1

Homework Assignment #3

Due 11/20 at 5:00pm

EE122 Fall 2012

Please submit your solutions using BSpace (https://bspace.berkeley.edu/). Log in to BSpace with your CalNet ID, click on the EL ENG 122 Fa12 tab, and click on Assignments under Course Tools. Your assignment should be submitted as hw3.txt, using the format

described on the class web page (http://www-inst.eecs.berkeley.edu/~ee122/fa12/homework/hw3). The answers will be graded

automatically, but please show your work so we can check your reasoning. This work can be included in the hw3.txt file, or in a separate hw3work.txt or hw3work.pdf file.

Assignments with only answers and no shown work will receive no credit.

Improperly formatted assignments will receive no credit.

Each question is worth a single point, and the total assignment is worth 30 points.

Some questions are similar to previous years’ mid-terms and/or homeworks, but you should work these out from

first principles. The old solutions may not be directly applicable, or could even be wrong!

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Packet Timings I

Consider a network with two links. Node X is trying to send data to node Z.

The first link (X-Y) has bandwidth B and latency (i.e., propagation delay) L, and the second link

(Y-Z) has bandwidth 2B and latency L/2. Consider two cases:

The network is circuit-switched: Assume that all path setup has already occurred and that

at time t=0 node X begins sending a 1MB file to Z. As soon as the last bit of the file has been put

on the wire, X sends a 2MB file to Z. The last bit of the 1MB file arrives at Z at time t=0.8sec,

and the last bit of the 2MB file arrives at Z at time t=1.8sec.

1. What is L (in msec)? 200

2. What is B (in Mbps)? 16

The circuit as a whole can offer service with end-to-end bandwidth B (we’ll call this throughput

T to avoid confusion with bytes), and latency of (L + L/2).

1MB / T + (L + L/2) = .8sec (1)

(1MB + 2MB) / T + (L + L/2) = 1.8sec (2)

2MB / T = 1.8sec - 0.8sec = 1sec (3) from (1) + (2)

T = 2MB/sec (4) from (3)

1MB / (2MB/sec) + (L + L/2) = 0.8sec (5) from (1) + (4)

0.5sec + (L + L/2) = 0.8sec (6) from (5)

L + L/2 = 0.3sec (7) from (6)

L = 0.2sec (8) from (7)

X Y Z

File 1: TransXZ 0.5s

PXY 0.2

PYZ 0.1

File 2: TransmissionXZ 1.0

PXY

0.2

PYZ

0.1

P = propagation time

3

The network is packet-switched: Assume that at time t=0 node X begins sending a 1KB

packet to Z. As soon as this transmission is completely “on the wire” X sends a 2KB packet to Z.

Assume there is no processing delay at any node. The first packet arrives at Z at time t=1.8msec,

and the second packet arrives at Z at time t=3.4msec.

3. What is L (in msec)? 0.56

4. What is B (in Mbps)? 12.5

Note: since the second link is faster, there is no queueing delay.

(1KB / T) + L + (1KB / 2T) + L/2 = 1.8ms (1)

(1KB / T) + (2KB / T) + L + (2KB / 2T) + L/2 = 3.4ms (2)

(5KB / 2T) = 3.4ms – 1.8ms = 1.6ms (3) from (1) + (2)

T = 5KB / 2 / 1.6ms = 12.5Mbps (4) from (3)

0.64ms + L + 0.32ms + L/2 = 1.8ms (5) from (1) + (4)

L + L/2 = 0.84ms (6) from (5)

L = 0.56ms (7) from (6)

File 1: TransXY 0.64

PXY 0.56

PYZ 0.28

TrYZ 0.32

File 2: TransmissionXY 1.28

PXY

0.56

PYZ

0.28

TrYZ

0.64

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Packet Timings II

Now consider a different network with only one link, of bandwidth B and latency L.

At time t=0, node V sends two data packets back-to-back to W, which ACKs them. Assume that

the data packets are 240 bytes each and the ACK packets are 360 bytes each (this is larger than

the data packet!1). The first ACK packet arrives at t=5.6ms, and the second arrives at t=5.96ms.

5. What is L (in msec)? 2.5

6. What is B (in Mbps)? 8

Note: the ACK packet is larger than the data packet, hence the difference between when the first

and second ACK packets arrives is due to the transmission time of the ACK packet.

5.96ms – 5.6ms = 360 bytes / T (1)

T = 360 bytes / 0.36ms (2) from (1)

T = 1KB / 1ms = 8Mbps (3) from (2)

L + 240 bytes / T + L + 360 bytes / T = 5.6ms (4)

L + 0.24ms + L + 0.36ms = 5.6ms (5) from (2) + (4)

L = 2.5ms (6) from (5)

1 e.g., imagine that the “data packet” is an HTTP request, and the ACK packet is an HTTP response (with pigged-

backed ACK).

V W

5

Plugging in the formula from last year’s mid-term solution will get the wrong answer!

The following diagram (not to scale) depicts what happened in last year’s mid-term

(Question 7. iii):

In the time it took for data packet 2 to be transmitted, ACK packet 1 was already

transmitted. Thus, when data packet 2 was received, it was possible to commence

transmitting ACK packet 2 immediately.

The following is the result of naïvely plugging-in last year’s formula:

The time (.36 + .36 + 2.35 + .54 + 2.35) == 5.96ms would assume that A2 is being

transmitted immediately, even though A1 is still being transmitted.

(If we did want to use the formula from the mid-term, we could revise it to be “T2 =

(max{P,A}/B) + T1”.)

In our revised question, there is queueing delay, because D2 is received before A1 is fully

sent:

Prop D1: 1.5ms TransD1: 0.48

Prop A1: 1.5ms TransA1: .32

TransD2: 0.48 Prop D2: 1.5ms

TransA2: .32 Prop A2: 1.5ms

TransA1: .54

Prop D1: 2.35ms TransD1:.36

Prop A1: 2.35ms

TransA2: .54

TransD2:.36 Prop D1: 2.35ms

Prop A2: 2.35ms

TransA1: .36

Prop D1: 2.5ms TransD1:.24

Prop A1: 2.5ms

TransA2: .36 TransD2:.24 Prop D1: 2.5ms Prop A2: 2.5ms

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ICMP

Consider the following network, which contains a client A, two routers B and C, and a webserver

D that listens on port 80 and does not listen on any other port:

The links AB, BC and CD have a maximum transmission unit (MTU) of 9000 bytes, 1500

bytes, and 1480 bytes respectively.

For every packet, we are using TCP (with no options) and IPv4 (with the DF flag set on every

packet, and no options). The TCP payload contains data that is within the receiver window, and

not previously acknowledged. All the checksums are valid, and no data corruption occurs.

In the following, the sender gets a response to every packet it sends. In the questions below, list:

Which node responds (select one of B, C, D); and,

The type of response (select either ICMP or TCP)

For example, a response might be “B,ICMP” or “D,TCP”. Please explain each answer.

7. The first packet from A to D has a TCP payload length of 1490, a TTL of 149, and a

destination port of 1490. B,ICMP

The packet is 20 (IP header) + 20 (TCP header) + 1490 (TCP payload) bytes == 1530 bytes. This

is large enough to pass from A to B, but not from B to C. Since the DF flag is set, B will send an

ICMP “Need Fragmentation” message to A.

N.B. the TTL is large enough to have traversed the link A-B.

8. The second packet from A to D has a TCP payload length of 222, a TTL of 2, and a

destination port of 22. C,ICMP

When the packet has reached router C, the TTL would be 0. It would send an ICMP “TTL

Expired” message to A.

N.B. the payload is small enough, and the TTL large enough, to have traversed links A-B and B-

C. The port is only relevant to the webserver D.

A B C D

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9. The third packet from A to D has a TCP payload length of 150, a TTL of 150, and a

destination port of 150. [This is a trick question; look at the section notes!] D,TCP

This is Question 2d) of the section 9 worksheet: “On the event of an unreachable port, the return

of a TCP RST message already indicates this.” We would generally not expect an ICMP

“Destination Port Unreachable” message!

N.B. the payload is small enough, and the TTL large enough, to have traversed links A-B, B-C

and C-D.

10. The fourth packet from A to D has a TCP payload length of 100, a TTL of 80, and a

destination port of 80. D,TCP

D would send an ACK.

N.B. the payload is small enough, and the TTL large enough, to have traversed links A-B, B-C

and C-D, and D has a webserver listening on the destination port.

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ARP

The following diagram shows three subnets (all with netmasks 255.255.0.0), each with a switch

(unlabeled squares). These subnets are connected by a router (unlabeled circle, with interfaces A,

B, and C). The MAC addresses of these three interfaces are listed below (in shortened form; in

the real-world, all MAC addresses are 48-bit), along with their IP addresses. The switches also

have MAC addresses on their interfaces, but we can ignore them in this question.

Initially, all the interfaces have no ARP table entries. Any ARP replies received by an interface

will be cached (i.e., can be used, if relevant, for subsequent questions).

11. The host 1.2.3.4 wants to send a message to 1.2.3.15. Specify the MAC addresses of all

the host and/or router interfaces (if any) that will receive an ARP request packet.

23-76-E5,48-DE-AD,35-67-90

1.2.3.4 and 1.2.3.15 are in the same subnet (1.2.0.0 – 1.2.255.255). However, 1.2.3.4 does not

have the MAC address of 1.2.3.15, so it sends an ARP request packet, which is broadcast on the

subnet i.e., the interfaces corresponding to 1.2.3.6, 1.2.3.15 and 1.2.3.2.

32-F4-DD

6.7.4.3

91-F0-F9

6.7.4.15

11-11-17

1.9.1.1

CA-BE-93

1.9.1.7

11-11-E5

1.2.3.4

23-76-E5

1.2.3.6

48-DE-AD

1.2.3.15

A

35-67-90

1.2.3.2

B

13-1F-F3

6.7.4.23

C

92-83-B0

1.9.2.7

A

B C

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12. Next, the host 1.2.3.6 wants to send a message to 1.2.3.15. Specify the MAC addresses of

all the host and/or router interfaces (if any) that will receive an ARP request packet.

11-11-E5,48-DE-AD,35-67-90

1.2.6 and 1.2.3.15 are in the same subnet. However, 1.2.3.6 did not learn about 1.2.3.15’s MAC

address from the previous question, because ARP replies are not broadcast. Thus, 1.2.3.6 must

broadcast an ARP request packet on the subnet, which is received by the interfaces

corresponding to 1.2.3.6, 1.2.3.15, and 1.2.3.2.

13. Finally, the host 1.2.3.4 wants to send a message to 1.9.1.1. After all the appropriate ARP

queries and replies occur, what destination MAC address should the sending adapter on

1.2.3.4 insert into its Ethernet header?

35-67-90

1.2.3.4 realizes that 1.9.1.1 is not on the same subnet, so it must send the packet to the router

(interface A, which is on 1.2.3.4’s subnet).

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BGP Policy I

Consider a set of non-transit ASes A, B, C, D, and a set of transit ASes P, R, X, Y, Z. They

follow standard selection and export policies, and have the following relationships:

A is customer of X.

B is customer of X and Y and peer of C.

C is customer of Y and peer of B.

D is customer of Z.

X is customer of P and peer of Y

Y is customer of P and R, and peer of Z.

Z is customer of R and peer of Y.

P and R are peers.

Assume that if there are two policy compliant paths, the shorter one is chosen.

For each of the following source/destination pairs, list the path that would be chosen, or “None”

if no policy-compliant path exists:

Recall from lecture 16 (p. 37): that standard policies for transit networks are:

“Selection: prefer customer to peer to provider

Export:

o Let customers use any of your routes

o Let anyone route through you to your customer

o Don’t export route to someone on that route (poison reverse)

o Block everything else”

These policies effectively mean that each AS is willing to carry traffic only if:

they are the source or destination; or,

they get to charge someone (i.e., a customer) money

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Hence, for the answers that follow, we can reason with the principle that “routing follows the

money” (lecture 16, p. 27).

It’s also helpful to draw a diagram:

14. from A to B: AXB

All parties are happy with the traffic flow (X gets to charge its customers A and B, and B is the

destination).

There is no shorter policy compliant paths (A and B are not directly connected).

15. from A to C: AXYC

All parties are happy with the traffic flow (X gets to charge its customer A, X sends directly to

its peer Y, Y gets to charge its customer C, and C is the destination).

There is no shorter policy compliant path.2 Also note that AXBC is not compliant (B would

object, because B has to pay X, but B does not gain anything).

2 For brevity, we don’t explicitly state this claim in subsequent answers.

R

X Y

P

Z

A B C D

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16. from A to D: AXPRZD

With AXPRZD, X charges A, P charges X, P sends directly to its peer R, R charges Z, and Z

charges D, which is the destination.

Note that AXYZD would not be policy compliant: Y would not get to charge anyone, so it does

not tell X about its route to Z or D.

17. from B to C: BC

Policy dictates that peers are chosen over providers.

18. from B to D: BYZD

Same reasoning as A to C.

19. from C to D: CYZD

Same reasoning as A to C (and B to D).

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For each of the following routing change questions, list your answer in the following form: if

you think that the route from A to B is now AXYZZYB (probably not a good answer!) then list

“A-B,AXYZZYB”.

20. Assume that P and R stop peering. What route changes? What is the new route?

A-D,None

The network topology is now:

Any route from A to D must go through Y, but Y will not allow it because it cannot

charge R or Z (the only plausible next hops).

R

X Y

P

Z

A B C D

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21. Ignoring Q20 (i.e., assume P and R are peering), assume instead that Z becomes a

customer of Y (rather than a peer). What route changes? What is the new route?

A-D,AXYZD

With this new route, everyone’s a winner:

A is able to send data to D (indirectly)

X is able to charge its customer A

X sends directly to its peer Y

Y is able to charge its customer Z

Z is able to charge its customer D (presumably more than Z got charged by Y!)

D is able to receive data from A

R

X Y

P

Z

A B C D

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BGP Policy II

Consider domains A, B, C, D, E, F connected as shown below.

Assume that all domains have the policy that given a choice of paths, they prefer the one starting

with the earliest letter in the alphabet. That is, A-*-* is preferred over B-*-*.

B C

D

A

F E

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22. Assume that all nodes export all selected routes to all peers (except when those routes

would create loops).

List all routes from A to D that are possible policy-compliant routes (i.e. that could be

steady-states of a BGP convergence process), comma-separated. For example, if you

think ABBCD and ACED are possible routes, write “ABBCD,ACED”.

Consider all the (non-looping) routes from A-D individually:

ABCD: This is possible – the following diagram shows the path taken by traffic from

any node to D.

It is a steady-state because all the route advertisements about D will not cause any

changes. (B would rather route through A, but A will not tell B about its route [it

would create a loop]; similarly, C would prefer to route through B, but B will not tell

C about its route. All other nodes are already doing as they best can with regards to

choosing the node with the earliest letter in the alphabet.)

Furthermore, it is a reachable state, under the scenario:

1. D exports its routes, and all neighbors receive and process the messages.

2. then C exports it routes, and all neighbors receive and process the messages.

3. then B exports its routes, and all neighbors receive and process the messages.

4. then A exports its routes, and all neighbors receive and process the messages.

B C

D

A

F E

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ABED: Ok

It is a steady-state because: B would prefer to go through A, but A will not give B a

route; similarly, E would prefer to route to B, but B will not give its route to E.

It is a reachable state, under the scenario:

1. D exports its routes, but only E receives and processes the message.

(i.e., the message to C is lost; since BGP runs over TCP, it will eventually be

resent)

2. E exports its routes, and all neighbors receive and process the messages.

3. B exports its routes, and all neighbors receive and process the messages.

4. C receives the message from D (it is ignored because C already has a better

route)

5. A exports its routes, and all neighbors receive and process the messages.

AFED: Ok

B C

D

A

F E

B C

D

A

F E

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It is a steady-state because: A would prefer to go through B, but B will not give A a

route; similarly, E would also like to route through B, but that would result in a loop,

so E cannot use that route.

It is a reachable state, under the scenario:

1. D exports its routes, but only E receives and processes the message.

2. E exports its routes, but only F receives and processes the message.

3. F exports its routes, and all neighbors receive and process the message.

4. A exports its routes, and all neighbors receive and process the message.

5. B exports its routes, and all neighbors receive and process the message.

6. The lost messages (i.e., D to C, and E to B) are received, but ignored because

the recipients already have better routes.

AFEBCD: Not possible

B routes through C, so it would happily tell A about its route. A would then choose to

route through B (rather than F), and F would then choose to route through A i.e., this

state is not stable, and would turn into the ABCD case.

B C

D

A

F E

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23. Keeping the same selection policies, but changing the export policies, now assume that

all nodes only export routes to letters that follow them in the alphabet (e.g., B does not

tell A about its routes, but it tells C and E).

List all routes from D to A (this is not the same as the previous question) that are possible

policy-compliant routes (i.e. that could be steady-states of a BGP convergence process).

The export policies effectively mean nodes can only direct traffic to nodes that are earlier in the

alphabet.

DCBA: Ok

DEBA: Not possible – E does not tell D about E’s routes

DEFA: Not possible – e.g., E does not tell D about E’s routes

DCBEFA: Not possible – e.g., E does not tell B about E’s routes

B C

D

A

F E

20

CDNs, proxies, and all that....

Web content is often replicated at multiple locations to minimize the time it takes a user to

download a webpage. This problem explores two basic approaches for replicating web content:

HTTP forward proxies and CDNs. CDN-based approaches rely on a content distribution network

to serve content from multiple locations; there are (at least) three ways to implement the CDN

distribution model. Thus, we have four different options at our disposal:

a) HTTP Caching with a Forward Proxy: A local network routes all HTTP requests through

a HTTP forward proxy. The forward proxy first checks if the requested page is already

cached, in which case the proxy returns the cached page to the user. If the page is not

cached, the proxy requests the page from the site, caches the result, and returns it to the

user.

b) CDN using DNS: When the authoritative server for the site receives a DNS request, it

returns one of many IP addresses, each of which corresponds to a CDN server hosting the

web content.

c) CDN using IP Anycast: DNS requests for the site always return the same, single IP

address. Multiple CDN servers (at various different locations) that host the web content

advertise this same IP address.

d) CDN based on rewriting media URLs: All requests for the base page of a site are served

from a single server hosted by that site. The site’s server re-writes the URLs for media

content (e.g., images) to be URLs corresponding to CDN servers, so that media content is

served from the CDN.

Your new web startup has gone viral, and you want to purchase services from a CDN in order to

minimize the load on your servers. List one of the options, or ‘None’ if appropriate.

24. Which approach allows the finest-grained control over which users are directed to which

servers? In particular, which approach allows a site to dictate exactly how many users are

sent to each server, and to instantaneously re-balance load between servers? D

a) Not applicable (not a CDN!)

b) Once a user has received an IP address from the DNS server, they will continue using it

for a long time (if cached, or if the user decides to browse using the IP address!).

c) Anycast would direct users to the nearest server, even if that server is overloaded.

d) With option D (CDN based on rewriting media URLs), the centralized server can look at

the incoming traffic and send a different URL.

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25. You delegated this decision to your business partner, who chooses one of the CDN-based

approaches. After purchasing services from the CDN, traffic to your servers decreases

substantially, but you continue to see requests for the content on your website (including

both base pages and media) for a few hours. After a full day, traffic to your server dies

out completely. Which approach did your business partner choose? B or C or None

There are three possible answers (this is where you reasoning section counts a lot!):

o b): DNS records are cached, so users will continue visiting the old IP address for some

time

o c): We didn’t state that the CDN servers take over the startup’s old IP address. If the

CDN uses a new IP address for anycast, and the old IP address is linked to the startup’s

servers), since users will have cached the old IP address, the same reasoning as option b)

applies.

o None: A CDN performs document caching, so ultimately it must request some pages

from your webserver, and it must make requests periodically3 to keep the cache up to

date. Thus, none of the CDN options would entirely eliminate requests to the website.

o None: It is also possible to argue that the DNS records may be cached in excess of a day,

so neither b) or c) would apply.

For completeness:

o a) is not applicable (not a CDN). Furthermore, proxies still need to contact the original

server (with conditional GET requests).

o d) is not correct, because requests for base pages would still be served from the original

server.

26. Which approach can guarantee that customers will be served from the CDN server

closest to the customer? C

See Q24.

27. Which approach leads to additional load on your local DNS server?

If “your” refers to users of the startup: D. Options b) and c) give them an IP address, almost as if

there was no CDN involved. With option d), the media content URLs may point to many

different CDN servers, and each of those URLs will need to be resolved.

If “your” refers to the startup owner: None. The owner’s local DNS server is irrelevant.4

Note that “local DNS server” is distinct from the authoritative DNS servers.

3 It’s likely “push” because the web startup is frequently accessed, but the same reasoning applies to a “pull” approach. 4 Unless they browse their own web site!

22

Now we consider network conditions at Berkeley. List one of the options, or ‘None’ if

appropriate.

28. UC Berkeley is suffering major budget cuts, and turns to you for help in lowering the cost

of providing Internet for students. UC Berkeley pays its Internet provider based on the

amount of bandwidth it uses, which is dominated by students downloading external web

content. Which option, if any, can help UC Berkeley reduce the amount it pays its

provider? A

The forward proxy is within UC Berkeley’s local network, so if students download previously

viewed content, Berkeley does not need to send the requests to its ISP.

29. AirBears wireless is terrible, and periodically drops connections, which is frustrating

when you are trying to stream the Giants game during EE122 lecture (we know who you

are!). Which option, if any, can help ensure that you don’t miss any of the game due to

AirBear flakiness? None

The proxy and CDN options will speed up getting traffic from the Internet to the nearest

AirBears router, but they can’t magically fix the connection between the AirBears router to your

laptop. (Put another way: suppose your laptop is connected to Berkeley’s network via Ethernet,

and someone pulls out your Ethernet cable.)

Note (if we wanted a super-tricky question): our best option, assuming we can’t change

AirBears, probably is a CDN, if it performs transcoding. If the bitrate is low, and AirBears only

drops out for short periods, the buffer in the video player should hide the losses.

30. You do a DNS lookup for www.google.com from your laptop connected to AirBears, and

compare the results with your friend. Your friend did the same DNS lookup from India,

and got a different IP address. Which option, if any, could Google have been using? B

o a) only affects HTTP requests, not DNS requests (besides, the forward proxy is at

Berkeley’s discretion i.e., not something that Google chooses!)

o b) gives “one of many IP addresses”

o c) would give everyone “the same, single IP address”

o d) would also give everyone the same IP address (“servers […] advertise this same IP

address”)