GATE Overflow Test Series | Mixed Subjects | Test 4 | Question: 12

Question

A path in a digital circuit switched network has a data rate of $2\; \text{Mbps}.$ The exchange of $2000\;\text{bits}$ is required for the setup and teardown phases. The distance between two parties is $10000\;\text{km. }$ If the propagation speed is $2\times 10^8\;\text{m/s}$, the total delay in milliseconds if $20000\;\text{bits}$ of data are exchanged during the data transfer phase will be __________

Answer 1

Total time = Time for setup and teardown + Time for data transfer.
    
    Time for setup and teardown $ = 3 \times (\text{Propagation Delay} + \text{Transmission Time})$
    
    $\qquad = 3 \times \left(\dfrac{10^4 \times 10^3}{2\times 10^8}+ \dfrac{2000}{2\times 10^6}\right) = 3 \times (50+ 1)\;\text{ms} = 153\;\text{ms}.$
    
   Time for data transfer $= \text{Propagation Delay} + \text{Transmission Time}$
    $\qquad = 50 + \dfrac{20000}{2\times 10^6} = 60\;\text{ms}.$
    
    So, total time $= 153+60 = 213\;\text{ms}.$

Comments

How time for setup and teardown is

3×(Propagation Delay+Transmission Time) ?

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Teardown Phase:

When one of the parties needs to disconnect, a signal is sent to each switch to release the resources.
 

 

As in the above figure, there is no waiting time at each switch. The total delay is due to the time needed to create the connection, transfer data, and disconnect the circuit. 

So, Time for setup and Teardown = 3 x (Propogation time + Transmission Time)

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@gatecse Sir, it is given that an exhange 2000 bits are required for setup and teardown phase. So doesn’t this mean 2000 bits is for the entire setup and teardown? If so, then

Time for setup and teardown = $(3 \times Tp)  \ + \ Time \ to \ send \ 2000 \ bits$

= 150ms + 1ms

= 151ms

Time for data transfer = 60ms

Total time = $211ms$

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Please correct me if I’m wrong.

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Hey @gatecse, please confirm the above doubt. Rohith’s reasoning seems valid if the total number of bits exchanged in setup and transmission is 2000 bits