Test: Computer Networks - 2 - Computer Science Engineering (CSE) MCQ

Given:

Message size = 2400 bits.

The header of the transport layer = 150 bits.

Segment size at transport layer = (Header + Message size) = 150 + 2400 = 2550 bits.

Maximum, transmission unit at destination network = 900 bits

 So, At a time, data supported by the destination network = 900 – 26 = 874 bits.

Hence, 2550 bits are divided into packets having a maximum of 874 bits. So:

Length of packet 1 = 874 bits.

Length of packet 2 = 874 bits.

Length of packet 3 = 802 bits.

Total  = 2550 bits.

Network layer packets are transmitted via two networks, each of which uses a 26-bit header. So, The number of bits, including headers delivered to the destination network is:

Packet 1 (Data + Header) size = 26 + 26 + 874 = 926 bits.

Packet 2 (Data + Header) size = 26 + 26 + 874 = 926 bits.

Packet 3 (Data + Header) size = 26 + 26 + 802 = 854 bits.

 Total = 2706 bits. 

Concept:

Maximum number of errors that can be corrected= ⌊ (Minimum Hamming distance – 1) ÷ 2 ⌋ 

Hamming distance: Number of positions at which each corresponding bits are different

Explanation:

Minimum Hamming distance is 4 more than one combination is here one of them is distance between 11001100 and 00000000 = 4

Maximum number of errors that can be detected =  ⌊ (Minimum Hamming distance – 1) ÷ 2 ⌋  =  ⌊( 4 – 1) ÷ 2 ⌋ = 1

Data = 4400 Byte

Router MTU = 900 byte

Data + header = 900

Data = 900 – 20

∴ Data = 880

Number of fragments = 4400/880 = 5
Since initial fragment offset is 100.
Fragment offset of 2^nd fragment = 100 + 880/8 = 210
Fragment offset of 3rd fragment =  210 + 880/8 = 320
Fragment offset of 4th fragment =  320 + 880/8 = 430
Fragment offset of 5^th fragment = 430 + 880/8 = 540
penultimate fragment = 2nd last fragment = 430
Shortcuts:
penultimate fragment = Initial fragment offset + 110 × (n - 2)
where n is number of fragments

PAT (Port Address Translation):

• Allows a one to many approaches to network address translation (NAT)
• Permits multiple devices on a LAN to be mapped to a single public IP address.
• Goal of PAT is to conserve IP address
• Most home networks use PAT.
• Thousands of users can be connected to the internet by using one public address.

Static NAT:

• A single private IP address is mapped with single public IP address.
• Used in web hosting.
• Private IP address is translated to a public IP address.

Dynamic NAT:

• Multiple private IP address are mapped to a pool of public IP address.
• Used when we know the number of fixed user wants to access the internet at any time.

Explanation:

From above concepts we can say that PAT is used can be used to have one IP address allow many users to connect to the global internet.

Given:

Maximum payload = 1200 bytes per frame of first network.

Maximum payload = 400 bytes per frame of second network.

Per packet IP overhead = 26 bytes

Now,

1^st network,

2100 bytes will be divided into 2 packets of size 1200 and 900 bytes.

∴ IP overhead of 1^st network (2 × 26) = 52 bytes.

2^nd network,

2100 bytes will be divided into 6 packets of 5 packets having size 400 bytes and 1 packet having size 100 byte.

∴ IP overhead of 2^nd network (6 × 26)

= 156 bytes

Total IP overhead = (52 + 156) = 208 bytes

In Pure Aloha, Efficiency = 18.4%

Usable bandwidth for 18.2 kbps = 18.2 × 0.18 = 3.276 kbps

Therefore, the maximum throughput of Pure Aloha

= 1/2e × 3.276

= (18.4 × 3.276) / 100

= 0.6027

The maximum efficiency of slotted aloha formula is G * e^-G.

Given, G = 1

= 1 x e-1

= 1 / e

= 0.368

= 36.8%

Concept: 

The Routers only contains 3 layer that's the reason sometimes they also called as 3-layer switch.

The Router is a physical or virtual internetworking device that is designed to receive, analyze, and forward data packets between computer networks. A router examines a destination IP address of a given data packet, and it uses the headers and forwarding tables to decide the best way to transfer the packets.

At Intermediate Router R the packet will not go beyond the network layer. Hence Network layer will only be visited once at both the routers.
But the Data link layer will be visited twice.
Hence Network layer will be visited 4 times and the Data link layer will be visited 6 times (1 + 2 (at R) + 2 (at R) + 1 = 6 )

Data:

Number of packets = n =1000

Size of each Packet = L = 1000 bits

The link bandwidth on each link is = BW = 1Mbps =10⁶ bits/s

Each link be of length = d= 100 km

Signals travel over each link at a speed = v= 108m/s

Transmission delay = T

Propagation delay = P

Formula:

Calculation:

Propagation delay for the first packet to transmit S to R₁

So, the propagation delay for the first packet to transmit S to D

P =1 × 3 = 3ms

Transmission delay for the first packet to transmit S to R₁

So, the transmission delay for the first packet to transmit S to D

T=1 × 3 = 3ms

So, total time is taken to transmit the first packet from S to D =P +T=6ms

Using pipelining, the remaining packet follows the first packet

then remaining every packet take 1ms to transmit from S to D

So, the total time is taken to transmit all 1000 packets from S to D is

⇒ time for the first packet + time for the remaining 999 packets

⇒ 6ms + 999ms

⇒ 1005ms

So, the total time is taken to transmit all 1000 packets from S to D is 1005ms.

The maximum efficiency of pure aloha formula is G * e^-2G.

Given, G =1/2

=1/2 x e^-2 x 1/2

= 1 / 2e

= 0.184

= 18.4%