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Consider a network path with three links.
Link 1 – Speed : 10 Gbps, Propagation Delay : 40ms
Link 2 – Speed : 2 Gbps, Propagation Delay : 5ms
Link 3 – Speed : 2 Gbps, Propagation Delay : 10ms
Assume processing delay to be zero.
A data packet of 500 KB is sent from a source to a destination via the path : link 1 – link 3 – link 2.
Link 1 – Speed : 10 Gbps, Propagation Delay : 40ms
Link 2 – Speed : 2 Gbps, Propagation Delay : 5ms
Link 3 – Speed : 2 Gbps, Propagation Delay : 10ms
Assume processing delay to be zero.
A data packet of 500 KB is sent from a source to a destination via the path : link 1 – link 3 – link 2.
Q. What is the total delay in transmitting the packet?
- a)54.90 ms
- b)50.20 ms
- c)55.55 ms
- d)59.40 ms
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Consider a network path with three links.Link 1 – Speed : 10 Gbp...
Total delay = Propagation delay + Transmission Delay + Queuing delay + Processing delay
Total delay = 40ms + 10ms + 5ms + 500KB X (1/10Gbps + 1/2Gbps + 1/2Gbps)
Total delay = 59.40 ms
NOTE : Here, ‘B’ is bytes (in data) and ‘b’ is bits (in speed)
Total delay = 40ms + 10ms + 5ms + 500KB X (1/10Gbps + 1/2Gbps + 1/2Gbps)
Total delay = 59.40 ms
NOTE : Here, ‘B’ is bytes (in data) and ‘b’ is bits (in speed)
Most Upvoted Answer
Consider a network path with three links.Link 1 – Speed : 10 Gbp...
Given information:
- Link 1: Speed = 10 Gbps, Propagation Delay = 40 ms
- Link 2: Speed = 2 Gbps, Propagation Delay = 5 ms
- Link 3: Speed = 2 Gbps, Propagation Delay = 10 ms
- Data packet size = 500 KB
To calculate the total delay in transmitting the packet, we need to consider the propagation delay and transmission delay for each link.
Transmission Delay:
The transmission delay is the time taken to transmit the entire packet over a link, which is calculated using the formula:
Transmission Delay = Packet Size / Link Speed
For Link 1:
Transmission Delay = 500 KB / 10 Gbps
= (500 * 8) / (10 * 10^9) seconds
= 0.00004 seconds
For Link 3:
Transmission Delay = 500 KB / 2 Gbps
= (500 * 8) / (2 * 10^9) seconds
= 0.002 seconds
For Link 2:
Transmission Delay = 500 KB / 2 Gbps
= (500 * 8) / (2 * 10^9) seconds
= 0.002 seconds
Propagation Delay:
The propagation delay is the time taken for a bit to travel from the source to the destination over a link, which is given in the problem.
For Link 1: Propagation Delay = 40 ms
For Link 3: Propagation Delay = 10 ms
For Link 2: Propagation Delay = 5 ms
Total Delay:
The total delay is the sum of the transmission delay and propagation delay for each link.
For Link 1: Total Delay = Transmission Delay + Propagation Delay
= 0.00004 seconds + 40 ms
= 40.00004 ms
For Link 3: Total Delay = Transmission Delay + Propagation Delay
= 0.002 seconds + 10 ms
= 10.002 ms
For Link 2: Total Delay = Transmission Delay + Propagation Delay
= 0.002 seconds + 5 ms
= 5.002 ms
The total delay in transmitting the packet is the sum of the total delays for each link.
Total Delay = 40.00004 ms + 10.002 ms + 5.002 ms
= 59.00404 ms
≈ 59.40 ms
Therefore, the correct answer is option 'D' - 59.40 ms.
- Link 1: Speed = 10 Gbps, Propagation Delay = 40 ms
- Link 2: Speed = 2 Gbps, Propagation Delay = 5 ms
- Link 3: Speed = 2 Gbps, Propagation Delay = 10 ms
- Data packet size = 500 KB
To calculate the total delay in transmitting the packet, we need to consider the propagation delay and transmission delay for each link.
Transmission Delay:
The transmission delay is the time taken to transmit the entire packet over a link, which is calculated using the formula:
Transmission Delay = Packet Size / Link Speed
For Link 1:
Transmission Delay = 500 KB / 10 Gbps
= (500 * 8) / (10 * 10^9) seconds
= 0.00004 seconds
For Link 3:
Transmission Delay = 500 KB / 2 Gbps
= (500 * 8) / (2 * 10^9) seconds
= 0.002 seconds
For Link 2:
Transmission Delay = 500 KB / 2 Gbps
= (500 * 8) / (2 * 10^9) seconds
= 0.002 seconds
Propagation Delay:
The propagation delay is the time taken for a bit to travel from the source to the destination over a link, which is given in the problem.
For Link 1: Propagation Delay = 40 ms
For Link 3: Propagation Delay = 10 ms
For Link 2: Propagation Delay = 5 ms
Total Delay:
The total delay is the sum of the transmission delay and propagation delay for each link.
For Link 1: Total Delay = Transmission Delay + Propagation Delay
= 0.00004 seconds + 40 ms
= 40.00004 ms
For Link 3: Total Delay = Transmission Delay + Propagation Delay
= 0.002 seconds + 10 ms
= 10.002 ms
For Link 2: Total Delay = Transmission Delay + Propagation Delay
= 0.002 seconds + 5 ms
= 5.002 ms
The total delay in transmitting the packet is the sum of the total delays for each link.
Total Delay = 40.00004 ms + 10.002 ms + 5.002 ms
= 59.00404 ms
≈ 59.40 ms
Therefore, the correct answer is option 'D' - 59.40 ms.
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Consider a network path with three links.Link 1 – Speed : 10 Gbps, Propagation Delay : 40msLink 2 – Speed : 2 Gbps, Propagation Delay : 5msLink 3 – Speed : 2 Gbps, Propagation Delay : 10msAssume processing delay to be zero.A data packet of 500 KB is sent from a source to a destination via the path : link 1 – link 3 – link 2.Q. What is the total delay in transmitting the packet?a)54.90 msb)50.20 msc)55.55 msd)59.40 msCorrect answer is option 'D'. Can you explain this answer?
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Consider a network path with three links.Link 1 – Speed : 10 Gbps, Propagation Delay : 40msLink 2 – Speed : 2 Gbps, Propagation Delay : 5msLink 3 – Speed : 2 Gbps, Propagation Delay : 10msAssume processing delay to be zero.A data packet of 500 KB is sent from a source to a destination via the path : link 1 – link 3 – link 2.Q. What is the total delay in transmitting the packet?a)54.90 msb)50.20 msc)55.55 msd)59.40 msCorrect answer is option 'D'. 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 Consider a network path with three links.Link 1 – Speed : 10 Gbps, Propagation Delay : 40msLink 2 – Speed : 2 Gbps, Propagation Delay : 5msLink 3 – Speed : 2 Gbps, Propagation Delay : 10msAssume processing delay to be zero.A data packet of 500 KB is sent from a source to a destination via the path : link 1 – link 3 – link 2.Q. What is the total delay in transmitting the packet?a)54.90 msb)50.20 msc)55.55 msd)59.40 msCorrect answer is option 'D'. 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 Consider a network path with three links.Link 1 – Speed : 10 Gbps, Propagation Delay : 40msLink 2 – Speed : 2 Gbps, Propagation Delay : 5msLink 3 – Speed : 2 Gbps, Propagation Delay : 10msAssume processing delay to be zero.A data packet of 500 KB is sent from a source to a destination via the path : link 1 – link 3 – link 2.Q. What is the total delay in transmitting the packet?a)54.90 msb)50.20 msc)55.55 msd)59.40 msCorrect answer is option 'D'. Can you explain this answer?.
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ample number of questions to practice Consider a network path with three links.Link 1 – Speed : 10 Gbps, Propagation Delay : 40msLink 2 – Speed : 2 Gbps, Propagation Delay : 5msLink 3 – Speed : 2 Gbps, Propagation Delay : 10msAssume processing delay to be zero.A data packet of 500 KB is sent from a source to a destination via the path : link 1 – link 3 – link 2.Q. What is the total delay in transmitting the packet?a)54.90 msb)50.20 msc)55.55 msd)59.40 msCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice Computer Science Engineering (CSE) tests.
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