Multiple Links between S & D1.1

Question

Consider a source computer (S) transmitting a file of size 106 bits to a destination computer (D) over a network of three routers (R1 , R2 and R3) and four links (L1, L2, L3 and L4). L1 connects S to R1; L2 connects R1 to R2; L3 connects R2 to R3; and L3 connects R3 to D. Let each link be of length 100 km. Assume signals travel over each link at a speed of 108 meters per second. Assume that the link bandwidth on each link is 1 Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. What is the total sum of transmission and propagation delays in transmitting the file from S to D?
1) 1007

2) 1010

3) 3000

4) 3003

Answer 1

Option (A) 1007ms

propogation delay from Source computer (S) to R1 = (Distance) / (Link Speed) = 10^5/10^8 = 1ms
Total prorogation delay to travel from S to D = 4*1 ms = 4ms ( since four links are in between L1, L2, L3 and L4) 

transmission delay for 1 packet = (Number of Bits in one packet) / Bandwidth = (1000/10^6) = 1ms.

total transmission delay for 1 packet = 4 * 1ms = 4ms  ( since four links are in between L1, L2, L3 and L4) 

The first packet will take 8ms to reach Destination Computer (D). While first packet was reaching D, other packets must have been processing in parallel. So D will receive remaining packets 1 packet per 1 ms from R3. So remaining 999 packets will take 999 ms. And total time will be 999 + 8 = 1007 ms

Comments

@Sushant any reference related this topic??

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@reena. Sorry no reference but you can consider it similar to instruction pipelining in CO.

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@Reena.

Dont consider above complicated formulaes :)

Just remember the pipeline concept which says that stage delay in pipeline is dictated by the delay of the longest stage in non-pipelined system.

See my answer below.

Answer 2

Propagation delay  $\approx$ 1ms

Transmission delay for packet of 1000 bits = 1ms

Now, consider a pipeline (as in Computer organization) with 8 stages(5  transmissions at source, destination and routers; 3 for propagation delays) with single stage delay of 1 ms(stage delay dictated by longest stage).

So, total time = 999 stages(1 stage of each instriction except the last) + 8 stages(due to last instruction)

                        = 1007 ms