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We have three stations $P, Q, R$ connected in serial manner. $P$ is connected to $Q$ through a $3Gbps$ fibre optic link and length is $500Km$. $Q$ is connected to $R$ through $60Mbps$ link and length is $15Km$.All the links are full duplex in nature. A file is sent from station $A$ to $C$. Packet size is $1Kb$.We use sliding window protocol such that $\text{SWS=RWS}$. Find the optimal $\text{SWS}$.

  • $30$ packets
  • $32$ packets
  • $28$ packets
  • $29$ packets
asked in Computer Networks by Loyal (6.3k points)
retagged by | 140 views
0
what is propagation speed ?
0
nothing given in the question about pd.......can u tell by assuming it
0
we can find propagation delay .length is given and propagation speed of fiber optic is approx 3*10^8
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Tp for P -> Q and Q->R is same ?
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shivanisrivarshini nothing mentioned in question

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Packet Size must be 1KB
I'm getting 38 though
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 @gauravkc even I am getting 38 packets

1 Answer

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1Kb = 1 * 1024 * 8 bits = 8192 Bits 

SWS (Sender Window Size) = 1+2a or Number of bits in transmission channel (RTT * bandwidth )

Using second 

RTT  = transmission time + 2* propgaiton delay

 = 8192 / 3 * 10^9 + 8192 / 60 * 10^6 + 2* ( 5* 10^5/ 2*10^8 + 15*10^3/2*10^8) 

 = 5.29 ms

FIber Optic Velocity ~ 2 * 10^8. ~ 75-80%

SWS = RTT * Bandiwidth ( Taking Min(3Gbps, 50Mbps) ) 

 = 5.29 * 60Mbps = 31.74 

~~ 32 Packets. 

 

answered by (11 points)
0
I am getting 38 packets using 1+2a

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