ISRO CS-2023
February 7, 2024Question 15127 – DSSSB TGT 2021
February 8, 2024Computer-Networks
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Question 47
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A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1 km cable with no repeaters. The signal speed in the cable is 2 × 108 m/sec. The minimum frame size for this network should be
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10000 bits
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10000 bytes
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5000 bits
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5000 bytes
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Question 47 Explanation:
For CSMA/CD protocol we know that minimum frame size required is,
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
10000 bits.
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
10000 bits.
Correct Answer: A
Question 47 Explanation:
For CSMA/CD protocol we know that minimum frame size required is,
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
10000 bits.
L ≤ 2×Tp×B
L ≤ 2×(d/v)×B
d = 1Km = 1000m
v = 2×103 m/s
B = 109 bps
By solving the above equation we will set the value of L as,
10000 bits.
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