GATE Overflow Test Series | Mixed Subjects | Test 4 | Question: 14

Question

Two hosts are connected via a packet switch with $2\times 10^7$ bits per second links. Each link has a propagation delay of $30$ microseconds. The switch begins forwarding a packet $40$ microseconds after it receives the same. If $13000$ bits of data are to be transmitted between the two hosts using a packet size of $6000$ bits. The time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is ___________

Answer 1

For $13000$ bits of data using packet size of $6000$ bits we need $\left\lceil \dfrac{13000}{6000} \right\rceil = 3$ packets where the last packet is of size $1000$ bits and others of size $6000$ bits.

The last packet will start transmitting from switch as soon as the transmission of the other packets are done (without waiting for them to reach the destination)

So, total time = Transmission times of first packet +Propagation time for first link
+ Switch Delay + Sum of transmission times of all but the last packet (for transmission from switch) + Transmission time of last packet (from switch)+
+ propagation time for second link.

Transmission time for $6000$ bits packet $ = \dfrac{6000}{2\times 10^7} = 300\; \mu s.$

Transmission time for $1000$ bits packet $ = 300/6 = 50 \;\mu s.$

So, total time $ = 300+30+40+ 2\times  300 +50+ 30  = 1050\; \mu s. $

Comments

So, total time = Transmission times of first packet +Propagation time for first link
+ Switch Delay + Sum of transmission times of all but the last packet (for transmission from switch) + Transmission time of last packet (from switch)+
+ propagation time for second link.

In this Sum of transmission times of all but the last packet (for transmission from switch)→

Transmission time of first packet is already added so why it is 2*300 ?

It should be [300 + 30] (for first packet) + 40 + [300] (for second packet) + [50 + 30] (for last packet)

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No. The packets are transmitted from the source as well as from the switch.

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But it is not mentioned that packet size is flexible ? If packet size is fixed (Last packet is filled with padding bits), answer comes out to 1300 micro seconds.

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Reason for that total time formula → 

(300 + 30  +  40  + 2*300 + 50 + 50)

A -------------------- S(switch) ----------------------- B

t = 300 + 30 + 40 →  370 + 300 + 30 = 700(1^st packet arrival time)

t = 600 + 30 + 40 + 300 + 30 → 1000(2^nd packet arrival time)

t = 850(chosen in such a way that 3^rd packet don’t collide) + 50 + 30 + 40 + 50 + 30 → 1050(3^rd packet arrival time)