Why Choose Us?
0% AI Guarantee
Human-written only.
24/7 Support
Anytime, anywhere.
Plagiarism Free
100% Original.
Expert Tutors
Masters & PhDs.
100% Confidential
Your privacy matters.
On-Time Delivery
Never miss a deadline.
A group of stations share a 16 kbps slotted Aloha channel
A group of stations share a 16 kbps slotted Aloha channel. Each station outputs a 80 Bytes frame and buffers any outgoing frames until they can be sent. Collectively (recall that this means original frames plus resending of frames that suffered a collision previously) the stations attempt to send 30 frames/sec. What is the throughput in frames/sec?
Expert Solution
Available bandwidth = 16kbps = (16/8) kBps = 2 kBps
Size of each frame = 80B
Since a frame is synonymous with a time slot in slotted Aloha protocol,
number of available slots per second = 2kBps / 80B = (2*1024)/80 = 25.6
Stations attempt to send 30 frames/sec,
so average number of frames generated per slot (G) = 30/25.6 = 1.171875
average number of frames sent successfully per slot (S) = G / e^G
= 1.171875 / e^1.171875
= 0.3631 (rounded off to 4th place after decimal)
e is the mathematical constant 2.718 (base in natural logarithm) in the formula used above.
Hence throughput in frames/sec = S * 25.6 = 9.3 (rounded off to one place after decimal)
Archived Solution
You have full access to this solution. To save a copy with all formatting and attachments, use the button below.
For ready-to-submit work, please order a fresh solution below.
