Computer Networks Test 4

The relation between the MF and fragment offset is given below:

Statement 1:  False.

• A router need not necessarily implement any (RIP or OSPF) routing protocol .
• If no routing protocol is specified for a router, default routing with address 0.0.0.0 is implemented.

Statement 2:  False

• Re-assembly of IP fragments into a single IP packet is done only at the receiver site.

Therefore both statement are not true

Data:

Packet size = 1850

Maximum transmission unit= 160 bytes

Header length = (1010)₂ = (10)₂ = 10 ×4(scaling factor) = 40bytes 

Calculation: 

Data length = 1850 - 40 = 1810 bytes

Data of one packet = 160 - 40 = 120

Number of fragments = \(\lceil\frac{{1810}}{{120}}\rceil=16\)

Concept:

Record route record the routers an IP packet takes as it propagates from source to destination device.

Format:

In IPv4, for options, 40 bytes are reserved. 3 bytes are already used 1 for code, 1 for length and 1 for the pointer. So only 37 bytes are remaining.

An IP address is of 4 bytes (32 bits)

Router addresses that can be listed in record route option field = \(floor\left( {\frac{{37}}{4}} \right) = 9\)

Statement P: Incorrect

RIP uses distance vector routing (DVR) protocol which employs the hop count as a routing metric and RIP uses the UDP as its transport protocol with port no 520.

Statement Q and R: Incorrect

OSPF uses link-state routing (LSR) protocol works within a single Autonomous System. OSPF encapsulates its data directly into IP Packets and does not use either TCP or UDP. Therefore, OSPF is not send using UDP packet

File size (data) = 5000 byte

Header size = 100 byte

Bandwidth (BW) = 2 × 10⁶ byte/second

First case,

\(Transmission\;time\left( {{T_t}} \right) = \frac{{data\; + \;header}}{{BW}}\;\)

\({T_t}\; = \;\frac{{5000 + 100}}{{2\; \times \;{{10}^6}}}\; = \;2.55\;ms\)

A → R1 → R2 → R3 → B

4 hops are needed

T1 = 4 × 2.55 ms

∴ T1 = 10.2 ms

Second case,

Data per packet = 5000 ÷ 20 = 250 B

Header size = 100 B

\(Transmission\;time\left( {{T_t}} \right) = \frac{{data\; + \;header}}{{BW}}\;\)

\({T_t}\; = \;\frac{{250 + 100}}{{2\; \times \;{{10}^6}}} = 0.175\;ms\)

X → R1 → R2 → R3 → Y

4 hops are needed

∴ time needed for 1^st packet = 4 × 0.175 ms = 0.7 ms

And time needed for remaining 19 packets = 19 × 0.175 = 3.325 ms

∴ T2 = 0.7 + 3.325 = 4.025 ms

Third case,

Data per packet = 5000 ÷ 40 = 125 B

Header size = 100 B

\(Transmission\;time\left( {{T_t}} \right)\; = \;\frac{{data\; + \;header}}{{BW}}\;\)

\({T_t} = \frac{{125 + 100}}{{2\; \times \;{{10}^6}}} = 0.1125\;ms\)

X → R1 → R2 → R3F → Y

4 hops are needed

∴ time needed for 1^st packet = 4 × 0.1125 ms = 0.45 ms

And time needed for remaining 19 packets = 39 × 0.1125 = 4.3875 ms

Data:

Length of packet = 1000 byte

Bandwidth (BW) = 2 Gbps

Propagation delay = Tp = 200 μs

d1 = 10 km, d2 = 5 km

Propagation speed = v m/s

Formula:

\(\frac{{{{\rm{d}}_1}}}{v} + \frac{{{d_2}}}{v} = {T_p}\)

Calculation:

\(\frac{{10 × {{10}^3}\;m}}{v} + \frac{{5 × {{10}^3}}}{v} = 200 × {10^{ - 6}}\)

\(v = \frac{{15 × {{10}^3}}}{{200 × {{10}^{ - 6}}}} = 75 ×10^6 m/s\)

v = 75 × 10³ km/s

Link state routing:

• Link state routing is based on Dijkstra’s algorithm.
• In Link state routing, if a packet with a sequence number higher than the highest one seen so far ever arrives, then it has more recent data, so it is used to update. If the sequence number is lower than the highest then it is obsolete and so rejected.
• Flooding technique is used in Link State Routing

Distance vector routing:

• Distance vector routing is based on the bellman ford algorithm or ford Fulkerson algorithm. 
• In distance vector routing only neighbors are updated. Distance vector routing is an adaptive routing algorithm which changes routing decision to reflect changes in topology and traffic.

Therefore statements S₁, S₂,S4, and S5, are false and only S₃ is true.