1. What is the difference between a characteristic and a requirement?
2. What characteristics are included within the concept quality of service (QoS), in a broad sense?
1. What QoS characteristics are of interest only to end users? Only to providers? Both to service providers and to users?
2. What are QoS characteristics, in the narrow sense of this term?
In the narrow sense, QoS characteristics reflect the negative effect of the queuing mechanism on traffic transmission, such as the temporary decrease of traffic delivery speed, variable delays in packet delivery, and packet loss because of overloading of the switch buffers.
3. What performance characteristics are of interest only to service providers?
4. What parties conclude Service Level Agreements?
5. Suggest a set of characteristics that you would like to include into a SLA, provided that you need to transmit the traffic of an IP telephony application through the network.
Maximum delay, maximum delay variation, maximum packet loss.
6. What type of information representation is used for the results of measuring packet delays?
7. What is the advantage of using characteristics such as coefficient
Coefficient of variation is a dimensionless quantity and therefore it charachterises the degree of delay variation in more universal way, without taking into account a range of delays. For example, delays of bus arrivals are in minutes range and packet delays are in milliseconds range but both can be described by coefficient of variation and the same value of the characteristic means that both processes are equal in terms in delays variation.
8. Which component is not taken into account when defining Round Trip Time?
Round Trip Time (RTT) is the net time required for data transportation from the source node to the destination node and back again, without accounting for the time required by the destination node for generating a response.
9. Is it possible to transmit traffic with long delays but without jitter?
Yes, when all delays are long and equal.
10. List the burst parameters. Are these parameters independent?
Traffic burst coefficient is a ratio between the traffic average rate measured on a long time interval and the maximum rate measured at some small time period. Peak Information Rate (PIR) is the maximum rate the user traffic flow is allowed to reach during the agreed, short time period T. This period is usually called the burst period. Burst size (usually designated B) is used for evaluating the switch buffer volume required for temporary data storage during congestion periods. Burst size is equal to the total volume of data that arrives to the switch during the allowed period of peak load. The three latter parameters are dependent:
B = PIR ´ T
11. Does average flow rate depend on packet delays?
12. Which characteristic of the transport service reliability is used in the short-term range, and which is used in the medium-term range?
In the short-term range it is ‘a ratio of lost packets to a total amount of packets sent’; in medium-range term ‘a service availability’ is used.
13. Describe two main approaches to ensuring network reliability.
First approach is a use of reliable network elements, such as links, routers, switches.
Second approach exploits redundancy of network elements so that in case of failure of an element a redundant element can carry on the necessary functionality.
14. How many methods of using alternate routes are available for improving the reliability of traffic transmission? What are their advantages and drawbacks?
There are three basic methods of using alternate routes:
1) The network determines the alternative route only after the failure of the main route.
2) The network finds two routes beforehand and uses them both.
3) The network finds two routes beforehand, but uses only one of them. When the main route fails, the transition to the alternative goes faster than when the first method is used.
15. What are two components of information security?
Computer security and network security
16. What is the difference between scalability and extensibility?
1. Two switches are connected by two physical links to improve reliability (Fig.6.7). Evaluate the volume of lost data in case of link failure for two variants of using these links as alternative routes, according to method 2, "Network finds two routes beforehand and uses them," and according to method 3, "Network finds two routes beforehand, but uses only one." The length of each link is 5,000 km, the data transmission rate is 155 Mbps, and the signal propagation speed in the link is 200,000 km/sec. In both cases, the S2 switch detects the link failure and switches to the backup link within 10 ms.
Fig. 6.7. Alternative routes
In the first case the loss of information will take place during 10 ms period when the switch S2 receives distorted information from the failed link. Loss of information is equal to: 155 x106 x 10 x 10 -3 = 1 550 000 bits.
In the second case the loss of information will take place during the same 10 ms period plus during next 25 ms that are needed to propagate the correct information along a redundant link. Loss of information is equal to: 155 x106 x 35 x 10 -3 = 5 425 000 bits.
2. Evaluate the link utilization coefficient if the data are transmitted in it using the protocol based on the idle source algorithm. The transmission rate is equal to 100 Mbps, the Round Trip Time (RTT) is 10 ms, and packets are not lost and do not get corrupted. Packet size is fixed and is equal to 1,500 bytes. Acknowledgment size can be neglected.
The time of packet transmission is: (1500ő8 /100 x 10 +6)= 12000/10+8=0.12 ms. The summary period of packet transmission plus interpacket gap is: 10 + 0,12 = 10.12 ms. Thus the link utilization is equal to 0.12 0.12/10.12 = 0.012
3. Determine the minimum size of the window that allows the transmission of packets using the link without the source running idle. The transmission rate is 100 Mbps, the RTT time is 10 ms, and packet are not lost and do not get corrupted. Packet size is fixed and equal to 1,500 bytes. Acknowledgment size can be neglected.
The transmitter will work without idle periods if it receives the acknowledgement of the first packet of the window by the moment when the transmission of the last packet of the window has finished.
The period of time for receiving of the first packet acknolodgement (T first ack) consists of four components (see the figure below):
- transmitting the first packet (T packet): 1500 x 8/(100 x 1000000)= 0,12 x 10-3 (s)= 0.12 (ms);
- transferring the first packet across the network: half RTT = 5 ms;
- receiving the first packet: the same time as transmitting, i.e. 0.12 ms;
- transferring the acknowledgement across the network; half RTT = 5 ms.
Thus T first ack = 10.24 (ms)
The period of time for transmission of all the window packets is equal to: W x T packet = 0.12 x W.
The condition of having transmitter without idle periods is:
0.12 x W > 10.24 or W > 85.33.
The nearest integer is 85 which is the answer.