LIVE Course for free

Rated by 1 million+ students
Get app now
JEE MAIN 2023
JEE MAIN 2023 TEST SERIES
NEET 2023 TEST SERIES
NEET 2023
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
77 views
in Computer by (30.0k points)
closed by
A network has a data transmission bandwidth of 20 × 106 bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is ______bytes.

1 Answer

0 votes
by (48.1k points)
selected by
 
Best answer

Data:

L = size of frame

BW → bandwidth = 20 × 106 b/s

Tp → propagation time = 40 μs = 40 × 10-6 s

Tt → transmission time

Formula:

For CSMA/CD,

Tt ≥ 2 × Tp

\({{\rm{T}}_{\rm{t}}} = \frac{{\rm{L}}}{{{\rm{BW}}}}\)

Calculation:

\(\frac{{\rm{L}}}{{{\rm{BW}}}} \ge 2 \times 40 \times {10^{ - 6}}\)

L ≥ 2 × 40 × 10-6 × 20 × 106

L ≥ 1600 bits

L ≥ 200 bytes

Lmin = 200 bytes

Therefore, the minimum size of a frame in the network is 200 bytes. 

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...