ECEN4533    Exam #1    2 March 2007   

1) An inverse multiplexer has two 10 Mbps output lines.  The IAT for inbound packets to the multiplexer has a mean time of 221.6 μ seconds/packet and is known to be exponentially distributed.  The average packet size is known to be 410 bytes and is also exponentially distributed.
[10] Compute the trunk load on this system. [Answer = 0.7401]
[15] Compute the probability a packet waits longer than 4 msec in the system's queue. [0.001112]

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2) If the P(Bit Error) = 0.0001 and bit errors occur independently...
[15] Compute the probability that 1 or more errors occur in a 410 byte packet.  [0.2796]
[10] What's the maximum packet size that can be transmitted if P(Packet Error) is desired to be < 0.1?  If there are 1 or more bit errors in the packet, a packet error occurs. [1,053 bits = 131 bytes]

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3) A 700,000 byte application file is to be transmitted from your PC over a switched 10BaseT Ethernet data network.  IPv4 and TCP are higher layer protocols that should be included in each packet.  On Ethernet, packets actually cannot be transmitted back-to-back.  Instead a 9.6 μ second gap is required between each packet.  Assuming no other traffic is on the network...
[15] Compute the time it would take your PC to completely inject this file onto the network. [0.5898 seconds]
[10] If IPv6 is used instead of IPv4, compute the time it would take your PC to completely inject this file onto the network  [0.5980 seconds]

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4) You are designing a system to transmit voice signals over a packet network from a company's small branch office in Texas to the home office in Stillwater, Oklahoma.  You have decided to use a fixed rate 8 Kbps compressed voice coder that outputs 10 bytes for transmission every 10 msec.  To reduce the % overhead, you plan on generating one packet every 20 msec.  Each packet will therefore contain 20 bytes of compressed voice and 47 bytes of layer 2-6 overhead.  The system over which the packets will travel is shown below.  Assuming the following...
*You want audio to pop out of the distant end < 150 msec after the voice capture commences at time t = 0 msec.                                
*The coder at phone #1 requires 35 msec to generate 20 bytes of compressed voice.
*The decoder at the phone #2 requires 10 msec to generate audio for playback.
*Each packet spends 3 msec in every store and forward router in a queue, waiting to be served.
*Only one voice conversation at a time will be stored..
[25] Complete the time-distance drawing shown and compute the slowest possible trunk line speed that will allow an end-to-end audio delivery of < 150 msec.



[The drawing should include a service time and 7 msec propagation delay for a 17 Kbps link, a 3 msec delay at router one, an unknown service time and 10 msec propagation delay for the router 1 to router 2 link, a 3 msec delay at router 2, a service time and 7 msec propagation delay for the last 17 Kbps link, and a 10 msec delay at phone two.  This equation can be used to figure out that the trunk speed should be > 44.89 Kbps.]

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