Supernetting
October 7, 2024Access-Control-Methods
October 8, 2024Access-Control-Methods
| Question 12 |
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 12 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=1⁄4 and S=G*e -G
=0.195 (or) 19.5%
System produces=250*0.195
=48.75
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=1⁄4 and S=G*e -G
=0.195 (or) 19.5%
System produces=250*0.195
=48.75
Correct Answer: A
Question 12 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=1⁄4 and S=G*e -G
=0.195 (or) 19.5%
System produces=250*0.195
=48.75
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=1⁄4 and S=G*e -G
=0.195 (or) 19.5%
System produces=250*0.195
=48.75