Question

In this problem we consider sending real-time voice from Host A to Host B over a packet-switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-byte packets. There is one link between Hosts A and B; its transmission rate is 4 Mbps and its propagation delay is 18 msec. As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packet's bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decodes (as part of the analog signal at Host B)?

Answer 1

Answer:

The total time elapsed from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B is 25.11 ms

Explanation:

Host A first converts the analog signal to a digital 64kbps stream and then groups it into 56-byte packets. The time taken for this can be calculated as:

time taken 1= $\frac{Packet Size in Bits}{Bit Rate}$

                 = (56 x 8) bits / 64 x 10³ bits/s

                 = 7 x 10⁻³s

time taken 1= 7 ms

The transmission rate of the packet from Host A to Host B is 4 Mbps. The time taken to transfer the packets can be calculated as:

time taken 2= (56 x 8) bits / 4 x 10⁶ bits/s

                    = 1.12 x 10⁻⁴ s

time taken 2= 112 μs

The propagation delay is 18 ms.

To calculate the total time elapsed, we need to add up all the time taken at each individual stage.

$Time_{total}$ = Time taken 1 + Time taken 2 + Propagation Delay

                 = 7 ms + 112 μs + 18 ms

                 = 0.025112 s

$Time_{total}$ = 25.11 ms