Ethernet
September 8, 2024Switching
September 9, 2024Computer-Networks
| Question 59 |
A link of capacity 100 Mbps is carrying traffic from a number of sources. Each source generates an on-off traffic stream; when the source is on, the rate of traffic is 10 Mbps, and when the source is off, the rate of traffic is zero. The duty cycle, which is the ratio of on-time to off-time, is 1:2. When there is no buffer at the link, the minimum number of sources that can be multiplexed on the link so that link capacity is not wasted and no data loss occurs is S1. Assuming that all sources are synchronized and that the link is provided with a large buffer, the maximum number of sources that can be multiplexed so that no data loss occurs is S2. The values of S1 and S2 are, respectively,
| 10 and 30 | |
| 12 and 25 | |
| 5 and 33 | |
| 15 and 22 |
Question 59 Explanation:
For S1,
Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.
Since data should not be lost, we calculate for the extreme case when all sources are ontime (that is transmitting).
10Mbps × n-station ≤ 100Mbps
n = 10 = S1
In the next part of question, it is given that the link is provided with large buffer and we are asked to find out large no. of stations.
For that we calculate expected value of bandwidth usage,
E = 1/3 × 10 + 1/3 × 10 + …….+ ……. n-station times ≤ 100Mbps
⇒ 1/3 × 10 × n-station ≤ 100Mbps
⇒ n-station = 30 = S2
So option (A) is answer.
Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.
Since data should not be lost, we calculate for the extreme case when all sources are ontime (that is transmitting).
10Mbps × n-station ≤ 100Mbps
n = 10 = S1
In the next part of question, it is given that the link is provided with large buffer and we are asked to find out large no. of stations.
For that we calculate expected value of bandwidth usage,
E = 1/3 × 10 + 1/3 × 10 + …….+ ……. n-station times ≤ 100Mbps
⇒ 1/3 × 10 × n-station ≤ 100Mbps
⇒ n-station = 30 = S2
So option (A) is answer.
Correct Answer: A
Question 59 Explanation:
For S1,
Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.
Since data should not be lost, we calculate for the extreme case when all sources are ontime (that is transmitting).
10Mbps × n-station ≤ 100Mbps
n = 10 = S1
In the next part of question, it is given that the link is provided with large buffer and we are asked to find out large no. of stations.
For that we calculate expected value of bandwidth usage,
E = 1/3 × 10 + 1/3 × 10 + …….+ ……. n-station times ≤ 100Mbps
⇒ 1/3 × 10 × n-station ≤ 100Mbps
⇒ n-station = 30 = S2
So option (A) is answer.
Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.
Since data should not be lost, we calculate for the extreme case when all sources are ontime (that is transmitting).
10Mbps × n-station ≤ 100Mbps
n = 10 = S1
In the next part of question, it is given that the link is provided with large buffer and we are asked to find out large no. of stations.
For that we calculate expected value of bandwidth usage,
E = 1/3 × 10 + 1/3 × 10 + …….+ ……. n-station times ≤ 100Mbps
⇒ 1/3 × 10 × n-station ≤ 100Mbps
⇒ n-station = 30 = S2
So option (A) is answer.