We have two hosts, host A and B connected by a single link of rate R bps. The di

Question

We have two hosts, host A and B connected by a single link of rate R bps. The distance between the hosts is m meters and the propagation speed on the link between them is s meters/sec. Host A attempts to send a packet of size L to host B.

a- What will be the propagation delay, d_prop , in terms of m and s.

b- What will be the transmission time d_trans , in terms of L and R.

c- If you ignore other types of delay, what will be the total delay?

d- If host A starts to transmit the packet at t = 0, where will be the last bit of packet at time t = d_trans

e- If d_prop < d_trans , at time t = d_trans where will be the first bit of the packet?

f- Assume that s = 2.5*10^8 , L = 100 bits, R = 28 kbps. What is the distance m if d_prop = d_trans.

Explanation / Answer

a.Propagation delay is the time taken by apacket to reach from source to destination

d_prop=distance btw hosts/propogation speed=m/s   sec

b.the transmission time, t is the time from the first bit until the last bit of a message has left the transmitting node.

d_trans = Packet size/Bit Rate=L/R sec

c.Toatal delay includes processing and queuing delays and propogation delay,transmission.Ignoring other.

T= d_prop+ d_trans =(m/s)+(L/R) sec

d.If host A starts to transmit the packet at t = 0,d_trans is the transmissio time,

The last bit will be at the begning of A ready to be transmitted.

e.

Given

d_prop > d_trans which implies

m/s   > L/R

L < (m*R)/s --Modifing the above inequation in terms of Lenght

is the first bit of the packet on the link will be.

f.

Given d_prop=d_trans

==>m/s = L/R

m=(L*s)/R=(100* 2.5*10^8)/28000=

m=892.8m

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.