CHAPTER 24
Congestion Control and
Quality of Service
Solutions to Odd-Numbered Review Questions and ExercisesReview Questions
1. In congestion control, the load on a network is prevented from exceeding thecapacity. Quality of service refers to the characteristics that a flow of data seeks toattain. If there is good congestion control, then the QoS is also good and viceversa.
3. The average data rate is always less than or equal to the peak data rate.
5. Open-loop congestion control policies try to prevent congestion. Closed-loop con-gestion control policies try to alleviate the effects of congestion.
7. Congestion can be alleviated by back pressure, a choke point, and explicit signal-ing .
9. Frame Relay uses the BECN bit and the FECN bit to control congestion.
11. Scheduling , traffic shaping, admission control, and resource reservation can
improve QoS.
13. Differentiated Services was developed to handle the shortcomings of IntServ. The
main processing was moved from the core of the network to the edge of the net-work. Also, the per-flow service was changed to per-class service.
15. The attributes are access rate, committed burst size, committed information rate,
and excess burst size.
Exercises
17. The bit pattern is 10110000 0001011. The FECN bit is 0 and the BECN bit is 1.
There is no congestion in the forward direction, but there is congestion in the back-ward direction.
19.
Input: (100/60) × 12 + 0 × 48 = 20 gallons
Output: 5 gallons
Left in the bucket: 20 − 5 = 15
1
2
21.
a. The access rate is the rate of T-1 line (1.544 Mbps) that connects the user to thenetwork. Obviously, the user cannot exceed this rate.
b. The user data rate cannot exceed the access rate, the rate of the T-1 line thatconnects the user to the network. The user should stay below this rate (1.544Mbps ).
c. The CIR is 1 Mbps. This means that the user can send data at this rate all thetime without worrying about the discarding of data.
d. The user can send data at the rate of 1.2 Mbps because it is below the accessrate . However, the user sends 6 million bits per 5 seconds, which is above Bc (5million per 5 seconds), but below Bc +Be (6 million per 5 seconds). The networkwill discard no data if there is no congestion, but it may discard data if there iscongestion.
e. The user can send data at the rate of 1.4 Mbps because it is below the accessrate . However, the user sends 7 million bits per 5 seconds, which is above Bc and above Bc +Be (6 million per 5 seconds). In other words, the user rate isbeyond its share. The network will discard some data to limit the data rate. f. To be sure that the network never discard her data, the user should stay at orbelow CIR rate all the time, which means below or at 1 Mbps.
g. If the user can accept possible data discarding in case of congestion, she cansend at a higher rate if the number of bits is below Bc +Be (6 million per 5 sec-onds in this case). This mans that the user can send at 1.2 Mbps all the time ifshe accepts this risk.
23. CTD is the average cell transfer delay. If each cell takes 10 μs to reach the destina-
tion, we can say that CTD = [(10 μs × n ) / n ] in which n is the total number of cellstransmitted in a period of time. This means that CTD = 10 μs
CHAPTER 24
Congestion Control and
Quality of Service
Solutions to Odd-Numbered Review Questions and ExercisesReview Questions
1. In congestion control, the load on a network is prevented from exceeding thecapacity. Quality of service refers to the characteristics that a flow of data seeks toattain. If there is good congestion control, then the QoS is also good and viceversa.
3. The average data rate is always less than or equal to the peak data rate.
5. Open-loop congestion control policies try to prevent congestion. Closed-loop con-gestion control policies try to alleviate the effects of congestion.
7. Congestion can be alleviated by back pressure, a choke point, and explicit signal-ing .
9. Frame Relay uses the BECN bit and the FECN bit to control congestion.
11. Scheduling , traffic shaping, admission control, and resource reservation can
improve QoS.
13. Differentiated Services was developed to handle the shortcomings of IntServ. The
main processing was moved from the core of the network to the edge of the net-work. Also, the per-flow service was changed to per-class service.
15. The attributes are access rate, committed burst size, committed information rate,
and excess burst size.
Exercises
17. The bit pattern is 10110000 0001011. The FECN bit is 0 and the BECN bit is 1.
There is no congestion in the forward direction, but there is congestion in the back-ward direction.
19.
Input: (100/60) × 12 + 0 × 48 = 20 gallons
Output: 5 gallons
Left in the bucket: 20 − 5 = 15
1
2
21.
a. The access rate is the rate of T-1 line (1.544 Mbps) that connects the user to thenetwork. Obviously, the user cannot exceed this rate.
b. The user data rate cannot exceed the access rate, the rate of the T-1 line thatconnects the user to the network. The user should stay below this rate (1.544Mbps ).
c. The CIR is 1 Mbps. This means that the user can send data at this rate all thetime without worrying about the discarding of data.
d. The user can send data at the rate of 1.2 Mbps because it is below the accessrate . However, the user sends 6 million bits per 5 seconds, which is above Bc (5million per 5 seconds), but below Bc +Be (6 million per 5 seconds). The networkwill discard no data if there is no congestion, but it may discard data if there iscongestion.
e. The user can send data at the rate of 1.4 Mbps because it is below the accessrate . However, the user sends 7 million bits per 5 seconds, which is above Bc and above Bc +Be (6 million per 5 seconds). In other words, the user rate isbeyond its share. The network will discard some data to limit the data rate. f. To be sure that the network never discard her data, the user should stay at orbelow CIR rate all the time, which means below or at 1 Mbps.
g. If the user can accept possible data discarding in case of congestion, she cansend at a higher rate if the number of bits is below Bc +Be (6 million per 5 sec-onds in this case). This mans that the user can send at 1.2 Mbps all the time ifshe accepts this risk.
23. CTD is the average cell transfer delay. If each cell takes 10 μs to reach the destina-
tion, we can say that CTD = [(10 μs × n ) / n ] in which n is the total number of cellstransmitted in a period of time. This means that CTD = 10 μs