Computer Networks

The network layer decides the path a packet takes from network to network, so (a)-(I) Packet routing. The transport layer works end to end, carrying data between the source and destination hosts, so (b)-(III) Host to host communication. The data link layer packs bits into frames and detects and handles errors on each single link, so (c)-(II) Framing and error handling.

Answer: B

In the 3-way handshake the client opens with a SYN segment (P1: SYN = 1, ACK = 0). The server replies with SYN + ACK (P2: SYN = 1, ACK = 1), and the client finishes with a pure acknowledgement (P3: SYN = 0, ACK = 1).

• (A) False: P3 has SYN = 0.
• (B) True: P2 acknowledges the client’s SYN and carries the server’s own SYN.
• (C) False: P2 must have SYN = 1.
• (D) True: P1 is the connection request, so SYN = 1.

Answer: B, D

A router uses longest prefix match: among the entries that match the destination, the one with the longest mask wins. The first two octets (145.36) agree with every listed subnet, so only the third octet matters: \(109 = 01101101_2\).

• /16 (145.36.0.0): matches.
• /17 (145.36.128.0): needs the third octet to begin with bit 1 (128 to 255); 109 begins with 0, so no match.
• /18 (145.36.64.0): needs the third octet to begin with 01 (64 to 127); 109 does, so it matches.
• /24 (145.36.255.0): needs the third octet to be 255, so no match.

Of the matches (/16, /18 and the default route), /18 is the longest, so the packet is sent on E3.

Answer: A

The message arrives error-free only if none of its 5 bits flips. Each bit survives with probability \(1 - 0.01 = 0.99\), and the bits are independent, so the probabilities multiply.

\[P = (0.99)^5 = 0.95099 \approx 0.951\]

Answer: 0.951

Use the standard 20-byte IPv4 header. The data in every fragment except the last must be a multiple of 8 bytes, and routers never reassemble: each router only splits the packets it receives.

• Router-1 (MTU 5000): at most \(5000 - 20 = 4980\) data bytes, i.e. 4976 as a multiple of 8. The 15000 bytes become \(4976 + 4976 + 4976 + 72\): 4 fragments.
• Router-2 (MTU 3000): at most \(3000 - 20 = 2980\), i.e. 2976 data bytes. Each 4976-byte fragment splits into \(2976 + 2000\), but the last fragment (72 data bytes, 92 bytes with its header) already fits and passes through unchanged.

Fragments delivered \(= 3 \times 2 + 1 = 7\). The tempting \(4 \times 2 = 8\) wrongly splits the small fourth fragment again; even ignoring the multiple-of-8 rule (4980, 4980, 4980, 60) the count is 7.

Answer: 7

(i) is TRUE. A host that knows the IP address of a machine on its local network still needs that machine’s link-layer (MAC) address to build a frame. ARP broadcasts a request for that IP address and gets back the matching hardware address.

(ii) is FALSE. TCP is full duplex, and every segment header has an acknowledgement number field, so a segment carrying data from S to R can also acknowledge data received from R. This is called piggybacking, and it is the normal case when both sides are sending.

Answer: B

In distance vector routing (used by RIP), a router learns only the distance vectors of its neighbours. It keeps improving its own estimates with the Bellman-Ford relation \(D_x(y) = \min_v \{c(x, v) + D_v(y)\}\): the cost to a neighbour \(v\) plus that neighbour's distance to \(y\). So (i) matches (a).

In link state routing (used by OSPF), every router floods the costs of its own links, so each router learns the whole topology and runs Dijkstra's shortest path algorithm from itself. So (ii) matches (b).

The other options pair at least one protocol family with the wrong algorithm.

Answer: A

The protocol field names the protocol whose message sits directly in the payload of the IP datagram, so X must be a protocol that runs immediately on top of IP.

• ICMP (protocol number 1) and IGMP (2) are carried directly in IP datagrams.
• OSPF also runs directly over IP, with protocol number 89.
• RIP does not. Its messages travel inside UDP datagrams on port 520, so a datagram carrying RIP has 17 (UDP) in its protocol field, and RIP is identified only by the UDP port.

So RIP cannot be X.

Answer: D

IP fragments the whole UDP segment, header included: \(8 + 7488 = 7496\) bytes of IP data.

A fragment has a 20-byte IPv4 header and must fit in 1500 bytes, so it carries at most 1480 data bytes. Since 1480 is a multiple of 8, as fragment offsets require, every fragment except the last carries exactly 1480 bytes.

• Five full fragments carry \(5 \times 1480 = 7400\) bytes.
• The remaining \(7496 - 7400 = 96\) bytes form a sixth fragment of \(96 + 20 = 116\) bytes.

Leaving out the UDP header gives 88 data bytes (108 bytes with the IP header), the trap in options B and C.

Answer: D

In Stop-and-Wait the sender transmits one frame and then waits idle until the acknowledgement comes back, so one cycle lasts \(T_t + 2T_p\) (the ACK and processing times are negligible).

• Transmission time: \(T_t = \frac{3000}{2000} = 1.5\) s.
• Round-trip propagation: \(2T_p = 2 \times 0.1 = 0.2\) s.

Utilization \(= \frac{T_t}{T_t + 2T_p} = \frac{1.5}{1.7} \approx 0.8824\), about 88.2%, which is option A. Option B (93.75) comes from counting the propagation delay only one way: \(\frac{1.5}{1.6} = 0.9375\).

Answer: A