Consider a slotted ring of length 20km with a data rate of


1. Consider a slotted ring of a slotted ring of length 10km with a data rate of 10 mbps and 500 repeaters. each of which introduces a 1-bit dealy. Each slot contains room for one source address byte, one destination address byte, two data bytes, and five control bits for a total length of 37 bits. How many slots are on the ring?

Solution would be: Tprop = 10x10^3/2x10^8 + 500/10^7

My Question is: How exactly did this come out to 100 x 10^-6 sec? This was never explained.

The rest would be: 100 x 10^-6 x 10^7bps = 1000bits.

1000/37 = 27.03 =

Answer:

27 slots

The 100 x10^-6 sec part really gets me.

Again here:

Consider a slotted ring of length 20km with a data rate of 100Mbps and 1000 repeaters, each of which introduces a 1-bit delay. Each slot contains room for one source address byte, one destination address byte, two data bytes and 8 control bits for a total length of 40 bits. Comes out to:

20 x 10^3/2x10^8 + 1000/10^8 = 1.1 x 10^-4

Please explain how did that come out to 1.1 x 10^4.

For the first question, why is one 10x10^3/2x10^8 + 500/10^7? How did it get to 10^7?

Second Question: 20x10^3/2x10^8 + 1000/10^8? How did it get to 10^8?

Solution Preview :

Prepared by a verified Expert
Business Management: Consider a slotted ring of length 20km with a data rate of
Reference No:- TGS01713231

Now Priced at $30 (50% Discount)

Recommended (99%)

Rated (4.3/5)