ASSIGNMENT-1 700630354 Vadlathala vijay ram reddy
ADVANCE COMPUTER NETWORKING & SECURITY
Problems;
6 Problem)
A Sol) Propagation delay can be expressed as: Dpropagation = m/s. Where m and s refers to
“m” is defined as distance between two loads and where “ s “is defined as propagation speed.
In terms of “m” and “s” Dpropagation is derived as: Dprop=m/s.
B Sol) It can be expressed as in terms of L and R. Dtransmission =L/R=(bits)/(bits/sec). Where…show more content… Dtrans = L/R
= m/s = L/R m = L/R*s
= 120/(56*10^3)*[2.5*10^8]
= 535Kms
25 problem)
A Sol) From the given data.
S= 2.5*10^8 meters/sec
L= 20,000 Kilometres = 20,000*10^3 meters.
R= 2 mbps = 2*10^5 bits/sec.
Where dprop=L/R
=S*L/R
=2.5*10^8 m/s* (20,000*10*3)/2*10*5)bits/sec
=160000 bits. = 1.6*10^5 bits.
Hence we calculated band width delay product.
B Sol)
From the above problem we determined BDP
Where BDP = Band width delay product
i.e = 1.6*10^5 bits.
C Sol)
The band width delay product of the link is maximum number of bits that can be in the link.
D Sol) From the given data propagation speed = 2.5*10^8 meters/sec So that one bit takes = 2.5*10^8/2*10^6= 125 meters.
The width of bit in a link is 125 meters.
E sol)
Width of the bit = S/R
WHERE
S= Propagation speed
R=Transmission rate.
10 Problem) a) end- to- end delay: Tpi (Propagation delay) {s} = di / si {m / m/s} Tti (Transmission time) {s} = L / Ri {bits/packet / bits/ s}
Total end to end = Tpi + Tti + Tdproc1+ Tdproc = Tp1 + Tp2 + Tp3 + Tt1 + Tt2 + Tt3 + Tdproc1 + Tdproc2
b) Size of packet : 1500 bytes x 8 = 12000bits (or L ) Tdproc 0.003 seconds d1 = 5,000,000…show more content… 31 Problem)
A sol)
Time taken to send the message from the source host to the first packet switch will be;
Time taken = 8*10^6/2*10*6 = 4 seconds.
By taking consideration with store and forward switching process,
We can calculate the total tome to move message from source host to destination hosts for 3 links will be:
i.e 4sec*3hops = 12 seconds.
B Sol)
Time taken to send first packet switch from source hosts to first packet switch.
= 10*10^3/2*10^6 = 0.005 seconds.
C Sol)
Time at which first packet is reached at the destination host will be.
=0.0055*3links = .0015 seconds
The time at which last packet is reached
=.0015=79980.001= 4.01 sec
We can say that delay in using message segmentation is very less.
D Sol)
The greatest advantage is that it seems to save a lot of time, when compared to send the file as a host. If we find their Is error or loss of a packet, only it would get transmitted instead of having to send the whole thing yet again.
E Sol) Drawbacks:
1.) The overall file cannot be read, If one segmented packet is missed.
2.) The amount for the header bytes will be more.
3.) At the destination, packets will have to be placed in sequence