A data message of 10 ms duration having 4800 bits crosses 9 nodes (10 hops) to reach its destination. The data rate and total delay for circuit switched connection (assuming node delays as 1 ms) are respectively.

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Question

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ESE Electronics Prelims 2021 Official Paper

Answer 1

Concept:

$D a t a r a t e = \frac{T o t a l d a t a}{D a t a d u r a t i o n}$

Delay = No. of hops × node delay + Data duration

Calculation:

Given : Data = 4800 bits

Data duration = 10 msec

No. of hops = 10

Node delay = 1 msec.

$D a t a r a t e = \frac{4000 b i t s}{10 \times 10^{- 3} s e c} = 480 k b p s$

Delay = 10 × 1 + 10 = 20 msec