ACE Test Series

Question

Comments

4 sec ?

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answer is 0.4 sec.My doubt is why are we taking into consideratio the transmission time of the current packet. QD=0 for first packet then for second it should be 0.1 sec and not 0.2 sec since the packet hs already been taken for transmission.

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Heh, yeah 0.4 sec, did division mistake. We have to consider transmission time of only 4 packets.

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to transmit one packet it takes 0.1sec.

for first packet router directly send the packet without delay.

for 2nd packet it should wait until transmission of first packet(means it is delayed because of first packet)

same as 3rd packet aslo waits for 2nd packet.

..so last packet waits for first (n-1) packets.

Answer 1

Transmission rate T=10kbps= 10×1000bps=10⁴bps

Last packet will be in the queue till the transmission of 4^th packet.

Size of 4 packets: 125×4 Bytes = 500 Bytes. = 500×8 bits. = 4000 bits.

Queueing delay = time for transmitting 4 packets.

= 4000 bits / 10⁴ bps = 0.4 second.

Answer = 0.4 second