IT 420 Spring 2017 Homework 3 Solutions

IT 420

Spring 2017

Homework 3 Solutions

1. Answer the following questions concerning error detection:

• In cyclic redundancy checking, what is shared ahead of time between the sender and the receiver?

1. the divisor b.  the quotient                        c.  the dividend           d.  the remainder

• A divisor is 6 bits long. How long is the CRC?

1. 5 bits b.  6 bits                      c.  7 bits                       d.  cannot be determined

           

• Which error detection method cannot always detect a multi-bit error?

1. a. single bit parity check checksum
2. CRC d.  all of the above can detect a multi-bit error

1.4  The checksum on 500 bytes of data is 16 bits long. How long is the checksum on 1000

bytes of data?

1. 160 bits b.  32 bits        c. 16 bits          d. cannot be determined

2. Assume 8-bit units of data. Calculate the checksum for 11010000  10111010

1 1 0 1 0 0 0 0

1 0 1 1 1 0 1 0

                 1     1 0 0 0 1 0 1 0

                  +1

1 0 0 0 1 0 1 1

Checksum = 01110100

1. Assume a bit sequence of 101110 and a divisor of 1001. 
1. Generate the CRC

           1001  101110000

                     1001

                      0101

                      0000

                        1010 

                        1001

                        0110

                         0000

                          1100

                          1001

                            1010

                            1001

                              011                        CRC = 011     

1. Check the CRC

           1001  101110011

                     1001

                      0101

                      0000

                        1010 

                        1001

                        0110

                         0000

                          1101

                          1001

                            1001

                            1001

                              000                        CRC checks correctly 

1. A CRC divisor is 1101001 . What is the polynomial?

x⁶ + x⁵ + x³ + 1

1. What kind of modulation or keying is shown below?  All three graphs together describe one modulation or keying method:  the top graph is the carrier wave, the middle graph is the data, and the bottom graph is the output signal.

     

Frequency Modulation

1. Draw constellation diagrams for:

1. 4-PSK

1. 8-QAM with at least two amplitudes

                                                                                       

                                                                                          

1. If you are using 256-QAM, how many bits does each point on the constellation diagram represent?

# bits = log₂ (256)

# bits = 8

1. Answer the following questions:

7.1  What characteristic of the carrier wave is changed to represent the data in FSK?

1. amplitude               b.  frequency               c.  phase                d.  period

7.2  What converts analog data to an analog signal?

1. multiplexing           b.  PCM           c. analog modulation        d.  keying

7.3  What kind of keying changes both the amplitude and the phase of a carrier wave to represent data?

1. ASK           b.  FSK            c. PSK                              d.  QAM

7.4  In what kind of multiplexing does each sender use some of the bandwidth of the link for all time?

1. FDM   TDM          c. both of these                 d.  neither of these

    7.5  What scheme multiplexes digital signals?

1. FDM             b.  TDM          c.  both of these                d.  neither of these

1. Three FDM multiplexers are connected as shown below.  If each input has a bandwidth of 20 MHz, what is the minimum bandwidth of the output from each multiplexer?  Assume a 2 MHz guard band between channels.

For the top first-layer multiplexer:

            BW_link = (2 * BW_channel) + (1 * BW_guard_band)

                            = (2 * 20 MHz) + (1 * 2 MHz)

                            = 40 MHz + 2 MHz

                            = 42 MHz

For the bottom first layer multiplexers:

            BW_link = (3 * BW_channel) + (2 * BW_guard_band)

                            = (3 * 20 MHz) + (2 * 2 MHz)

                            = 60 MHz + 4 MHz

                            = 64 MHz

For the second-layer multiplexer:

            BW_link  =  42  MHz + 64 MHz + 2 MHz

                            = 108 MHz

1. Design a hierarchical TDM system to multiplex 7 channels, each with a 2 Mbps bandwidth, together.  Assume each multiplexer can combine, at most, 3 channels.  Sketch your design, showing the minimum bit rate of the output from each multiplexer.

There are many possible solutions. Sample solution shown below.

1. Answer the following questions concerning network performance:

• What is propagation delay?
  1. The time required for a signal to travel across a transmission medium
  2. The time needed to obtain access to a transmission medium
  3. The time a packet spends waiting to be processed by a router
  4. The time required for a server to respond to a request

• What is jitter?
  1. the percentage of the network capacity being used
  2. the time between when the first bit of a message is sent and when it is received
  3. rate at which bits are transferred between a sender and receiver
  4. change in delay

      10.3 What is goodput?

1. The total round-trip delay
2. The channel capacity of a single channel calculated using Nyquist’s or Shannon’s

                    formula

1. Amount of data transferred per unit time
2. Change in delay

       10.4  Which is correct about Quality of Service (QoS)?  (Choose two.)

1. Routers choose routes to provide specific levels of service, such as throughput,

                    delay, and/or jitter

1. Most common QoS protocol is IP
2. Most common QoS protocol is MPLS (Multiprotocol Label Switching)
3. It is not possible to provide a specific QoS in the internet

12. Two hosts are separated by 2000 meters. Propagation speed in the link is 2x10⁸m/sec. What is the propagation delay?

Prop Delay = d / s

                    = 2000 m / 2 x 10⁸ m/sec

                   = 0.00001 sec = 10 µsec

1. A user downloads a 500 Mbyte file in 2 seconds.  What is the average throughput in Mbits per second? 

500 Mbytes * 8 bits = 4000 Mbits

                                              byte

                  throughput = data size / time

                                     = 4000 Mbits / 2 sec

                                     = 2000 Mbps

1. What is the round-trip delay for a packet that passes through 4 routers on the way to the server and 4 routers on the way back?  Assume each router has a queuing delay of 2µsec,  each router has a switching delay of 1µsec, and the server delay is 50µsec. Neglect propagation and access delays.

Total delay = 8(2µsec +1µsec)  + 50µsec

                    = 74 µsec

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