In this part, you will perform effective data rate calculations with two scenarios. Each scenario gives you the maximum data rate of the circuit, along with details of the protocol(s) involved, and asks you to calculate the transmission efficiency and effective data rate. For full credit, the calculations you use to derive the answers must be shown.

 

Scenario 1: 22,000 ASCII characters are being transmitted over a 2 Mbps circuit. The protocol uses 8-bits for each ASCII character, as well as 2 parity bits (used for error detection and correction) for each ASCII character.

 

a. Calculate the transmission efficiency of sending these characters using this protocol.

Transmission Efficiency: number of information bits / total number of transmission bits

 

T.E. = (22,000 char * 8 bits/char) / (22,000 char * 8 bit/char) + (22,000 char * 2 parity bits/char)

      = 176,000 / (176,000 + 44,000)

      = 0.8 or 80%

 

b. Determine the effective data rate of sending these characters using this protocol.

Effective Data Rate = 80% of 2Mbps

        = 2Mbps  *  .8  =  1.6 Mbps

 

Scenario 2: A web browser sends an HTTP packet using TCP over IP over Ethernet. The web browser request consists of 125 bytes. The HTTP protocol adds an additional 25 bytes of overhead to the request as part of the HTTP packet. The transmission occurs over a 1 Gbps circuit.

 

a. Calculate the transmission efficiency of sending this packet using the specified protocol. All protocols involved add some overhead bytes to the request. Unless the scenario otherwise specifies a specific number of bytes, use the nominal header sizes for the protocols.

Number of overhead bits = 25 bytes (HTTP) + 20 bytes (TCP) + 20 bytes (IP) = 65 bytes

Transmission Efficiency = 125 / (125 + 65) = 0.657 = 65.7% = 66%

 

b. Determine the effective data rate of sending this packet over this circuit.

Effective Data Rate = 66% of 1 Gbps = 82.5 MBps

 

Please review my answers and correct as needed.

SophiaPretty

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