1. Explain the difference between network extensibility and network scalability using the example of the Ethernet.
2. Compare random and deterministic methods of access to a shared medium.
3. Why are data link-layer protocols of WAN technologies not divided into MAC and LLC sublayers?
Because WANs don’t use shared media.
4. What functions are carried out by the LLC layer?
5. What is collision?
What are the functions
of the preamble and the start-of-frame delimiter in the
7. Which network tools carry out jabber control?
8. Why was the interpacket gap introduced into the Ethernet?
This pause is used to restore in initial state of network adapters and to prevent shared media to be monopolized by a node.
9. What are the values of the following characteristics of the 10Base-5 standard?
· Nominal bandwidth (bps) – 10 Mbps
· Effective bandwidth (bps) – up to 9.76 Mpbs for maximum length frames
· Throughput (fps) – 14880 fps for minimal length frames
· Intrapacket transmission rate (bps) – 10 Mbps
· Interbit gap (sec) – 0.1 μs
10. Why was minimum frame size in the 10Base-5 standard set at 64 bytes?
To provide a stable collision detection for a network of maximum diameter
11. Why have 10Base-T and 10Base-FL/FB pushed coaxial Ethernet standards practically out of use in beginning of 90s?
They provide higher level of reliability and manageability of a network
12. Explain the meaning of each field of the Ethernet frame.
13. There are four standards for Ethernet frame format. From the list provided here, choose the names of these standards, taking into account that some standards have several names:
A. Novell 802.2
B. Ethernet II
D. Novell 802.3
E. Raw 802.3
F. Ethernet DIX
H. Ethernet SNAP
What may happen in a
network in which Ethernet frames of different formats
Some nodes might not be able to communicate
15. How does a packet size value influence network operation? What problems are related to too long frames? Why are short frames inefficient?
Too long frames monopolise a network and result in big transmission delays. Too short frames are not effective as they decrease a speed of user data transmission as a number of header bits gets comparable to a number of user data bits.
16. How does the utilization coefficient influence Ethernet network performance?
As soon as utilization coefficient exceeds a certain limit, queues become building up very fast, in non-linear way. A network gets stalled as a result.
17. How does the data transmission rate of a shared medium Ethernet network influence the maximum network diameter?
The higher the data transmission rate the lesser is a network maximum diameter.
18. What considerations influence the choice of the maximum length of the physical segment in the Ethernet standards?
To provide a reasonable power of a transmitter for the maximum length of the physical segment and maximum attenuation of medium given.
19. What allowed the maximum segment length to be increased during the switch from the FOIRL standard to 10Base-FL?
Due to increase of a transmitter power
20. What is the reason that caused the limitation known as the four hubs rule?
Every hub causes a delay in bit transmission.
21. Why is full-duplex Ethernet mode not supported in concentrators?
Concentrator assumes that only one node has an access to medium at any moment; whereas in full-duplex mode two nodes use a medium at the same time.
P1. Do the network fragments shown here represent collision domains?
Fig. 12.18. Possible collision domains
Yes, all of them but domain e)
long can the workstation wait until its frame will be discarded by the
A network adapter discards a frame when 16 consecutive attempts to transfer the frame resulted in collision. An interval between successive collisions consists of frame transmission period until collision detection moment plus random pause time until the next attempt to transfer the frame. In the worst case scenario an adapter detects a collision in the end of 576-bit period, i.e. after transmission of a frame during 57.6 μs. As an adapter makes 16 of such attempts this component of waiting time is 57.6 õ 16 = 921.6 μs. Again in the worst case scenario an adapter chooses a value of random pause of 2N õ 51.2 for its first 10 attempts of frame transmission; this gives us 51.2 õ (210+1 – 2) = 104755.2 μs. The each of the rest of 5 pauses will equal to the maximum value of 51.2 õ 210 = 52428.8 μs, so that the total amount of waiting random pauses is equal to 5 x 52428.8 = 262144 μs. Therefore a total maximum waiting time is 104755.2 + 262144 = 367820 μs.
P3. What will happen in a network built on concentrators if there are closed circuits (loops) in it, such as the ones shown in Fig. 12.19?
A. The network will operate normally.
B. Frames will not reach the destination node.
C. Collisions will happen when attempting to transmit any frame - correct
D. Frames will get looped.
Fig. 12.19. Loops in an Ethernet network built on concentrators
P4. Evaluate the performance drop of an Ethernet network when transmitting a file of 240,000 bytes if the level of lost or corrupted frames increases from 0% to 3%. Network operation is illustrated in Fig. 12.20.
The file is being transmitted using the following protocols: Ethernet, IPX (network layer) and NCP (application layer of the file service). Header sizes of the protocol headers are as follows:
· Ethernet — 26 bytes (with a preamble and an FCS field)
· IPX — 30 bytes
· NCP — 20 bytes
The file is transmitted in 1,000 byte segments. Only NCP, operating according to the Idle Source method, restores lost or corrupted frames. Time-out for waiting for positive acknowledgments is fixed at 500 msec. (This is not the only mode of NCP operation: It can also operate using the Sliding Window algorithm, but in this case, this mode is not used). Acknowledgment size is 10 bytes. Processing time of a single packet on the client side is 650 msec; on the server side, it is 50 msec.
Fig. 12.20. Ethernet network operation in the course of file transmission
Tip: The problem comprises two parts. First, it is necessary to determine the actual rate of file transmission under conditions of ideal network operation, when the percentage of lost or corrupted Ethernet frames is zero. The second part of the problem requires determining the file transmission rate when frames begin to get lost or corrupted.
File transmission will require 240 packets in total. The size of an Ethernet frame carrying 1,000 bytes of the file being transmitted will be 1,000 + 20 + 30 + 26 = 1,076 bytes, or 8,608 bits.
The size of the Ethernet frame carrying the acknowledgment is 86 bytes (with the preamble), or 688 bits.
Under these conditions, the time of a single cycle of transmission of the next part of the file in ideal network will be 860.8 + 68.8 +650 + 50 = 1,629.6 msec.
Time required to transmit a file of 240,000 bytes will be 240 ´ 1,629.6 = 0.391 sec, and the information rate will be 240,000/0.391 = 613,810 bps.
Now, it remains to find the information rate when frames start to get lost or distorted.