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In a token ring, if $a<1$, i.e  $\frac{T_p}{T_t} < 1$ , then will not the data get corrupted because it will come in contact with the data that is still being transmitted ?
in Computer Networks by Active (3.8k points) | 359 views

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yes . it will corrupt the data thats why if we want to apply such type of topologies bit delayer were introduced in between so that the data may not collide . so we always calculate first . whether a is less than 1 then we calculate the bit delayer required and then apply it in somewhere in the middle . it delays the data for some time.
by Boss (16.1k points)
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token ring is not in syllabus now right??
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yes . out of syllabus now .
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