# Minimum length of Token Ring

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 Shud 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 * 10^{8}

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 * 10^{8})) * (4 * 10^{6}) = 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.

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