Super easy question. But I have a doubt

accoding to this problem:

Formula :

Formula for Queuing Delay : [NL+ (L- x)]/R

Where

N indicates the number of packets are already in the queue.

L is ength.

x indicates number of bits currently transmitted packet.

R is transmission rate.

From the question :

N=3, L=1000 , R=1 MBPS

Since half packets transmitted x=1000/2=500

Substitute coorosponding values in [NL+ (L- x)]/R.

=[3* 1000 + (1000-500)]/1 MBPS

=[3000 + 500]/1 MBPS

=[3500]/1 MBPS

Convert bytes to bits multiply with 8 on numerator and 1024 on denominator.

1 MBPS=1048576 bps

=[3500*8]/1048576 (MY QUESTION IS HOW DO YOU GET 1048576)

=28000/1048576

= 0.0267 seconds

=26.7 milli second

Queuing Delay is 26.7 milli second.

In the following example I got

A packet switch receives a packet and determines the outbound link to which the packet should be forwarded. When the packet arrives, one other packet is halfway (50%) done being transmitted on this outbound link and three other packets are waiting to be transmitted. Packets are transmitted in order of arrival. Suppose all packets are 800 bytes and the link rate is 2 Mbps. What is the queuing delay for the packet? More generally, what is the queuing delay when all packets have length L, the transmission rate is R, x bits of the currently-being-transmitted packet have been transmitted, and n packets are already in the queue?

[3*800+(800-400)]/2MBPS ( I WONDER IF THE N=3 IS BECAUSE OF THE "three other packets are waiting to be transmitted")

[2400+400]/2MBPS

[2800*8]/2097152

0.0106 SEC

10.6 MS

THE FINAL ANSWER SHOULD BE 11.2MS

i WONDER IF MY ANSWER IS CORRECT?