GATE Exam  >  GATE Questions  >  A sender uses the Stop - and - Wait ARQ proto... Start Learning for Free
A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.
Assuming no frame islost,the sender through put is ___________ bytes/second.
  • a)
    2400.0 : 2400.0
  • b)
    2500.0 : 2500.0
  • c)
    2600.0 : 2600.0
  • d)
    2700.0 : 2700.0
Correct answer is '2500'. Can you explain this answer?
Verified Answer
A sender uses the Stop - and - Wait ARQ protocol for reliable transmis...
B = 80 kbps
L = 1000 bytes
Tp = 100 ms
Tx = L/B = 100 ms
Tax = ack size/ bandwidth = 100 ms
Efficiency = tx/(tx +2tp+tax)
Throughput = efficiency * bandwidth = .25 *104 bytes
= 2500 bytes
View all questions of this test
Most Upvoted Answer
A sender uses the Stop - and - Wait ARQ protocol for reliable transmis...
Calculation of Frame Transmission Time
- Size of frame = 1000 bytes
- Transmission rate at sender = 80 Kbps
- Transmission rate = (80 * 1000) bits/second = 80,000 bits/second
- Time to transmit one frame = (1000 * 8) / 80000 = 0.1 seconds

Calculation of Acknowledgement Transmission Time
- Size of acknowledgement = 100 bytes
- Transmission rate at receiver = 8 Kbps
- Transmission rate = (8 * 1000) bits/second = 8000 bits/second
- Time to transmit acknowledgement = (100 * 8) / 8000 = 0.1 seconds

Calculation of Round Trip Time (RTT)
- One-way propagation delay = 100 milliseconds
- Round trip time = 2 * propagation delay = 2 * 100 milliseconds = 0.2 seconds

Calculation of Throughput
- Throughput = (number of frames transmitted successfully) / (total time taken)
- In Stop-and-Wait ARQ protocol, only one frame is transmitted at a time and the sender waits for an acknowledgement before transmitting the next frame
- Therefore, the maximum number of frames that can be transmitted in one RTT is 1
- Total time taken for one RTT = frame transmission time + acknowledgement transmission time + RTT
- Total time taken for one RTT = 0.1 + 0.1 + 0.2 = 0.4 seconds
- Throughput = (1 * 1000) / 0.4 = 2500 bytes/second

Therefore, the correct answer is option (b) 2500.0: 2500.0
Explore Courses for GATE exam

Similar GATE Doubts

A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer?
Question Description
A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer?.
Solutions for A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE. Download more important topics, notes, lectures and mock test series for GATE Exam by signing up for free.
Here you can find the meaning of A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer?, a detailed solution for A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer? has been provided alongside types of A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice A sender uses the Stop - and - Wait ARQ protocol for reliable transmission of frames.Frames are of size 1000 bytes and the transmission rate at the sender is 80 Kbps (1Kbps=1000 bits/second). Size of an acknowledgement is100 bytes and the transmission rate at the receiver is 8 Kbps.The one-waypropagation delayis100milliseconds.Assuming no frame islost,the sender through put is ___________ bytes/second.a)2400.0 : 2400.0b)2500.0 : 2500.0c)2600.0 : 2600.0d)2700.0 : 2700.0Correct answer is '2500'. Can you explain this answer? tests, examples and also practice GATE tests.
Explore Courses for GATE exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev