Skip to content
Related Articles

Related Articles

Improve Article
Minimum length of Token Ring
  • Difficulty Level : Medium
  • Last Updated : 01 Jun, 2021

Prerequisite – Token Ring frame format, Efficiency Of Token Ring, Problems with Token Ring 
As we have already familiar with the Token Ring. Now, in this article, we are going to discuss What should be the minimum length of the Token Ring. 

In the worst case, it may be possible that all the system will Shut Down and the only monitor will remain active. Therefore the Token Ring should be able to hold one Token completely to avoid the collision. 
We know that the Size of Token = 24 bits. 

 

 



Hence, the minimum length of the Token Ring 
= Length of wire that can hold 24 bits 

Let’s see how to calculate the length of wire that can hold 24 bits:  

Capacity of wire = Propagation Delay * Bandwidth
Capacity >= 24
Propagation Delay * Bandwidth >= 24

Propagation Delay = Length of wire / velocity

(length of wire/velocity) * Bandwidth >= 24
Length of wire >= ( 24 * velocity ) / Bandwidth 

Hence, size of the ring must be greater than (24 * velocity) / Bandwidth. 

Example: Given,  

Bandwidth = 4 Mbps, 
Velocity = 2 * 108 

What should be the minimum length of the Ring? 

Explanation: 

length >= ( 24 * velocity ) / Bandwidth
length >= (24 * 2 * 10^8) / (4 * 10^6)
length >= 1200 meter 

In case, if we don’t have this much amount of wire, then we can use Delay device
Let’s assume for the given example we have only 1000 meter of wire then Delay will be equal to 200 meter. To convert the meter into bits: 

Meter --/velocity--> Second --*Bandwidth--> bit 

So, the bit delay we require for 200 meter is 

= (200/(2 * 108)) * (4 * 106) 
= 4 bit time 

 

Hence, a Delay device of 4 bit is required for working the ring without collision, in 1000 meter of wire where 1200 meter is required.
 

Attention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :