1

A. (6p)

What is the purpose of the link layer header?
What is the purpose of the network layer header?

The link layer (layer 2) header is used for physical addressing, as it contains the MAC address of the end device. This header exists only within a single link, it gets stripped and re-added each time (each hop)

The network layer (layer 3) header is used for e.g. routing and subnetting. It contains (most commonly) the IP address of the destination, which may be a (default) gateway or the actual end device

B. (4p)

Compare the Ethernet (802.3) header and the WiFi (802.11) header.
Why does one header contain only two addresses while the other has space for four?

The ethernet header only needs two addresses, as one can always send directly to the destination (or the gateway, in case of default routes)

In wireless networks, more addresses are needed, as we need (in the case of frames to/from APs) to:

• Indicate the source address
• Address a certain AP to pick up the packet
• Tell the AP where to actually send it to

2

Your company has two departments, A and B.
The machines belonging to these departments are spread across two locations: 1 and 2

Design a network that:

1. Allows all nodes from dep. A to communicate to one another
2. Allows all nodes from dep. B to communicate to each other
3. Allows node from departement A to only communicate with nodes from department B at the same location and vice versa

A.

Draw your envisioned topology

B. (10p)

Explain how your topology, the devices you used and the configurations you made accomplished what has been requested

R: Router
S: Switch + router on a stick

Devices from each dept. get added to their respective VLAN
VLANs are spread across locations by using trunk ports to the "global router"

• Req. 1 can be achieved by configuring the global router to allow traffic on VLAN A to go across locations
• Req 2. can be achieved in the same way
• Req 3. can be achieved by setting the router on a stick in each location to allow traffic between VLANs on it's local interfaces, but not towards the global router

3

In the given network, assume that R1 is the default gateway for every node in both subnets
Static addresses are omitted from the ARP tables

A (4p)

i) What is the Target IP of the ARP request generated by H1 when it pings H2? Which nodes receive this request
ii) What is the target IP of the ARP request when H1 pings S1? Which nodes receive this request
iii) Fill in the ARP table of H1 after the 2 pings

i) The target IP is 10.1.1.2. It is broadcasted, with the router as a barrier, so H2 and R1 receive it
ii) Since this is outside the subnet, it will use it's default gateway, therefore the IP would be 10.1.1.99. It would still be received by H2 and R1
iii)

• 10.1.1.2 -> 0055.ffff.bbbb
• 10.1.1.99 -> 0055.ffff.9999

B (6p)

i) Assume now that H2 tries to ping S2. What is the Target IP of the ARP request?
ii) You observe that that ping does not succeed. Why?
iii) How could you fix this?
iv) Fill-in the figure the ARP table of H2 after you have fixed the problem with the pings.

i) H2 has it's subnet length misconfigured, thus thinks it's in the same subnet and will target it directly instead of using the default gateway. So the IP would be 10.1.5.4
ii) H2 tries to ping S2 directly, but ARP broadcasts will not cross the router, thus it's ARP request will never be answered, meaning it cannot use ICMP to perform a ping
iii) You can either fix the subnet settings on H2, or alternatively configure the router to either forward ARP requests, or impersonate other hosts and respond to ARP requests itself
iv)

• If we fix the subnet mask then it will use the default gateway, thus:
  • 10.1.1.99 -> 0055.ffff.9999
• If we turn on ARP forwarding:
  • 10.1.5.4 -> 0055.ffff.dddd
• If we turn on ARP impersonation:
  • 10.1.5.4 -> 0055.ffff.9999

C (2p)

An admin changes the IP of H1 to 10.1.1.3. Which ARP-related mechanism makes sure that H2
updates the respective entry in its ARP table?

Gratuitous ARP can be used to notify other devices in the network of this change

4

A (3p)

In the topology above, STP is enabled in all bridges. Which one is the root bridge and why?

Root bridge is elected by:

• Lowest bridge priority
• Lowest MAC address

SW1 and SW2 have the same (lowest) priority, but SW3 has the lower MAC and will thus become the root bridge

B. (5p)

i) What is the role of each port, use "R" for root port, "D" for designated port and "B" for blocked port. Explain your rationale
ii) What purpose do blocked ports serve?

i)

1. All switches are directly connected to the root bridge, since all links have equal cost, this is the shortest path to the root bridge and will thus be used as root port
2. The networks directly connected to the root bridge will use that port as their designated port
3. The link on the left and the diagonal link will have equal cost paths either through SW 1 or 2, so it will choose based on bridge ID, aka priority & MAC, which SW1 wins
4. All other ports will be blocked

ii) Blocked ports prevent loops, which in turn prevent, for example, broadcast storms

C. (2p)

The priority of SW4 is changed to 28621. Does this bring a change in the root bridge? Explain.

Yes, since it now has the shared lowest priority with SW1 and SW3, but it has the lowest MAC, SW4 will become the root bridge

This change will cause the STP protocol to recompute, and thus RDB state of ports will change

5 (10p)

Two machines in the same LAN happen to have the same MAC address.
Which problems would occur in this case? List and explain at least two.

Switches will get confused about on which port this MAC address lives, due to the fact that frames with the MAC in question arrive on different ports, the forwarding table will flap back and forth, meaning that which device will receive the traffic also flaps back and forth.

This leads to the following problems:

• Traffic delivery flaps back and forth between the two devices, meaning it sometimes ends up at the wrong host. Leading to spotty connectivity
• The delivery of traffic to the wrong host can also be seen as a security concern
• The constant switch updates lead to higher resource usage

6 (10p)

i) Explain why fragmentation can occur in IPv4
ii) What are the disadvantages of fragmentation
iii) Why is this not the case in IPv6?

Fragmentation can occur in IPv4 when some link on the path has a lower MTU than the frame that is being sent. In order to still deliver it, it gets fragmented into multiple smaller frames that DO fit in that MTU

Some of the disadvantages of fragmentation are the performance overhead and the complexity that comes with re-assembling them on the other side, especially when part of the fragment gets lost, as that makes the entire frame unusable

Fragmentation is not possible in IPv6 as, instead of fragmenting, a router on the path will send back an ICMP message telling the origin that it's packet is too big, and what MTU to use instead. The origin will then have to adjust it's frame size to said MTU.
This can happen repeatedly until the frame size is small enough to traverse the entire path

7 (10p)

In 802.11 networks, you can use RTS/CTS mechanisms to solve the problem of hidden terminals.

List the pros and cons of this mechanism, and which considerations you make to decide to use it or not.

Request to Send and Clear to Send have the following pros and cons:
++ Avoids collisions, by making sure that even hidden terminals become aware that a host wants to start sending, by having the receiver broadcast a CTS, which the hidden terminal CAN hear

-- Adds more delay (and therefore decreases bandwidth), by first having to send RTS&CTS before actually being able to send, even if you were to be the only 2 hosts in the vicinity

I would personally choose to use it if:

• The devices are used far apart, making the hidden terminal problem more probable
• Reliable sending is more important than raw throughput or latency, for example in case of big frames, where retransmission would be costly

8

A.

Use the Dijkstra’s Algorithm to calculate the shortest path from S to every other node (Shortest Path Tree - SPT). Show in the labels next to each node the order of the node’s introduction to the SPT and the final distance from the source.

B.

List the edges that would participate in the Minimum Spanning Tree (MST) of the network above.
What is the total length of the MST?

I will use Kruskal:
Sort edges in order, pick the ones that do not form a loop until there are V-1 edges

With a total length of 116

C

i) Does the source node affect the form of an SPT? Explain your answer.
ii) Is an SPT unique? How about an MST? Explain.

Yes the source node affects the form an SPT:

E.g. A -- 5 -> B -- 2 -> C
\---------5---------^

Here if we use A as the source node, the SPT would include AC and AB, which add up to 10
But if we use B or C as the source node, the SPT would include the edges AB or AC and BC, which adds to 7 and is also an MST

ii)
If all weights on edges are unique, then there is a single SPT (per source) and MST
But if some edge weights are not unique, then there can be multiple

9

A (2p)

Which router in the above OSPF topology is going to be the DR and which one the BDR for the
common 10.1.1.1/24 network? Explain why.

Election for DR/BDR goes by:

• Highest priority
• Highest router ID
  Thus, R1 would become DR and R4 the BDR

B (2p)

The priority of R2 is changed to 20. Which router is now the DR and which one is the BDR? Explain.

So long as R1 does not fail it will remain the DR, and R4 will remain BDR, since routes are sticky to prevent unnecessary re-runs of the protocol

C (2p)

The connection is suddenly lost between SW1 and SW2, which causes the separation of the network in two. After the Dead-Timers expire, what changes do we expect in the DR and BDR roles?

Since the two halves can no longer communicate, each will get their own DR and BDR

With R1 the DR and R2 the BDR for the left side
And R4 the DR and R3 the BDR for the right side

D (2p)

In your opinion, what will the roles look like if the connectivity is restored? Why?

If the connection is restored, both sides can communicate again
Since there is now a conflict, due to multiple routers believing themselves to be DR/BDR, the one with the lower priority/ID will have to yield
The same goes for the clashing BDRs
There is no complete election process, the choice is made only between the existing DRs and BDRs

10

Hot and Cold potato routing are practices used to exchange traffic between BGP peers. Refer to slide 34-37 of Lecture 12.

A

What could you do to ’force’ your peer to use hot potato routing?

First of all, you cannot 'force' incoming traffic to do anything, you can suggest certain treatment, but in the end it's up to your peer how to route their traffic

Hot potato routing is the default behaviour, thus simply advertising your routes equally across your peering points should be enough in most cases.

B

What could you do to ’force’ your peer to use cold potato routing?

Again, one cannot force their peer to do anything
Assuming we know their topology, or infer based on geography

There are several approaches to this issue:

1. You can make an agreement with your peer and use communities such that your routes will be routed with cold potato routing
2. Alternatively, you can use MED to suggest them to use cold potato routing for your routes
3. Or advertise certain routes with a longer prefix (more specific)
4. Or with duplicate AS_PATH entries to discourage these routes