Question 9668 – Linked-List
March 13, 2024Computer-Networks
March 14, 2024Access-Control-Methods
| Question 5 |
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 :
| 49 | |
| 368 | |
| 149 | |
| 151 |
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