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Sr Systems Engineer Interview Questions

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Senior Software Systems Engineer - Professional Services at Aspera was asked...

Jul 5, 2012
 How would you peform an SSH connection between these two PCs. I can't give you any information on these PCs, just do it.4 AnswersCouldn't do it. Not enough information to complete.ssh into localhost ;PSimply SSH in to localhostShow More Responsesssh username@localhost A ssh server must be installed on both systems. Any PCs running Windows will require an SSH client, such as PuTTy.

Senior Systems Engineer at Qualcomm was asked...

Apr 25, 2012
 Given a wireless channel with loss rate 0.1, what's the throughput one can get with retransmission.8 AnswersIt can modeled as binary symmetric channel. As to my understanding, channel capacity can be acheived with perfect feedback and simple retransmission scheme, so i guess the answer is 1-H(0.1).10/(1.23456...)=8.1 packets per secInterviewer was correct. With probability 0.1 you have one retransmission, with probability (0.1)^2 you have two, etc., since you can also lose the retransmitted packets, reducing the throughput to approximately 0.89.Show More ResponsesIt is equivalent to simply a binary erasure channel with erasure probability 0.1, whose capacity is 1-0.1 = 0.9. aa should re-learn information theory and stochastic processFirst of all I don't think it has anything to do with the capacity of BSC. Note that h(0.1) = 0.46 and that means 1 - h(0.1) is roughly 0.5. If the packet error rate is 10% then BER is in the order of 0.1 / N where N is the length of the packet in bits. For that, the capacity of BSC is almost 1. In any case, in the non-ergodic case I believe the throughput is less than 0.9. Assume you want to send packets p_1, p_2, ..., p_k and each one takes N_1, N_2, ..., N_k time slots. Then, define the k-Packet Throughput (I made it up) as follows k-packet Throughput = (k / (N_1 + N_2 + ... + N_k)). Note that this throughput is a random variable. We can define the throughput based on this as the expected value of k-packet Throughput with respect to random variables N_1 , N_2, ... E[k-packet throughput] = E[ k / (N_1 + N_2 + ... +N_k)] 8' from the law of large numbers we have N_1 + N_2 + ... + N_k -> k E[N] = 1 / 0.9 * k. This suggests that k-packet throughput is a random variable which converges to its mean for large k, but for a finite k, its average is less than 0.9 All in all, I would have answered the question the way you did. In the long run, out of N transmissions you have 0.9 received and therefore the throughput should be 0.9Model the problem based on the average number of transmissions it takes to deliver a packet. Let N be the number of transmissions. Then: Pr[N=1] = 0.9 Pr[N=2] = 0.1*0.9 Pr[N=3] = 0.1*0.1*0.9 ... and so on. The expected number of transmissions per packet is: E[N] = Sum{ k*Pr[N=k] } = 1*Pr[N=1] + 2*Pr[N=2] + 3*Pr[N=3] + ... where the summation index k goes from 1 to infinity. Then the throughput is 1/E[N] packets per transmission, which is 0.896...In the last solution, the number is calculated incorrectly. Accurate calculation gives the expected number of transmissions per packet E[N] = 1.11111, so that 1/E(N)=0.9, which corresponds to the reduction of transmission rate by exactly 10%, as suggested earlier.p=0.1 - probability of a packet transmission failure If we transmit a sequence of N packets with retransmission then an expected number of successfully transmitted packets will be E[N]=p*E[N-1]+(1-p)*(E[N-1]+1)=(1-p)+E[N-1], and as E[0]=0 it easy to solve the recursion: E[N]=N*(1-p) Expected number of successfully transmitted packets per one transmitted packet will be C[N]=E[N]/N=(1-p). The channel capacity is limit of C[N] as N approaching to infinity that is clearly equal to 1-p=0.9

Jun 7, 2010

Senior Systems Engineer at Yahoo was asked...

Feb 17, 2013
 If you could be any Linux command, which would you be?3 Answers"whomai" :)Was excited - correct one - "whoami"top sleep

Senior Systems Engineer at NJVC was asked...

Jun 19, 2010
 The most difficult question was "If you are not hired for this position, would you be willing to accept a position one step down and work up to this one? 2 AnswersNoyour website doesn't reflect intial instructions if you want a resume please advise

Dec 5, 2013

Senior Systems Engineer at TriZetto was asked...

Dec 22, 2010
 How to deal with a difficult customer1 AnswerEveryone knows the right answer, but it's easy to remember a difficult customer and actually go off track here. Stick to the game plan.

Senior Staff Systems Engineer at Vencore was asked...

Dec 31, 2013
 Do you balance your checkbook?1 AnswerYes - to the penny and I always have.

Senior Systems Engineer at ARRIS was asked...

May 3, 2013
 What would i NOT want to do as a systems admin1 Answerrun exchange

Senior Systems Engineer at Intelity was asked...

Apr 28, 2015
 Experience, specifics about different technologies, asked about client interactions.1 AnswerI answered as I would casually speaking to anyone I knew. Described my experience and wishes for the position.
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