A link of capacity 100 Mbps is carrying traffic from a number of sources. Each…
2006
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,
- A.
10 and 30
- B.
12 and 25
- C.
5 and 33
- D.
15 and 22
Attempted by 49 students.
Show answer & explanation
Correct answer: A
Key facts and notation: each source sends 10 Mbps when on; duty cycle (on:off) = 1:2 so fraction of time on = 1/3. Link capacity = 100 Mbps.
S1 (no buffer, sources synchronized):
When synchronized with no buffer, all sources are on simultaneously during their on-period. To avoid both data loss and unused capacity the instantaneous aggregate on-rate must equal the link capacity: number_of_sources × 10 Mbps = 100 Mbps.
Solve: number_of_sources = 100 ÷ 10 = 10. Therefore S1 = 10.
S2 (large buffer, sources may be asynchronous or buffering absorbs bursts):
With a sufficiently large buffer, what matters is the long-term average input rate. Average rate per source = 10 Mbps × (1/3) = 10/3 Mbps.
Maximum number of sources allowed without long-term loss solves: number_of_sources × (10/3) ≤ 100. So number_of_sources ≤ 100 ÷ (10/3) = 30.
Therefore S2 = 30.
Final answer: S1 = 10 and S2 = 30.