TCP Slow Start is used to prevent network overload, but in high speed, long distance networks, it can be too slow because it starts with a small data window and only increases a little with each RTT. This delay means it takes a long time to use the full network capacity.

In comparison, QUIC is a modern transport-layer protocol created by Google to fixes many of these issues. It runs its congestion control outside the operating system, so it can be updated more easily and adjusted for faster performance. Unlike TCP, it sends multiple streams without one being blocked by another if a packet is lost. QUIC can reach full speed faster and works better on fast but faraway networks.

NAT lets many devices share one public IP, which saves address space, but it causes problems for apps like p2p and VoIP that need direct connections.
That’s because NAT blocks incoming traffic unless it knows where to send it, which makes it hard for outside devices to reach someone inside the network. It also breaks the original idea of the internet where any two devices could talk directly. Some apps put IP addresses inside the message itself, and NAT can’t change those, leading to connection issues. In busy networks, there might not be enough ports for everyone to connect out.

3) Both CSMA and Slotted ALOHA are medium access control (MAC) protocols used for managing channel access in network communication. While Slotted ALOHA improves upon pure ALOHA by reducing collisions, CSMA introduces carrier sensing to further enhance efficiency. How do these protocols differ in terms of throughput, collision probability, and delay? Under what conditions would Slotted ALOHA be preferred over CSMA, if any?

Slotted ALOHA and CSMA are both ways to control when devices send data over a shared channel, but they work differently.

Slotted ALOHA sends only at fixed time slots to reduce collisions, but it still has a decent chance of two devices colliding if they pick the same slot.
CSMA is smarter, it listens first and only sends if the channel is free, which makes it more efficient and gives higher throughput with fewer collisions, especially when traffic is busy.
CSMA also usually has lower delay since it doesn’t wait for a time slot. However, Slotted ALOHA might be better for systems like satellite communication where delays are big and sensing the channel doesn’t help much. It’s also easier to implement, so it’s sometimes chosen for simpler setups even if it’s not the most efficient.

Token Ring networks avoid collisions by using a token that lets only one device send data at a time, but this also slows things down if the token is delayed or the device holding it has nothing to send, everyone else has to wait. It's also more complex to manage and more likely to break if any part of the ring fails.

In contrast, modern Ethernet uses switches connect each device directly, so there's no need to share the same path, and no collisions happen because each connection is its own separate link. This setup, called Switched Ethernet is faster, cheaper, easier to expand, and works with newer and older devices. Thanks to these improvements, Ethernet has become much better than Token Ring, even though Token Ring was designed to avoid collisions in the first place.

CRC is better than simple methods like parity bits or checksums because it’s much better at catching errors, especially when many bits get messed up together which is called burst errors. It uses math with polynomials that can spot even tricky patterns of errors, and it does this with very little extra data and fast processing.

CRC can’t fix errors by itself, it’s often used with other tools that can, like Forward Error Correction which adds extra data so the receiver can fix certain errors without asking for a resend.

In some systems, both CRC and FEC are used together, so small errors can be corrected right away, and bigger ones can be retried. This combo makes communication more reliable and efficient, especially in things like WiFi and video streaming.

4a. Show the forwarding table in router A, such that all traffic destined to host H3 is forwarded through interface 3.

Destination IP	Subnet Mask	Next Hop	Outgoing Interface
H3's IP (e.g., 10.0.0.3)	255.255.255.255	—	3
All Other Destinations	0.0.0.0	Some Router	(Default Interface)

4b. Can you write down a forwarding table in router A, such that all traffic from H1 destined to host H3 is forwarded through interface 3, while all traffic from H2 destined to host H3 is forwarded through interface 4?

To achieve this H1 → H3 interface 3, H2 → H3 interface 4, we would need: Policy-Based Routing : Allows rules based on both source and destination IPs Source Routing: Sender includes the path in the packet. Access Control Lists: Used with route-maps or filters to forward differently based on source.

• N' (visited) – the set of nodes whose shortest-path distances have been finalized .
• d(x) – the current known shortest distance from the source to node x.
• p(x) – the predecessor of x along the current shortest path

a. Compute the shortest path from t to all network nodes.

Step	N' (visited)	d(t), p(t)	d(u), p(u)	d(v), p(v)	d(w), p(w)	d(y), p(y)	d(z), p(z)
0	{t}	0, –	2, t	4, t	∞, –	7, t	∞, –
1	{t, u}	0, –	2, t	4, t	5, u	7, t	∞, –
2	{t, u, v}	0, –	2, t	4, t	5, u	7, t	15, v
3	{t, u, v, w}	0, –	2, t	4, t	5, u	7, t	15, v
4	{t, u, v, w, y}	0, –	2, t	4, t	5, u	7, t	15, v
5	{t, u, v, w, y, z}	0, –	2, t	4, t	5, u	7, t	15, v

b. Compute the shortest path from u to all network nodes.

Step	N' (visited)	d(t), p(t)	d(u), p(u)	d(v), p(v)	d(w), p(w)	d(y), p(y)	d(z), p(z)
0	{u}	2, u	0, –	3, u	3, u	∞, –	∞, –
1	{u, t}	2, u	0, –	3, u	3, u	9, t	∞, –
2	{u, t, v}	2, u	0, –	3, u	3, u	9, t	14, v
3	{u, t, v, w}	2, u	0, –	3, u	3, u	9, t	14, v
4	{u, t, v, w, y}	2, u	0, –	3, u	3, u	9, t	14, v
5	{u, t, v, w, y, z}	2, u	0, –	3, u	3, u	9, t	14, v

c. Compute the shortest path from v to all network nodes.

Step	N' (visited)	d(t), p(t)	d(u), p(u)	d(v), p(v)	d(w), p(w)	d(y), p(y)	d(z), p(z)
0	{v}	4, v	3, v	0, –	8, v	∞, –	11, v
1	{v, u}	4, v	3, v	0, –	6, u	∞, –	11, v
2	{v, u, t}	4, v	3, v	0, –	6, u	11, t	11, v
3	{v, u, t, w}	4, v	3, v	0, –	6, u	11, t	11, v
4	{v, u, t, w, y}	4, v	3, v	0, –	6, u	11, t	11, v
5	{v, u, t, w, y, z}	4, v	3, v	0, –	6, u	11, t	11, v

d. Compute the shortest path from w to all network nodes.

Step	N' (visited)	d(t), p(t)	d(u), p(u)	d(v), p(v)	d(w), p(w)	d(y), p(y)	d(z), p(z)
0	{w}	∞, –	3, w	8, w	0, –	∞, –	∞, –
1	{w, u}	5, u	3, w	6, u	0, –	∞, –	∞, –
2	{w, u, t}	5, u	3, w	6, u	0, –	12, t	∞, –
3	{w, u, t, v}	5, u	3, w	6, u	0, –	12, t	17, v
4	{w, u, t, v, y}	5, u	3, w	6, u	0, –	12, t	17, v
5	{w, u, t, v, y, z}	5, u	3, w	6, u	0, –	12, t	17, v

e. Compute the shortest path from y to all network nodes.

Step	N' (visited)	d(t), p(t)	d(u), p(u)	d(v), p(v)	d(w), p(w)	d(y), p(y)	d(z), p(z)
0	{y}	7, y	∞, –	∞, –	∞, –	0, –	∞, –
1	{y, t}	7, y	9, t	11, t	∞, –	0, –	∞, –
2	{y, t, u}	7, y	9, t	11, t	12, u	0, –	∞, –
3	{y, t, u, v}	7, y	9, t	11, t	12, u	0, –	22, v
4	{y, t, u, v, w}	7, y	9, t	11, t	12, u	0, –	22, v
5	{y, t, u, v, w, z}	7, y	9, t	11, t	12, u	0, –	22, v

f. Compute the shortest path from z to all network nodes.

Step	N' (visited)	d(t), p(t)	d(u), p(u)	d(v), p(v)	d(w), p(w)	d(y), p(y)	d(z), p(z)
0	{z}	∞, –	∞, –	11, z	∞, –	∞, –	0, –
1	{z, v}	15, v	14, v	11, z	19, v	∞, –	0, –
2	{z, v, u}	15, v	14, v	11, z	17, u	∞, –	0, –
3	{z, v, u, t}	15, v	14, v	11, z	17, u	22, t	0, –
4	{z, v, u, t, w}	15, v	14, v	11, z	17, u	22, t	0, –
5	{z, v, u, t, w, y}	15, v	14, v	11, z	17, u	22, t	0, –

When a node A wants to start a TCP connection with node B on an Ethernet LAN using a switch that initially has an empty forwarding table, a total of N+4 Ethernet frames will be sent during the TCP three-way handshake.

Reason: The first SYN packet from A is flooded by the switch to all other N−1 ports since the switch doesn’t yet know where B is, plus the original ingress frame from A , totaling N. Then B responds with a SYN+ACK, which goes to the switch and then directly to A 2 frames. Finally, A sends an ACK back to B, again through the switch as a unicast 2 more frames.

So in total, it's N (for the first SYN) + 2 (SYN+ACK) + 2 (ACK) = N+4 frames.

a. Each node knows only the distances to its immediate neighbors.

Node	d(·,A)	d(·,B)	d(·,C)	d(·,D)	d(·,E)	d(·,F)
A	0	∞	3	8	∞	∞
B	∞	0	∞	∞	2	∞
C	3	∞	0	∞	1	6
D	8	∞	∞	0	2	∞
E	∞	2	1	2	0	∞
F	∞	∞	6	∞	∞	0

b. Each node has reported the information it had in the preceding step to its immediate neighbors.

Node	d(·,A)	d(·,B)	d(·,C)	d(·,D)	d(·,E)	d(·,F)
A	0	∞	3	8	4	9
B	∞	0	3	4	2	∞
C	3	3	0	3	1	6
D	8	4	3	0	2	∞
E	4	2	1	2	0	7
F	9	∞	6	∞	7	0

c. Step (b) happens a second time.

Node	d(·,A)	d(·,B)	d(·,C)	d(·,D)	d(·,E)	d(·,F)
A	0	6	3	6	4	9
B	6	0	3	4	2	9
C	3	3	0	3	1	6
D	6	4	3	0	2	9
E	4	2	1	2	0	7
F	9	9	6	9	7	0