Computer Networking Midterm Makeup Problems

Extra Credit for test results

Due Thursday, November 17 at 11:55pm, on Moodle. Absolutely no late submissions will be considered.

The following problems can be completed to suppliment the grade you recieved on similar problems on the test. You may ONLY recieve credit back on problems you missed, and up to number of points missed on that problem.

  • All work MUST be typed or computer drawn, complete, and well organized to recoeve ANY credit.
  • The questions must be in order, and FOR EACH PART each question, you must specify exactly how many points you missed, and how many points you are going for.
All work MUST be done independantly. You MAY NOT discuss this work with anyone BUT the instructor, and must not show your work to any other student.

Multiple Choice

For each incorrectly answered multipe choice problem,you may get a maximum of 1/2 point back if you do the following:

  1. Restate the problem completely
  2. State wrong answer you gave, and fully explain why it is wrong.
  3. Give the correct answer and justify it with a complete explaination, and/or an example, as is appropriate for the problem,
You answer must be typed (or computer drawn), complete, understandable and correct to receive any extra credit.

Other Questions

For problems 33-37, you may recieve up to .75 ofd the missed points.

33. extra credit problem, worth up to 7.5 points toward problem 33

Create a series of nice drawing showing exactly how Collision Avoidance works step by step. Along side the diagrams, explain in text what is happening. Include examples of hiddden nodes explicitly in this presenation. Then show a similar example WITHOUT Collision Avoidance, and explain exactly what the problem is.

35. extra credit problem, worth up to 9 points toward problem 35

Show, and explain, all your work!

Calculate the total time time required to transfer a 5-MB file in the following cases, assuming a RTT of 300ns, a packet size of 2 KB data, and an initial 2xRTT of “handshaking” before data is sent:

a. The bandwidth is 10Gbps, and the data can be sent continuously, one packet after the another.

  • 5MB = 1048576*5*8 = 41,943,040bits Total Size
  • 41,943,040/10,000,000,000 = 0.004194304 sec = 4.194304ms
  • 0.004194304sec + 0.00000015 + 0.00000060 = 0.004195054sec
b. The bandwidth is 10Gbps, but after each packet is sent, a 10 byte ACK must be recieved by the sender before the next packet can be transmitted. Assume no errors.
  • RTT = 300 ns = .0000003 sec
  • packet size = 2KB = 2048 * 8 = 16,384 bits
  • 16,384/10,000,000,000 = 0.0000016384 sec per packet
  • 41,943,040/16,384 = 2,560 packets
  • (10*8)/10,000,000,000 = 0.000000008sec = 8 ns per ack
  • each packet + Ack + RTT = 0.0000016384 + 0.000000008 + .0000003 = 0.0000019464
  • 2,556 * 0.0000019464 = 0.0049749984 sec
c. Compute the maximum number of complete packets that can be on the link at the same time (delay x bandwidth). Using this number (call it w packets), assume we can sent w packets without waiting, but after sending w packets the sender must wait for a 10byte ACK would wait to acknowledge all w packets. Again, if the bandwidth is 10Gbps, what is the time to transmit?
  • Propagation delay = 150ns, 0.00000015 sec. 10,000,000,000* 0.00000015 = 1,500 bits. So 2 packets needed to completely fill the link.
  • So we transmitt 2 packets at a time.
  • 2 * 0.0000016384 = 0.0000032768 to transmit 2 packets.
  • 2 packets + Ack + RTT = 0.0000032768 + 0.000000008 + .0000003 = 0.0000035848
  • 1278 * 0.0000035848 = 0.0045813744
d. How does question c above change if we reduce the link speed to 100Mbps?
  • 100,000,000* 0.00000015 = 15 bits. Obviously this is a window size of only one packet!
  • 16,384/100,000,000 = 0.00016384 sec per packet
  • (10*8)/100,000,000 = 0.0000008sec = 8 ns per ack
  • each packet + Ack + RTT = 0.00016384 + 0.0000008 + .0000003 = 0.00016494
  • 2,556 * 0.00016494 = 0.42158664 sec
36. extra credit problem, worth up to 9 points toward problem 36

Suppose a 1-Mpbs point-to-point link is set up between the Earth and a lunar colony. The distance from Earth to the moon is approximately 385,000km, and the data travels over the link at the speed of light – 3 x 108 m/s.

a. Calculate the minimum RTT for the link.

  • The minimum RTT is 2 ◊ 385, 000, 000m / 3◊10^8m/s = 2.57 seconds.
b. Using the RTT as the delay, calculate the delay x bandwidth product for the link. Explain the significance of the product.
  • The delay◊bandwidth product is 2.57 s◊1Mbps = 2,570,000bits = 2,570,000/(8*2^10) = 313.7KB
  • This represents the amount of data the sender can send before it would be
    possible to receive a response.
c. A camera on the lunar base takes pictures of earth and saves them in digital format. If an image s 25MB,what is the minimum time it will take to transmit the picture is packets can be sent continously?
  • We require at least one RTT from sending the request before the first bit of the picture could begin arriving at the ground (TCP would take longer).
    25 MB = 25*8*2^20 = 209,715,200
    . Assuming bandwidth delay only, it would then take 209,715,200b/1Mbps = 209.7152 seconds to finish sending, for a total time of 209.7 + 2.57 = 212.27 sec until the last picture bit arrives on earth.
d. How long will the tranmission of the the picture from c take if the data is broken into packets 20 50KB, and we must wait 1 RTT after transmitting each packet before transmitting the next?
  • 50KB = 50*1024*8 = 409,600 bits/packet
  • 409,600/1,000,000 = 0.409600 sec transmission time per packet
  • 25MB = 25*8*(2^20) = 209,715,200 bits per pictures
  • 209,715,200/409,600 = 512 packets
  • 512 * (0.409600 + 2.57) = 1,525.5552 seconds
37. extra credit problem, worth up to 9 points toward problem 37

Suppose you are designing a sliding window protocol for a 1-Mbps point-to-point link to the moon, which has a one-way latency of 1.25 seconds.

a. Assuming that each frames carries 1 KB of data, what is the minimum number of bits you need for the sequence number? Why?

  • Bandwidth◊(roundtrip)delay is about 1MBps ◊ 1.25s = 1◊(2^20) ◊ 1.25 = 1,310,720bits = 1,310,720/(1024*8) = 160 packets.
  • The window size should be this large; the sequence number spacemust cover twice this range, or up to 320. 9 bits are needed.
b. What will be the expected time to transmit a 25MB file from the moon to the earth with this setup?
  • 25MB = 25*8*(2^20) = 209,715,200 bits for file
  • If we have a window size that is 160 packets, we can transmit continously (if there are no errors)
  • 209,715,200/1,000,000 = 209.7152 seconds
38. extra credit problem, worth up to 9 points toward problem 38

a. Describe a protocol using sliding windows with selective ACKs. Your protocol should retransmit promptly, but not if a frame simply arrives one or two positions out of order. Your protocol should also make explicit what happens if several consecutive frames are lost.

Here is one approach; variations are possible.

  • If frame[N] arrives, the receiver sends ACK[N] if NFE (next frame expected)=N; otherwise if N was in the receive window the receiver sends SACK[N].
  • The sender keeps a bucket of values of N>LAR for which SACK[N] was received;
  • note that whenever LAR slides forward this bucket will have to be purged of all N≤LAR.If the bucket contains one or two values, these could be attributed to out-of ordern delivery.
  • However, the sender might reasonably assume that whenever there was an N>LAR with frame[N] unacknowledged but with three, say, later
    SACKs in the bucket, then frame[N] was lost. (The number three here is taken
    from TCP with fast retransmit, which uses duplicate ACKs instead of SACKs.)
    Retransmission of such frames might then be in order.
b. Draw timing diagrams for you protocol showing the following cases:
  1. Everything is running normally, no errors.
  2. One packet is lost
  3. One ACK is lost
  4. Two packets in a row are lost
-- JimSkon - 2011-11-08
Topic revision: r11 - 2013-07-15 - JimSkon
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