GATE CSE 2012 | Question: 44

Question

Consider a source computer $(S)$ transmitting a file of size $10^{6}$ bits to a destination computer $(D)$ over a network of two routers $(R_{1}\text{ and }R_{2})$ and three links $(L_{1},L_{2},\text{ and } L_{3})$. $L_{1}$ connects $S$ to $R_{1}$; $L_{2}$ connects $R_{1}$ to $R_{2}$; and $L_{3}$ connects $R_{2}$ to $D$. Let each link be of length $\text{100 km}$. Assume signals travel over each link at a speed of $10^{8}$ meters per second. Assume that the link bandwidth on each link is $\text{1 Mbps}$. Let the file be broken down into $1000$ packets each of size $1000$ bits. Find the total sum of transmission and propagation delays in transmitting the file from $S$ to $D$?

• A. $\text{1005 ms}$
• B. $\text{1010 ms}$
• C. $\text{3000 ms}$
• D. $\text{3003 ms}$

Comments

@Sidd1425 one point i would like to add. The pipeline formula can only be used here when $T_t=T_p$.

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@rhl More like when $T_{t} <= T_{p}$ right ? Because in case of queuing delay, we need to add that also ?

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packet1  takes total 6msec to reach D  (3Tt + 3Tp)
packet2 is just behind packet1 so,  at 7msec it reaches D
similarily rest all 999 packets (packet no. 2  - 1000)  will take  999*1msec  
so Total Time  =  6 + 999 = 1005

Answer 1

Tt = 1ms, Tp = 1ms

Source~~Router1~~Router2~~Destination

Let’s think in terms of the last packet.

When the last packet reaches destination then all packets will reach and we will get the final answer. So we will count for last packet only. Now, the last packet will be transmitted on the first link L1 only when all previous packets have been transmitted.

So,

Time required = transmission time of 999 pkts from S + transmission time of last pkt From S + propagation from S to R  + transmission time of last pkt From R1 + propagation from R1 to R2 + transmission time of last pkt From R2 + propagation from R2 to D

= (999*1) + 1 + 1 + 1 + 1 + 1 +1

= 1005 ms

 

Answer 2

Transmission time=  1000 bits/ 1Mbps = 1 ms.

Propagation time= 1000km/10^8 m/s= 1ms

Now just focus on last 1000th bit. You need 999 ms to put each one of 999 packets on the link (as Tt is 1 ms), doesn't matter now how they travelling and how much they are travelling.

Total time= 3(links)*(1+1) {this time is for last bit to reach destination}+ 999 = 1005ms.

 

Answer 3

Answer: 1005 ms

Why we need to use Overlapping of Packet time to be considered and not going other way around. Okay Packets are transmitted on the link from sender or any other router one after another when  a packet loaded onto the link and it immediately start propagating , the next packet started loading on the link with previous packets propagation start and so it goes for every packet like that. 
Now a router is considered to be a store and forward device, but here no type of processing time not mentioned so we keep it negligible . Okay this question is simple and with given data we have to solve it simply without making it more complex (as usual of gate).

Here since both transmission and propagation time is the same, it may look like propagation time is controlling the time when the second packet reaches the destination. Take different values of Tx and Tp, and you will see it is taking max(Tx,Tp) time for 2^nd packet to reach destination . 

For data rate, we need to consider only the slowest part of the network as data will be getting accumulated at that sender (data rate till that slowest part, we need to add time if a faster part follows a slower part).

 

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Answer 4

Ans

Answer 5

 reminder to notice  difference 

Answer 6

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