CS601 Assignment No 02 Spring 2020 Solution & Discussion Due Date: 13-06-2020

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CS601 Assignment No 02 Spring 2020 Solution & Discussion Due Date: 13-06-2020

Question No. 1

 

After carefully analyzing the following figure that represents encoding scheme based on IEEE 802.3 Manchester approach, you are required to identify the bit pattern associated with the given figure:

 

Note: Provide your answer in the following table:

 

        

Question No. 2

 

Suppose a computer in an army headquarters is transferring a file of 10 Megabits to a computer placed in a check post situated at the distance of 2000 KM. Assuming that sender and receiver  computers are connected through a point to point link of 1 Gigabit and propagation speed as 2.4 × 10⁸ meters per seconds, you are required to calculate the following:

 

1. Propagation Time
2. Transmission Time

Note: You need to write all necessary formulas & calculation steps to find the final answers in “Microseconds”.

 

Spring%202020_CS601_2.doc

[+ M.Tariq Malik]

CS601 Data Communication Assignment 2 Solution & Discussion Spring 2020

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CS601 Assignment 2 Solution Idea:

 

Question No: 1

After carefully analyzing the following figure that represents encoding scheme based on IEEE 802.3 Manchester approach, you are required to identify the bit pattern associated with the given figure:

 

Solution:

 

According to instruction encoding scheme is given below in the table.

1

0

1

1

1

0

1

 

 

Question No: 2

Suppose a computer in an army headquarters is transferring a file of 10 Megabits to a computer placed in a check post situated at the distance of 2000 KM. Assuming that sender and receiver  computers are connected through a point to point link of 1 Gigabit and propagation speed as 2.4 × 10⁸ meters per seconds, you are required to calculate the following:

1. Propagation Time
2. Transmission Time

 

Solution:

1. Propagation time.

Distance = 2000 KM

Propagation Speed = 2.4 × 10⁸m/s =

                            = 240000000ms^-1

For conversion of meter to Kilometer

Propagation Speed = 240000000/1000

                            = 240000Kms^-1

Propagation time = ?

Propagation time = Distance/propagation time

Propagation time = 2000KM/ 240000Kms^-1

Propagation time = 0.008333 sec

we convert it into milli second and then into microsecond.

Propagation time = 0.008333/1000 milli sec

Propagation time = 8.333 milli sec

Propagation time = 8333.33 µs

 

 

2. Transmission Time:

Message Size = 10 Megabits

Bandwidth = 1Gigabit/s

Bandwidth = 1000 Megabits/sec

Transmission Time = ?

Transmission time = Message Size/Bandwidth

Transmission Time = 10 Megabits/1000Megabits sec^-1

Transmission Time = 0.01 sec

Transmission Time = 10 millisec

Transmission Time = 10000 µs

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CS601_Assignment_No_02_Solution_Spring_2020

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CS601_Assignment_No_02_Solution_Spring_2020

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CS601 assignment 2 solution spring 2020

CS601 assignment 02 solution spring 2020. 100% correct solution watch only one before submit your assignment.