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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,
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
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A link of capacity 100 Mbps is carrying traffic from a number of sourc...
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 on-time (that is transmitting)..
10 Mbps * n-station 100 Mbps
n-station = 10..
In the next part of the 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'll calculate expected value of bandwidth usage (if more data comes we store in buffer and due to expectation,
the buffer will be emptied soon):
E = 1/3 * 10 + 1/3 * 10 +...n-station times 100 Mbps [ total time is (1+2) = 3 then on time is 1 so 1/3 of BW]
=> 1/3 * 10 * n-station 100 Mbps
=> n-station = 30
so, option (A)
Since data should not be lost, we calculate for the extreme case when all sources are on-time (that is transmitting)..
10 Mbps * n-station 100 Mbps
n-station = 10..
In the next part of the 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'll calculate expected value of bandwidth usage (if more data comes we store in buffer and due to expectation,
the buffer will be emptied soon):
E = 1/3 * 10 + 1/3 * 10 +...n-station times 100 Mbps [ total time is (1+2) = 3 then on time is 1 so 1/3 of BW]
=> 1/3 * 10 * n-station 100 Mbps
=> n-station = 30
so, option (A)
Most Upvoted Answer
A link of capacity 100 Mbps is carrying traffic from a number of sourc...
Explanation:
Given:
- Link capacity = 100 Mbps
- Rate of traffic when a source is on = 10 Mbps
- Duty cycle (on-time : off-time) = 1 : 2
Minimum number of sources that can be multiplexed (S1):
To avoid wasting link capacity and ensure no data loss, we need to consider the scenario when all sources are on simultaneously. In this case, the total traffic rate should not exceed the link capacity.
Let's assume the number of sources that can be multiplexed is S1.
- On-time traffic rate of each source = 10 Mbps
- Total on-time traffic rate = S1 * 10 Mbps
Since the duty cycle is 1 : 2, the off-time traffic rate for each source is zero. Therefore, the total off-time traffic rate is also zero.
The total traffic rate is the sum of the on-time and off-time traffic rates. It should not exceed the link capacity:
Total on-time traffic rate + Total off-time traffic rate ≤ Link capacity
S1 * 10 Mbps + 0 Mbps ≤ 100 Mbps
S1 * 10 Mbps ≤ 100 Mbps
S1 ≤ 10
Therefore, the minimum number of sources that can be multiplexed (S1) without wasting link capacity and without data loss is 10.
Maximum number of sources that can be multiplexed (S2):
When there is a large buffer at the link, it can store the excess traffic during the off-time of each source. Therefore, the maximum number of sources that can be multiplexed is determined by the buffer capacity rather than the link capacity.
Since the duty cycle is 1 : 2, the average traffic rate for each source is 1/3 of the on-time traffic rate.
- Average traffic rate of each source = (1/3) * 10 Mbps = (10/3) Mbps
To ensure no data loss, the sum of the average traffic rates should not exceed the link capacity:
Total average traffic rate ≤ Link capacity
S2 * (10/3) Mbps ≤ 100 Mbps
S2 ≤ 30
Therefore, the maximum number of sources that can be multiplexed (S2) without data loss is 30.
Hence, the correct answer is option 'A' - 10 and 30.
Given:
- Link capacity = 100 Mbps
- Rate of traffic when a source is on = 10 Mbps
- Duty cycle (on-time : off-time) = 1 : 2
Minimum number of sources that can be multiplexed (S1):
To avoid wasting link capacity and ensure no data loss, we need to consider the scenario when all sources are on simultaneously. In this case, the total traffic rate should not exceed the link capacity.
Let's assume the number of sources that can be multiplexed is S1.
- On-time traffic rate of each source = 10 Mbps
- Total on-time traffic rate = S1 * 10 Mbps
Since the duty cycle is 1 : 2, the off-time traffic rate for each source is zero. Therefore, the total off-time traffic rate is also zero.
The total traffic rate is the sum of the on-time and off-time traffic rates. It should not exceed the link capacity:
Total on-time traffic rate + Total off-time traffic rate ≤ Link capacity
S1 * 10 Mbps + 0 Mbps ≤ 100 Mbps
S1 * 10 Mbps ≤ 100 Mbps
S1 ≤ 10
Therefore, the minimum number of sources that can be multiplexed (S1) without wasting link capacity and without data loss is 10.
Maximum number of sources that can be multiplexed (S2):
When there is a large buffer at the link, it can store the excess traffic during the off-time of each source. Therefore, the maximum number of sources that can be multiplexed is determined by the buffer capacity rather than the link capacity.
Since the duty cycle is 1 : 2, the average traffic rate for each source is 1/3 of the on-time traffic rate.
- Average traffic rate of each source = (1/3) * 10 Mbps = (10/3) Mbps
To ensure no data loss, the sum of the average traffic rates should not exceed the link capacity:
Total average traffic rate ≤ Link capacity
S2 * (10/3) Mbps ≤ 100 Mbps
S2 ≤ 30
Therefore, the maximum number of sources that can be multiplexed (S2) without data loss is 30.
Hence, the correct answer is option 'A' - 10 and 30.
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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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer?
Question Description
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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer?.
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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer?.
Solutions for 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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Computer Science Engineering (CSE).
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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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer?, a detailed solution for 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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of 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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 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 30b)12 and 25c)5 and 33d)15 and 22Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
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