UGC NET CSE | June 2016 | Part 3 | Question: 27

Question

A Pure ALOHA Network transmits 200 bit frames using a shared channel with 200 kbps bandwidth . If the system  (all stations put together) produces 500 frames per sec , then the throughput of the system is

• A. 0.384
• B. 0.184
• C. 0.286
• D. 0.586

Answer 1

The throughput for pure ALOHA is S = G x e^-2G , where G is the Load factor and take e = 2.71

The throughput is maximum , when G = 1/2 , which is S_max = 0.184 .

Transmission time of frame = 200 bits / 200 kbps = 1ms 

Now, system as taken altogether produces 500 frames per sec i.e , 1/2 frame per ms which is load factor of transmission.

Now, as load factor G = 1/2 , then throughput is maximum for pure aloha .

Hence, option B) is correct.

Comments

I am confused about G, I have read that G represents no. of stations who want to transmit frames in one time slot, but here G is taken as load factor. I am not able to understand ....what is load factor in aloha ?

---

http://bbcr.uwaterloo.ca/~lcai/ece418/5-2.pdf

Number of transmission attempts per X sec, where X is Transmission time.

Answer 2

Ans: B

ref: https://www.slideshare.net/WayneJonesJnr/ch12-3361657