Question 9668 – Linked-List
March 13, 2024Computer-Networks
March 14, 2024Access-Control-Methods
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Question 5
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A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200 Kbps bandwidth. Find the throughput of the system, if the system (all stations put together) produces 250 frames per second :
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49
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368
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149
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151
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Question 5 Explanation:
Step-1:
Given data,
Slotted ALOHA transmits = 200 bit frames
Bandwidth = 200 Kbps
System produces = 250 frames per second
Throughput = ?
Step-2:
Pure ALOHA formula S = G * e-2G
Slotted ALOHA formula S = G * e-G
Frame transmission time = 200/200 kbps = 1ms
Here,
G = ¼ and S = G * e-G
= 0.195 (or) 19.5%
System produces = 250 * 0.195
= 48.75
Given data,
Slotted ALOHA transmits = 200 bit frames
Bandwidth = 200 Kbps
System produces = 250 frames per second
Throughput = ?
Step-2:
Pure ALOHA formula S = G * e-2G
Slotted ALOHA formula S = G * e-G
Frame transmission time = 200/200 kbps = 1ms
Here,
G = ¼ and S = G * e-G
= 0.195 (or) 19.5%
System produces = 250 * 0.195
= 48.75
Correct Answer: A
Question 5 Explanation:
Step-1:
Given data,
Slotted ALOHA transmits = 200 bit frames
Bandwidth = 200 Kbps
System produces = 250 frames per second
Throughput = ?
Step-2:
Pure ALOHA formula S = G * e-2G
Slotted ALOHA formula S = G * e-G
Frame transmission time = 200/200 kbps = 1ms
Here,
G = ¼ and S = G * e-G
= 0.195 (or) 19.5%
System produces = 250 * 0.195
= 48.75
Given data,
Slotted ALOHA transmits = 200 bit frames
Bandwidth = 200 Kbps
System produces = 250 frames per second
Throughput = ?
Step-2:
Pure ALOHA formula S = G * e-2G
Slotted ALOHA formula S = G * e-G
Frame transmission time = 200/200 kbps = 1ms
Here,
G = ¼ and S = G * e-G
= 0.195 (or) 19.5%
System produces = 250 * 0.195
= 48.75