## Qualcomm Interview Question for Software Engineer / Developers

Country: United States
Interview Type: In-Person

Comment hidden because of low score. Click to expand.
2
of 2 vote

I think this is a good solution:

The guy is talking about turnaround time (or round-trip time RTT) so I guess he wants a protocol where sent data must be acknowledged.

So how much data does a server need to store while waiting for and ACK? all the data that has been sent before the arrival of an ACK!

buff = 1Gbps * 500usec. = 1024*Mbps * 5*10^-4 sec = 0.05Mb = 51.2KB

Now we need one buff to store the data that we sent and one to store the data that
we are sending so probably we need tot_buff = 2*buff

The protocol would require a packet ID, so that:

Send_packet: { void* data }
Response_packet: {bool ack }

The protocol should not use and ID because the ACK arrives always on time to be sure
that the ACK refers to the previously sent data frame.

Comment hidden because of low score. Click to expand.
0
of 0 vote

I think this is a good solution:

The guy is talking about turnaround time (or round-trip time RTT) so I guess he wants a protocol where sent data must be acknowledged.

So how much data does a server need to store while waiting for and ACK? all the data that has been sent before the arrival of an ACK!

buff = 1Gbps * 500usec. = 1024*Mbps * 5*10^-4 sec = 0.05Mb = 51.2KB

Now we need one buff to store the data that we sent and one to store the data that
we are sending so probably we need tot_buff = 2*buff

The protocol would require a packet ID, so that:

Send_packet: { void* data }
Response_packet: {bool ack }

The protocol should not use and ID because the ACK arrives always on time to be sure
that the ACK refers to the previously sent data frame.

Comment hidden because of low score. Click to expand.
0

Shouldn't it be 512Kb? 5*10^-1 = .5 Mbps = 512Kb/8 = 64KBps

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