Generalize the Formula for Sending K Such Packets Back-to-back Over the N Links. Explain

P2.

Equation 1.1 gives a formula for the end-to-end delay of sending one packet of length L  over N  links of transmission rate R . Generalize this formula for sending P  such packets back-to-back over the N  links.

The general case of sending one packet from source to destination over a path consisting of N links each of rates R. so,

end-to-end delay (d) = N(L/R)

1 packet = the end-to-end delay (d) = N(L/R)

P packets = d1

one packet x d1 = d x p

1 x d1 = [N(L/R)] x P

so as a result, the end-to-end delay is

d1 = [N(L/R)] x P

P5.

Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 100 km/hour.

a. Suppose the caravan travels 150 km, beginning in front of one tollbooth, passing through a second tollbooth, and finishing just after a third tollbooth. What is the end-to-end delay?

ans: total distance = 150 kms

the speed = 100 km/hr

so the total transmission delay = 150/100 = 1.5 hrs

the time taken by each toll booth to reach a single car = 12 sec

so the time taken to reach 10 cars is = 120 sec or 2 min

the time taken by 3 toll booth to reach 10 cars = 6 min

the end-to-end delay is = 1.5 hrs + 6 min = 1 hr 36 min

b. Repeat (a), now assuming that there are eight cars in the caravan instead of ten.

answer : the total transmission delay = 150/100 = 1.5 hrs

the time taken by each toll booth to reach a single car = 12 sec

the time taken to reach 8 cars = 8 x 12 = 96 sec

the time taken by three toll booth to reach 8 cars = 3 x 96 = 288 sec

end-to-end delay is = 1.5 hrs + 4 min 48 sec

so,the for end to end delay for 8 cars is = 1 hr 34 min 48 sec

P6.

This elementary problem begins to explore propagation delay and transmission

delay, two central concepts in data networking. Consider two hosts, A

and B, connected by a single link of rate R  bps. Suppose that the two hosts

are separated by m  meters, and suppose the propagation speed along the link

is s  meters/sec. Host A is to send a packet of size L  bits to Host B.

a) Express the propagation delay, in terms of m  and s .

The distance between the two hosts A and B = m meters

The propagation speed along the link = s meter/sec

So the propagation delay is = m/s or distance/s

b) Determine the transmission time of the packet, in terms of L and R .

The size of the packet = L bits

And the transmission rate of the link is R bps

The transmission of the packet = L/R sec

c) Ignoring processing and queuing delays, obtain an expression for the end-to-end delay.

The end to end delay = d(proc) + d(queue) + d(trans) + d(prop)

Which would result in = d(trans) + d(prop)

d(trans)= L/R and d(prop)=m/s

The end to end delay = (L/R)+(m/s)

d) Suppose Host A begins to transmit the packet at time t  = 0. At time t  = d(trans) ,

where is the last bit of the packet?

transmission delay is time it takes for the host to push the packet out.

so, at t = d(trans) the last bit of the packet has been pushed out or it was transmitted.

e) Suppose dprop  is greater than dtrans . At time t  = d(trans) , where is the first bit of

the packet?

the first bit is on the first packet

f)Suppose d(prop)  is less than d(trans) . At time t  = d(trans) , where is the first bit of the packet?

if d(prop) < d(trans)

at time t = d(trans) the first bit has reached destination B.

g) Suppose s  = 2.5 · 10^8 , L  = 120 bits, and R  = 56 kbps. Find the distance m so that d(prop) equals d(trans) .

s = 2.5 x 10^8 seconds

L = 120 bits R = 56 kbps

d(trans) = d(prop)

so, (L/R)=(m/s)

distance m = sL/R

= (2.5×10^8×120)/(56×1000)

m (distance) = 535.7 km

P7.

In this problem, we consider sending real-time voice from Host A to Host B

over a packet-switched network (VoIP). Host A converts analog voice to a

digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-byte

packets. There is one link between Hosts A and B; its transmission rate is 2

Mbps and its propagation delay is 10 msec. As soon as Host A gathers a

packet, it sends it to Host B. As soon as Host B receives an entire packet, it

converts the packet's bits to an analog signal. How much time elapses from

the time a bit is created (from the original analog signal at Host A) until the

bit is decoded (as part of the analog signal at Host B)?

host A coverts analog voice to digital bit stream = 64 kbps

host A grouping the bits  = 56 byte packets

transmission rate = 2 Mbps

propagation delay = .01 sec

consider that first but in a packet. before this bit can be transmitted, all of the bits in the packet must be generated. means,

= (56 x 8)/ ( 64×10^3)sec   = .007 sec

The time required to transmit this packet = (56 x 8) /(2 x 10^6) = 0.000224 sec

time elapses from the time a bit is created until the bit is decoded

= .007 sec + .000224 sec + .01 sec

= .017224 sec

P9.

Consider the discussion in Section 1.3 of packet switching versus circuit

switching in which an example is provided with a 1 Mbps link. Users are

generating data at a rate of 100 kbps when busy, but are busy generating

data only with probability p  = 0.1. Suppose that the 1 Mbps link is replaced

by a 1 Gbps link.

a. What is N,  the maximum number of users that can be supported

simultaneously under circuit switching?

The total transmission rate = 1 Gbps

Thedata generation rate of each user = 100 kbps

So the maximum number of possible users = 1 Gbps/ 100 Kbps = 10000

b. Now consider packet switching and a user population of M  users. Give a

formula (in terms of p , M , N ) for the probability that more than N  users are

sending data.

total user population = M

number of users transmitting = N

data generation probability of each user , P = 0.1

So you use the binomial distribution formula

= M (sigma) n=N+1 [MP^n(1-p)^M-n]

P10.

Consider a packet of length L  which begins at end system A and travels over

three links to a destination end system. These three links are connected by

two packet switches. Let di, si,  and Ri  denote the length, propagation speed,

and the transmission rate of link i,  for i  = 1, 2, 3. The packet switch delays

each packet by dproc.  Assuming no queuing delays, in terms of di, si, Ri,

(i  = 1,2,3), and L,  what is the total end-to-end delay for the packet? Suppose

now the packet is 1,500 bytes, the propagation speed on all three links is 2.5 ·

108  m/s, the transmission rates of all three links are 2 Mbps, the packet switch

processing delay is 3 msec, the length of the first link is 5,000 km, the length

of the second link is 4,000 km, and the length of the last link is 1,000 km. For

these values, what is the end-to-end delay?

packet length = L

link i Length = di

propagation speed = si

transmission rate = Ri

The first end system requires to transmit the packet onto the first link = L/R1

The first end system requires to transmit the packet onto the second link = L/R2

The first end system requires to transmit the packet onto the third link = L/R3

The packet propagates over the first link = d1/s1

The packet propagates over the second link = d2/s2

The packet propagates over the third link = d3/s3

end to end delay = L/R1+L/R2+L/R3+d1/s1+d2/s2+d3/s3+d(proc)+ d(pro)

packet size  = 1500 bytes

The propagation speed on both links = 2.5 x 10^8

transmission rate of all three links = 2 Mbps

packet switch processing delay = 3 msec

length of the first link = 5000 km

length of the second link = 4000km

length of the last link = 1000 km

the first end system requires to transmit the packet onto the first link = L/R1 = .006 sec

packet propagates over the first link = .02 sec

the packet switch requires to transmit the packet onto the first link = L/R1 = .006 sec

packet propagates over the second link = .016 sec

the packet switch  requires to transmit the packet onto the third link = L/R1 = .006 sec

packet propagates over the third link = .004 sec

end to end delay = .006+.006+.006+.02+.016+.004+.003+.003

= 0.064 sec

P11.

In the above problem, suppose R1  = R2  = R3  = R  and dproc  = 0. Further suppose

the packet switch does not store-and-forward packets but instead immediately

transmits each bit it receives before waiting for the entire packet to

arrive. What is the end-to-end delay?

transmission rate = R1=R2=R3

propagation delay = 0

packet =1500 bytes = 1500 x 8 bits

propagation speed = 2.5 x 10^8 m/s

the end system requires to transmit the packet onto the link =L/R = (1500×8)/(2×10^6)= 6 ms

length of the first link = 5000 km

the packet propagates over the 1st link delay = 5000×10^3  /  2.5×10^8 = 20 ms

length of 2nd link = 4000 km

the packet propagates over the 2nd link delay = 4000×10^3  /  2.5×10^8 = 16 ms

length of last link = 1000 km

the packet propagates over the 3rd link delay = 1000×10^3  /  2.5×10^8 = 4 ms

the end to end delay = 6+20+16+4 = 46 ms

P13.

(a) Suppose N packets arrive simultaneously to a link at which no packets

are currently being transmitted or queued. Each packet is of length L

and the link has transmission rate R. What is the average queuing delay

for the N packets?

(b) Now suppose that N such packets arrive to the link every LN/R seconds.

What is the average queuing delay of a packet?

P14.

Consider the queuing delay in a router buffer. Let I  denote traffic intensity;

that is, I  = La/R . Suppose that the queuing delay takes the form IL/R  (1 – I )

for I  < 1.

a. Provide a formula for the total delay, that is, the queuing delay plus the

transmission delay.

b. Plot the total delay as a function of L/R .

P15.

Let a  denote the rate of packets arriving at a link in packets/sec, and let μ

denote the link's transmission rate in packets/sec. Based on the formula for

the total delay (i.e., the queuing delay plus the transmission delay) derived in

the previous problem, derive a formula for the total delay in terms of a  and μ .

(L/R)/(1-I)=(L/R)/(1-aL/R)

= (1/μ)/(1-a/μ)

=1/(μ-a)

P18.

Perform a Traceroute between source and destination on the same continent

at three different hours of the day.

a. Find the average and standard deviation of the round-trip delays at each of

the three hours.

three trials the round trip delay between source and the router was

delay at 1st hour = 1.03 ms

delay at 2nd hour = .48 ms

delay at 3rd hour = .45 ms

-average = (1.03+.48+.45)/3 = .65 ms

-standard deviation = sq rt ((1/3)[(1.02-.65)^2+(.48-.65)^2+(.45-.65)^2])

= sq rt (.0711) = 0.267 ms

b. Find the number of routers in the path at each of the three hours. Did the

paths change during any of the hours?

-the number of routers is 9 between source and destination. the paths may have changed over the amount of time.

c. Try to identify the number of ISP networks that the Traceroute packets pass

through from source to destination. Routers with similar names and/or similar

IP addresses should be considered as part of the same ISP. In your experiments,

do the largest delays occur at the peering interfaces between adjacent ISPs?

-the number of ISP networks are 7

-the largest delays are most likely to occur between the adjacent ISPs.

d. Repeat the above for a source and destination on different continents.

Compare the intra-continent and inter-continent results.

P19.

(a) Visit the site http://www.traceroute.org and perform traceroutes from two different

cities in France to the same destination host in the United States. How many

links are the same in the two traceroutes? Is the transatlantic link the same?

(b) Repeat (a) but this time choose one city in France and another city in

Germany.

(c) Pick a city in the United States, and perform traceroutes to two hosts, each

in a different city in China. How many links are common in the two

traceroutes? Do the two traceroutes diverge before reaching China?

P20.

Consider the throughput example corresponding to Figure 1.20(b). Now

suppose that there are M  client-server pairs rather than 10. Denote Rs , Rc , and

R  for the rates of the server links, client links, and network link. Assume all

other links have abundant capacity and that there is no other traffic in the

network besides the traffic generated by the M  client-server pairs. Derive a

general expression for throughput in terms of Rs , Rc , R , and M .

P22.

Consider Figure 1.19(b). Suppose that each link between the server and the

client has a packet loss probability p,  and the packet loss probabilities for

these links are independent. What is the probability that a packet (sent by the

server) is successfully received by the receiver? If a packet is lost in the path

from the server to the client, then the server will re-transmit the packet. On

average, how many times will the server re-transmit the packet in order for

the client to successfully receive the packet?

figure: server—R1–0–R2–0——-0–RN–client where 0=links

each link between the server and the client has a packet loss probability=p.

probability of the packet is successfully received by the receiver(p1) = (1-p)^n

so, the server will transmit the packet in order for the client to successfully receive  the packet on average = 1/(p1)

the server will re transmit the packet in order for the client to succefully receive the packet on average = (1/(p1))-1

P23.

Consider Figure 1.19(a). Assume that we know the bottleneck link along the

path from the server to the client is the first link with rate Rs  bits/sec. Suppose

we send a pair of packets back to back from the server to the client, and there

is no other traffic on this path. Assume each packet of size L  bits, and both

links have the same propagation delay dprop.

a. What is the packet inter-arrival time at the destination? That is, how much

time elapses from when the last bit of the first packet arrives until the last

bit of the second packet arrives?

b. Now assume that the second link is the bottleneck link (i.e., Rc  < Rs ). Is it

possible that the second packet queues at the input queue of the second

link? Explain. Now suppose that the server sends the second packet T  seconds

after sending the first packet. How large must T  be to ensure no

queuing before the second link? Explain.

suppose the first packets as A and the second packet as B.

If the bottleneck link is the first link of the packet A, and the packet B is queued at the first link. Since it is waiting for the transmission of packet A. So, the packet arrival time at the destination is L/Rs.

Now we will assume that the second link is the bottleneck link (exampleR c <R s ). Since both packets are sent back to back, it's possible that the second packet will arrive at the input queue of the second link before the second link finishes the transmission of the first packet, which is:

L/Rs+L/Rs+d(prop) < L/Rs+d(prop)+L/Rc …….(1)

The left side of the equation above shows the time that is needed by the second packet to arrive at the input queue of the second link. And the right side of the equation shows the time needed by the first packet of finish its transmission onto the second link transmission onto the second link.

The equation 1 is possible as Rc < Rs  and it's clear that the second packet has to have a queuing delay at the input queue of the second link. If we send the second packet T sec later, then the delay for the second packet at the second link, then

(L/Rs)+(L/Rs)+(d(prop))+T < (L/Rs)+d(prop)+(L/Rc)

P24.

Suppose you would like to urgently deliver 40 terabytes data from Boston to

Los Angeles. You have available a 100 Mbps dedicated link for data transfer.

Would you prefer to transmit the data via this link or instead use FedEx overnight

delivery? Explain.

40 terabytes of data is a very large amount of data. But I am given a 100 Mbps link to transfer. So it takes a lot time to transfer the data through the link provided. Also there may be some chances of missing data during this huge transmission. So if the data is sent to be urgently than it would be better to do FedEx overnight delivery.

P25.

Suppose two hosts, A and B, are separated by 20,000 kilometers and are

connected by a direct link of R  = 2 Mbps. Suppose the propagation speed

over the link is 2.5   108  meters/sec.

a. Calculate the bandwidth-delay product, R  d(prop).

propagation delay = Distance/speed

= 2 x 10^7 / 2.5 x 10^8   = 0.08 sec

The transmission rate = 2 Mbps

So, bandwidth delay product = R x dprop = 2 x 10^6 x .08 = 16 x 10^4 bits

b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose

The file is sent continuously as one large message. What is the maximum

number of bits that will be in the link at any given time?

file size = 8e5  bits

If the file is being sent continuously as one message, a link can have the maximum number of bits at the same as band width delay product.

The maximum number of bits at a given time will be 16 x 10^4 bits

c. Provide an interpretation of the bandwidth-delay product.

Bandwidth delay product is the number of bits transmitted per/ sec when propagation delay is one second.

d. What is the width (in meters) of a bit in the link? Is it longer than a football

field?

propagation speed over the said link = 2.5×10^8 m/s

so 1 bit takes = (2.5×10^8)/(2×10^6) =125 m/bit

1 bit is 125 meters. so its longer than a football field.

e. Derive a general expression for the width of a bit in terms of the propagation

speed s,  the transmission rate R,  and the length of the link m .

The width of a bit is related with the propagation speed per bandwidth

then the general expression for the width of bit is = s/R

P27.

Consider problem P25 but now with a link of R  = 1 Gbps.

a. Calculate the bandwidth-delay product, R  dprop .

Bandwidth delay product = R x dprop = 10^9 x 0.08 = 80000000 bits

b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose

the file is sent continuously as one big message. What is the maximum

number of bits that will be in the link at any given time?

The max number of bits in the link at any given time

= min (bandwidth delay product,packet size)

=(80000000,800000)

so, 8×10^5 bits can be sent as continuous transmission.

c. What is the width (in meters) of a bit in the link?

width of a bit in the link = s/R

= 2.5e8/10^9  = 0.25 meters

P31.

In modern packet-switched networks, including the Internet, the source host

segments long, application-layer messages (for example, an image or a music

file) into smaller packets and sends the packets into the network. The receiver

then reassembles the packets back into the original message. We refer to this

process as message segmentation . Figure 1.27 illustrates the end-to-end

transport of a message with and without message segmentation. Consider a

message that is 8 · 106  bits long that is to be sent from source to destination in

Figure 1.27. Suppose each link in the figure is 2 Mbps. Ignore propagation,

queuing, and processing delays.

figure 1.27End-to-end message transport: (a) without message segmentation

= source—-packet switch(message)—-packet switch—-destination

figure 1.27(b) with message segmentation

= source(packet)—-packet switch—-packet switch—destination

a. Consider sending the message from source to destination without  message

segmentation. How long does it take to move the message from the source

host to the first packet switch? Keeping in mind that each switch uses

store-and-forward packet switching, what is the total time to move the

message from source host to destination host?

message sent from source ti destination = 8e6

transmission rate = 2 Mbps

time to send message from the source host to first packet switch = 8e6/2e6 = 4 sec

with store and forward switching, the total time to move message from source host to destination host for 3 links = 4 sec x 3 hops = 12 sec

b. Now suppose that the message is segmented into 800 packets, with each

packet being 10,000 bits long. How long does it take to move the first

packet from source host to the first switch? When the first packet is being

sent from the first switch to the second switch, the second packet is being

sent from the source host to the first switch. At what time will the second

packet be fully received at the first switch?

time to send first packet from source host to first packet switch

= (10×10^3)/(2×10^6) = 0.005 seconds

c. How long does it take to move the file from source host to destination host

when message segmentation is used? Compare this result with your

answer in part (a) and comment.

time at which first packet is received at the destination host = .005 x 3 link = .0015 sec

the time at which last packet is received

= .0015 sec + 799×0.001 sec  = 4.01 sec

the result is significantly less than it.

d. In addition to reducing delay, what are reasons to use message segmentation?

There are a lot of good reasons to use message segmentation. One is that if any failure of delivery message, only a small portion has to be retransmit, but not the whole message. and other thing is that segmented packet can be transmitting different routes depending on congestion.

e. Discuss the drawbacks of message segmentation.

Drawbacks of message segmentation

-if one segmented packet is missing, then the overall file cannot be read at all.

-more bandwidth of overhead

-the total amount of header bytes is more

-packets have to be put in sequence at the destination

Generalize the Formula for Sending K Such Packets Back-to-back Over the N Links. Explain

Source: https://culbertsonj.wordpress.com/homework-1/

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