Answer: 0.027 sec
Explanation:
Within the switch, waiting to be transmitted, we have 4 + ½ = 4.5 packets
1 packet = 1500 bytes
4.5 packets = 6750 bytes.
So, before any arriving packet can be delivered, it will wait for these 6750 bytes to be transmitted.
Data rate = 2 Mbps
1 byte = 8 bits
Data rate = 2.106 bits/sec = 250 KB/s
Time for a byte to be transmitted:
TB = 1/250 KB/s
Time to transmit 6750 Bytes:
T= 1/250 KB/s . 6750 Bytes = 0.027 sec = 27 msec.
In order to be transmitted, an arriving packet must wait at least 27 msec.
Queing delay = 0.027 sec. = 27 msec.