Chapter 1 Questions

Question #3: Consider an application that transmits data at a steady rate. Also, when such an application starts, it will continue running for a relatively wrong period of time.

A. Would a packet switched network or a circuit switched network be more appropriate for this application? Why?

In the case of an application in which N bits of data were being sent in k units of time (where k is small and fixed), and the application runs for a long period of time, I believe the best network to use to run the application would be a circuit switched network. A circuit switched networks ability to reserve time for this application would ensure its uninterrupted service. With K being small, with a circuit switched network, queuing delays and packet loss is unlikely because the application is using a dedicated circuit and does not have to wait for the packets of other network users.

 B. Suppose that a packet switched network is used and only traffic in this network comes from such an application as described above. Furthermore, assume that the sum of the application data rates is less than the capacities of each an every link. Is some from of congestion control needed? Why?

If it were a packet switched network, then I don’t believe a form of congestion control would be necessary. As stated in the problem the only traffic in the network comes from the above application the combined data rates is less than the capacity of each link. That would insinuate that the packets are moving through links faster than the can queue, which would not result in packet loss or congestion.

Question #4: Consider the circuit switched network in Figure 1.13. Recall that there are 4 circuits on each link. Label the 4 switches A, B, C, D, going in the clockwise direction.

A. What is the maximum number of simulataneous connections that can be in progress at one time on this network?

The maximum number of connections of the network is 16. Each line on the network can hold one circuit switched connection at a time. There are 4 nodes each connected by 4 wires. Four WiresxFour Connections is 16 possible circuits. The 16 possible circuits are shown in blue.

B. Suppose that all connections are between A and C. What is the maximum number of simultaneous connections that can be in progress?

There are 8 possible connections between A and C. They can travel on the 4 lines from A->B->C (shown in blue) and the 4 lines from A->D->C (shown in green).

C. Suppose we want to make four connections between A and C, and another 4 connections between B and D.Can we route these calls through the 4 links  to accomodate all 8 connections?

 Yes, all connections can be made 2 lines between each host can be used for the connection from A->C (shown in blue)  and 2 connections between each host can be used for the connection between B->D (shown in yellow).

Question #6: This elementary problem begins to explore the propagation delay and transmission delay, two central conceppts in data networking. Consider 2 hosts, A and B, connected by a single link of R bps. Suppose that the two hosts are seperated 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, dprop, in terms of m and s.

Dprop=m/s –Propagation delay is the result of how long it takes the data to travel down through the medium in meters/second. Therefore, the delay can be expressed as the distance (in meters) or m divided by s or the rate of travel (in meters/second).

B. Determine the transmission time of the packet, dtrans, in terms of L and R.

Dtrans=L/R –Transmission delay is a result of the packet waiting to be placed onto the wire. Before a packet can be put onto the wire, it must be entirely received by the host system. Therefore L=Bits and R=bits/sec give you the delay, in seconds,  that it takes for the host to place all the bits onto the wire.

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

Dend=Dprop+Dtrans –Without processing and queuing delays, the only delays a packet is susceptible to are propagation and transmission delays, which can be added together for the total delay it would take for a packet to reach its end destination.

D. Suppose host A begins to transmit the packet at time T=0. At time T=dtrans, where is the last bit of the packet?

At time=Dtrans, the last bit should be just leaving the transmitting host.  If the first bit started to be placed onto the wire at t=0, then when time=Dtrans, all bits should have been placed onto the wire. Dtrans is how long it takes a host to push all bits out onto the network, therefore when t is equal to Dtrans, all bits have just now been placed onto the wire.

E. Suppose that Dprop is greater than dtrans. At time T=dtrans, where is first bit of the packet?

Assuming that dprop is greater than dtrans, the first bit of the packet should be on the network but not yet at host B.  The packets have left because dtrans has pushed the packets out onto the network, but they haven’t started arriving yet because the distance (propagation delay) is greater than the time it takes for host A to start sending all the packets.

F. Suppose Dprop is less than Dtrans. At time T=Dtrans, where is the first bit of the packet?

The first bit has reached host B in  this scenario. If Dtrans is greater than Dprop, then the last packet has been placed onto the network by the transmitting host. The network can only hold so many packets, therefore, the first packets must have reached host B by the time the last packets are transmitted and placed onto the network by host A.

G.Suppose s= 2.5*10^8, L=120 bits, and R=56 kbps. Find the distance m so that Dprop equals Dtrans

The distance m=(L/R)*S.  M=535,714 M or 535.7 Km. The transmission delay is L bits/R kbps is the time it takes for the bits to travel according to the kbps speed. Then this number is multiplied by the speed of the transmission to see how many meters the bits have traveled in that number of seconds. See below for the math.

Question #7: 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 into 56-byte packets. There is one link between Host 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 the bit is created until the bit is decoded?

First of all, while the question only deals with a singular bit, all of the bits must be converted from analog to digital before the packet can be transmitted across the network. Therefore, in order to generate a packet, host A must convert 56 bytes at a rate of 64 kbps.  First translate bytes to bits to find the speed at which the bytes are converted. There are 8 bits in one byte. Therefore, to get the number of seconds it will take to translate the analog to digital media it is: (56*8)/64*10^3)=7 ms.

Next,  you must then decipher how long it takes for the packet to transmit at a rate of 2 Mbps. Again because the rate is in bits per second, you must first translate bytes to bits or 56*8. Then you divide the number of bits, by the transmission rate of 2*10^6 bits per second. (56*8)/(2*10^6)=.224 ms

Next you must factor in the propagation delay which in this case is 10 ms.

Therefore, the end-to-end delay is calculated by adding the conversion time+dtrans+dprop=7ms+.224ms+10ms=17.224ms

Question #10: Consider a packet of length L which begins at end system A and travels over three links to a destination end system. These 3 links are connected by 2 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 delay, in terms of di, Si, Ri  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 3 links is 2.5*10^8 m/s, the transmission rates on all 3 links are 2 Mbps, the packet switch 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?

If we consider a packet length L which Di equaling length, Si equivalent to the propagation  speed, and Ri being the transmission rate, then the total end-to-end delay of the packet from host A to host B with 2 different packet switches will equal the sum of the links and the processing delays that occur at each router as the packet is transmitted. In order to find the answer we must first note that  Di/Si will give the propagation delay on  a given link and that L/Ri will equal the time it takes for each host along the way to read and re-transmit the packet to the next link. Also because packet switch delays occur at each node between the transmitting and receiving host on the network, the answer must also factor 2*processing delay, to be referred to as Dproc. The 3 over the summation sign is due to the fact that the total propagation and transmission delays must be accounted for on all 3 links between the 2 hosts(therefore the 3 are added together).

End to end Delay=

3

∑    ((di/si)+(L/Ri)) + (2*proc)

i=1

Now, if we input actual values into the equation we must find the propagation and transmissions delays denoted by (di/si)+(L/Ri) above for all 3 links and add the processing delays at the 2 nodes between the hosts.

Therefore, when we input the actual data, the formula will look something like this.

end-to-end delay= (d(link1)/Si +L/Ri)+ (d(link2)/Si +L/Ri)+(d(link3)/Si +L/Ri)+ (2*dproc)

Note:  The processing delay is given as 3 ms.

Link1=26 ms, Link2=22ms, Link3=10ms, Processing Delay=6 ms

end-to-end delay= 26ms+22ms+10ms+6ms=64 ms

Question #16: Consider a router buffer preceding an outbound link. In this problem, you will use Little’s Fomula, a famous formula from queuing theory. Let N denote the average number of packets in the buffer plus the packet being transmitted. Let a denote the rate of packets arriving at the link. Let d denote the average total delay experienced by a packet. Littles formula is N= a*d. Suppose that on average, the buffer contains 10 packets, and the average queuing delay is 10 msec. The link’s transmission rate is 100 packets/sec. Using Little’s formula, what is the average packet arrival rate, assuming there is no packet loss?

For this problem, we are given Little’s Formula N=a*d. Therefore, a=N/d. N is the average number of of packets in a buffer plus the 1 packet currently being transmitted (N = 10packets queuing +1 transmitted = 11), d is the average total delay (in this case it will be the queuing delay plus the transmission delay). The queuing delay is given as 10 ms. We are also told that the link’s transmission rate is 100 packets/second, which means that a packet is transmitted every 10 ms, giving a total delay of 20 ms. As stated before, the average packet arrival rate can be denoted as a=N/d. Therefore, the average packet arrival rate is a=11packets/20 milliseconds or 550 packets per second.

Question #18: Perform a trace route between source and destination on the same continent at 3 different hours of the day.

Answer the following:

• Find the average and Standard deviation of the round trip delays at each of the 3 hours
• Find the number of routers in the path at each of the 3 hours. Did the paths change during any of the hours
• 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 part of the same ISPs. In your experiments, do the largest delays occur at the peering interfaces between adjacent ISPs?
• Repeat the above for a source and destination on different continents. In your experiments do the largest delays occur at the peering interfaces between adjacent ISPs?

For this question, I chose to use the Visual Trace route tool from yougetsignal.com. For my same continent trace route, I chose to  trace route www.netflix.com at 11 AM, 3 PM, and 10:30 PM respectively.

The average time of a round-trip delay for Netflix.com was 1.23 seconds with a standard deviation of about 0.6 seconds. The fastest return occurred at 3:00 PM at 0.6 seconds with the slowest return of 1.8 seconds at 10:30 PM. At each of the 3 hours, the trace route performed 10 hops, it did not deviate on the path between any of the times. Each path traced along 5 different ISPs from dreamhost.com to comcast.net, followed by an unnamed IP, linw.net, then finally to netflix.com. The largest delays occur between peering interfaces.  While most adjacent ISP’s delays only taking about 1 ms, where delay’s between peering ISP’s can delay up to 4 ms based on information gathered from http://traceroute.monitis.com at 10:30 PM.

Netflix @ 11 AM

Netflix @ 3 PM

Netflix @ 10:30PM

Netflix ISP delays

For my international trace route, I chose to route to Amazon.co.uk  based out of Germany again at 11 AM, 3 PM, and 10:30 PM.

The average time for a round-trip delay was much higher than the in country delay at 24.23 seconds with a standard deviation of 1.6 seconds. For this trace route, the number of routers did change for one of the trips. While 2 of the routes traced through 9 routers, from dreamhost.com to cognetico.com, tinet.net, then amazon.co.uk, the 3 PM route took only 6 hops through dreamhost.com, comcast.net, tinet.net then amazon.co.uk. Interestingly, the route with the fewest hops took the longest amount of time with the route taking a a total of 25.8 seconds. Either way, the path flowed through 4 different ISP’s with the longest delays coming between peer ISP’s according to http://traceroute.monitis.com at 10:30 PM. Adjacent ISP’s tended to only have delays of only a few milliseconds, while one of the hops between peer ISP’s took 49 ms.

Amazon @ 11 AM

Amazon @ 3 PM

Amazon @ 10:30 PM

Amazon ISP delays

The main difference between the continental and intercontinental trace routes was obviously the time is took for the routes to be returned. The continental routes took only an average of 1.23 seconds to return while the intercontinental routes took 23 seconds longer for an average of 24.23 seconds. However, surprisingly, the intercontinental routes took fewer hops than the continental routes. For both, the delays between peering ISP’s were longer than those between adjacent ISPs.

Question #22: Consider figure 1.19. 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 is successfully received by the receiver? If a packet is lost in the path from the server to the client, then the server will retransmit the packet. On average, how many times will the server retransmit the packet in order fo the client to successfully receive the packet?

Based on the figure above, we are told two hosts are connected by Rn links. The probability of packet loss at each link on the network is denoted as P. Therefore, the likelihood that the packet will not be dropped along a link can be stated as (1-P). Now, considering that the number of links is Rn, then the likelihood that the packet will make all hops from host A to host B can be shown as Ps=(1-P)^n where Ps is short for Probability of Success.

In order to calculate the average number of times the server will have to re-transmit the packet, we must first think about the probability of success versus transmissions.

As you can see from the chart about, the probability of success dictates the number of retransmits. If the packet succeeds in making all of the hops on the first transmission, it does not need to retransmit. If the probability of success is only 1/2, on average, the packet will fail once, and therefore need to be retransmitted once more. When probability of success goes down, the average number of retransmits increases. When only one packet in ever 8 transmits reaches the destination, probability of success falls to 1/8 and on average the packet must be transmitted 7 more times for the packet to successfully reach host B. Based on this information, the graph continues to 1/Ps. Where 1 is the successful packet and Ps is the number of attempts to get it to host B. However, one of these failures would have to occur with the first packet, meaning that it must be retransmitted one less time than the total number of retransmits. Thus  this can be described as (1/Ps)-1, where Ps is the total number of transmissions necessary for one successful packet.

Question #25: Suppose that 2 hosts, A and B, are seperated by 20,000 km and are connected by a direct link of R = 2Mbps. Suppose the propagation speed over the link is 2.5*10^8 meters/sec

A. Calculate the bandwidth delay product, R*Dprop

For question 25a, we are told to calculate the bandwidth-delay product. The bandwidth delay product according to wikipedia.org is the maximum amount of data on a network at any given time, measured in bits or bytes. We are given that the distance between host A and host B is 20,000 Km and that they are connected by a direct link of R=2 Mbps. Finally, we are told that the propagation speed is 2.5*10^8 meters/sec. Using this information, we can calculate the bandwidth delay product with the given formula, R*Dprop.  First, we must calculate Dprop which is the length of the connection or 20,000 Km divided by the propagation speed of 2.5*10^8 meters/second. Based on the aforementioned equation, Dprop=.08 seconds or 80 milliseconds. Dprop is then multiplied by R or 2 Mbps to get the final bandwidth-delay of 0.16 Mb or 160,000 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?

In question 25B, we must again consider that Dprop is the time that it will take for the first bit to get onto the wire. We found Dprop in our answer to question A to be 0.08 seconds or 80 milliseconds. Therefore, we can find the maximum number of bits on the wire by finding how many bits could be trailing the first bit at time=Dprop from host A to host B. The equation for this is once again R*Dprop or 160,000 bits as you can see from the equation in question 22A. Because it is then given that the file is 800,000 bits long, there are obviously enough bits within the file to fulfill the number of bits found on the wire within the equation.

C. Provide an interpretation of bandwidth delay product

 According to wikipedia, the bandwidth-delay product is the maximum amount of data that can be on a link at any given time. It is calculated by multiplying the links capacity, in this case R=2Mbps, by the propagation delay or the amount of time it takes for one bit of data to travel between hosts along the network. It can be expressed as R*(m/S) where m is the length of the link (20,000Km in problem 22) and S is the speed as which data travels over the link (2.5*10^8).

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

 In order to find the length of an individual bit on the link, we must divide the length of the link (20,000 Km) by the number of bits on the link at one time, or the bandwidth delay product. To do this, the equation will be m/(R*Dprop) again like the question above where m is the length of the link. Based on this information, we are asked to determine if each bit will be longer than a standard American football field of 100 yards.

Because the end result is 125 meters per bit, we can the show that the end result is longer than a football field. 125 meters is equivalent to 136.7 yards, well over the standard length or 100 yards per field or if including the end-zones 120 yards for the entire field (1.4x the length without end zones, 1.1x the length including end zones).

E. Derive a general expression for the width of a biit in terms of the propagation speed s, the transmission rate R, and the length of the link m.

 The general expression for the width of a bit as shown with exact numbers above can be expressed as (m/(R*(m/S) where m is the length of the link, R is the transmission rate of the link and S is the propagation speed measured in meters per second.

 Question #31: In modern packet switched networks, the source host segments long, application layer messages into smaller packets and send the packets into the network, the receiver then reassembles the packets into the original message. We refer to this as message segmentation. Figure 1.27 illustrates the end to end transport of message with and without message segmentation. Consider a message that is 8*10^6 bits long that is sent from source to destination in figure 1.27. Suppose each link in the figure is 2 Mbps. Ignore propagation, queuing, and processing delays.

A. Consider sending the message from source to destination without message segmentation. How long does it take to move the message from the source to 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 to destination host?

 In this problem we are told to consider sending a message of 8*10^6 bits over a packet-switched network. The packet must make 2 hops between source and destination and travel over 3 links. However, as part of the problem, we are told to ignore all queuing, propagation, and processing delays. In order to find the time it takes for the packet to reach the first packet switch, we must multiply the number of packets (P) by the Length of the packet divided by the transmission rate of the link (L/R). The total equation for one link will be P*(L/R). We also know that each switch on the link uses store and forward packet switching. Based on this model, the entire packet must be received before it can forwarded onto the next link. Essentially this means that we must account for the full-time it takes to reach each host without any crossover in the transmission time.

To find just the time it takes for the packet to reach only the first link, we will use the equation P*(L/R) or for this problem 1*((8*10^6 bits) /2Mbps)= 4 Seconds

Then, again, because it is a store and forward network, the entire packet must be received before it can be forwarded to the next link, this means that the full transmission time must be accounted for in each of the 3 links. Therefore, the total time it takes for one 8,000,000 bit packet to travel from host A to host B along 3 links separated by 2 switches is (P*(L/R))*3links or 4 sec*3links=12 seconds end-to-end.

B. Now suppose that each message is segmented into 800 packet, 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?

 While part A included a single packet, part B breaks up a packet of the same size (8,000,000 bits) into 800 smaller segments of 10,000 bits each. In order to find the time it takes for the first packet to reach the first switch it is once again denoted by L/R where L is the length of the packet and R is the transmission rate. 10,000bits/2Mbps=5ms for the first packet to reach the host.

After the first packet reaches the first switch and being sent from the first switch to the second switch, the originating host, host A, can begin to transmit the second packet to the first switch. Due to this equivalency, we can just say that the amount of time it takes for the second packet to travel across the first link is the same as the first packet traveling across the second link and therefore can be denoted as N*(L/R) where N is the number of links. The time it takes for the second packet to reach the first link is 2*(10,000bits/2Mbps)=10ms.

C.  How long does it take to move the files from source host to destination host when message segmentation is used? Compare this result with your answer in part A and comment.

Finally, in order to answer part C, we must first find the total time it would take the segmented file from part B to send all packets to host B. In order to do this, we must add the amount of time that it takes for the first packet to reach the destination host, and then add the amount of time it takes for the remaining packet as they then begin to send one after the other without delay. The travel time for the first packet can be denoted as N*(L/R) where N is the number of links that the packet must travel through. The travel time for the remaining packets can be shown as (P-1)*(L/R) where P is the total number of packets being sent. This is because following the receipt of the first packet, the remaining packets need only make one hop before they arrive as their previous hops have been made in conjunction with the time of the packet before them, as you can see from the chart below.

Based on this information, the final time for the entire file to reach the end host will be:

(N*(L/R))+(P-1)(L/R)= (3*((10^4bits)/2Mbps)+(799packets*((10^4bits)/2Mbps))=4.01 seconds

As you can see from the two different times, it is much faster to send a message when it is segmented into smaller parts. When the 8,000,000 bit file was sent without segmentation, the entire end-to-end delay was 12 seconds, however when sent in 800 segmented packets of 10,000 bits each, the file arrived in just 4.01 seconds, a difference of nearly 8 seconds, and it could be much longer with a larger file.

Question #33: Consider sending a large file of F bits from Host A to Host B. There are 3 links between A and B. The links are uncongested. Host A segments the file into segments of S bits each and adds 80 bits of header to each segment,  forming packets of L=80+S bits. Each link has a transmission rate of R bps. Find the value of S that minimizes the delay of moving the file from Host A to Host B. Disregard propagation delay.

For this question we are told to send a large file (F) from host A to host B. We are told that there are 3 links and 2 packet switches. We are also told that there are no queuing delay. The file (F) is segmented into pieces of S bits each and an 80 bit header is added to each segment, forming packets L= 80 +S. The transmission rate is also given as R bps. In order to minimize the delay of the file, without worrying about propagation delay we have to find S.  In order to do so we must first find the time it would take to get the packets from A to B. First, we must calculate the time it would take the first packet across the network: [(Lbits/R bps) * 3 hops]. Then we must add the time it takes for the rest of the packets to cross the network: [((F/S)-1)*(L/R)], where F/S equals the total number of packets formed. This formula is in place because much like question 31, once the first packet has arrived, packets will start to arrive in succession after making just one hop. Therefore, our total formula for question #33 is: [(Lbits/R bps) * 3 hops]+[((F/S)-1)*(L/R)].  This can be further expressed by plugging in the value 80+S for L as [((160+2S)/R) + ((F/S)*((80+S)/R))]. From here, using Wolfram Alpha, we can plug-in the equation to solve for S in order to minimize the total end-to-end delay. (shown below)