as file and record locking is an example of____
a. LAN intrinsic software
b. LAN aware software
c. Groupware
d. LAN ignorant software
Answer : a

14. The ______ portion of LAN management software restricts access, records user activities and audit data etc.
a. Configuration management
b. Security management
c. Performance management
d. Authentication Management
Answer : b

15. What is the max cable length of STP?
a. 100 ft
b. 200 ft
c. 100 m
d. 200 m
Answer : d

16. What is the max data capacity of STP?
a. 10 mbps
b. 100 mbps
c. 1000 mbps
d. 10000 mbps
Answer : b

17. Which connector STP uses?
a. BNC
b. RJ-11
c. RJ-45
d. RJ-69
Answer : c

18. What is the central device in star topology?
a. STP server
b. Hub/switch
c. PDC
d. Router
Answer : b

19. What is max data capacity for optical fiber cable?
a. 10 mbps
b. 100 mbps
c. 1000 mbps
d. 10000 mbps
Answer : c

20. Which of the following architecture uses CSMA/CD access method?
a. ARCnet
b. Ethernet
c. ARPAnet
d. Sancharnet
Answer : b
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