Computer networks miscellaneous
- Station A needs to send a message consisting of 9 packets to Station B using a sliding window (window size 3) and goback-n error control strategy. All packets are ready and immediately available for transmission. If every 5th packet that A transmits gets lost (but no acts from B ever get lost), then what is the number of packets that A will transmit for sending the message to B?
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Total 16 packets are sent. See following table for sequence of events. Since go-back-n error control strategy is used, all packets after a lost packet are sent again.
Correct Option: C
Total 16 packets are sent. See following table for sequence of events. Since go-back-n error control strategy is used, all packets after a lost packet are sent again.
- In a token ring network, the transmission speed is 107 bps and the propagation speed is 200 m/ms. The 1-bit delay in this network is equivalent to
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107 bits require time = 1 s
⇒ 1 Bit require = 1/107 s (unitary method)⇒ 1 = 10-6 / 10 s = 1 ms 10 × 106 10
Now, 1 ms = 200 m cable
⇒ 1/10 ms = 20 m cableCorrect Option: C
107 bits require time = 1 s
⇒ 1 Bit require = 1/107 s (unitary method)⇒ 1 = 10-6 / 10 s = 1 ms 10 × 106 10
Now, 1 ms = 200 m cable
⇒ 1/10 ms = 20 m cable
- A computer on a 10 Mbps network is regulated by a token bucket. The token bucket is filled at a rate of 2 Mbps. It is initially filled to capacity with 16 Megabits. What is the maximum duration for which the computer can transmit at the full 10 Mbps?
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Let the duration of transmission be x s.
Transmission Network of Computer by y s
Network initially filled with p Megabits
Initial rate of transmission be q per second
Now, from the above, we can easily formulate
y + qx = xy ⇒ y = x (y – q)
⇒ 16 = x(10 – 2) ⇒ x = 2Correct Option: B
Let the duration of transmission be x s.
Transmission Network of Computer by y s
Network initially filled with p Megabits
Initial rate of transmission be q per second
Now, from the above, we can easily formulate
y + qx = xy ⇒ y = x (y – q)
⇒ 16 = x(10 – 2) ⇒ x = 2
- Consider a source computer (S) transmitting a file of size 106 bit to a destination computer (D) over a network of two routers (R1 and R2) and three links (L1 , L2 and L3 ). L1 connects S to R1 ; L2 connects R1 to R2 and L3 connects R2 to D. Let each link be of length 100 km. Assume signals travel over each link at a speed of 108 m/s. Assume that the link bandwidth on each link is 1 Mbit/s. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D?
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Transmission delay for 1st packet from each of S, R1 and R2 will take 1 ms Propagation delay on each link l1 , l2 and l3 for one packet is 1 ms Therefore the sum of transmission delay and propagation delay on each link for one packet is 2 ms. The first packet reaches the destination at 6th ms The second packet reaches the destination at 7th ms So, inductively we can say that 1000th packet reaches the destination at 1005th ms.
Correct Option: A
Transmission delay for 1st packet from each of S, R1 and R2 will take 1 ms Propagation delay on each link l1 , l2 and l3 for one packet is 1 ms Therefore the sum of transmission delay and propagation delay on each link for one packet is 2 ms. The first packet reaches the destination at 6th ms The second packet reaches the destination at 7th ms So, inductively we can say that 1000th packet reaches the destination at 1005th ms.
- Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 106 bytes/ sec. A user on host A sends a file of size 103 bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?
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I Transmission
1000 + 100 = 1100 bytes (transmitted from A at a time)
In 1 second → 106 bytes∴ For 1100 bytes ⇒ 1100 × 8 = 8.8 m sec.(transmission time) 106
From A, R1 and R2 it takes T1 = 8.8 + 8.8 + 8.8 = 26.4 m sec. (to reach B)
II Transmission
100 + 100 = 200 bytes (transmitted 10 times from A)For 200 bytes ⇒ 200 × 8 = 1.6 m sec. 106
From A, transmission time to entire packet = 1.6 × 10 = 16 m sec.
From A, R1 and R2 it takes T2 = 16 + 1.6 + 1.6 = 19.2 m sec. (to reach B)
III transmission
50 + 100 = 150 bytes transmitted 20 times from A.For 150 bytes ⇒ 150 × 8 = 1.2 m sec. 106
From A, transmission time for entire packet = 1.2 × 20 = 24 m sec.
From A, R1 and R2 it takes T3 = 24 + 1.2 + 1.2 = 26.4 m sec. (to reach B)
∴ T1 = T3 = 26.4 m sec.
T2 = 19.6 m sec. < T3Correct Option: D
I Transmission
1000 + 100 = 1100 bytes (transmitted from A at a time)
In 1 second → 106 bytes∴ For 1100 bytes ⇒ 1100 × 8 = 8.8 m sec.(transmission time) 106
From A, R1 and R2 it takes T1 = 8.8 + 8.8 + 8.8 = 26.4 m sec. (to reach B)
II Transmission
100 + 100 = 200 bytes (transmitted 10 times from A)For 200 bytes ⇒ 200 × 8 = 1.6 m sec. 106
From A, transmission time to entire packet = 1.6 × 10 = 16 m sec.
From A, R1 and R2 it takes T2 = 16 + 1.6 + 1.6 = 19.2 m sec. (to reach B)
III transmission
50 + 100 = 150 bytes transmitted 20 times from A.For 150 bytes ⇒ 150 × 8 = 1.2 m sec. 106
From A, transmission time for entire packet = 1.2 × 20 = 24 m sec.
From A, R1 and R2 it takes T3 = 24 + 1.2 + 1.2 = 26.4 m sec. (to reach B)
∴ T1 = T3 = 26.4 m sec.
T2 = 19.6 m sec. < T3
