SQL
September 8, 2024Computer-Networks
September 8, 2024Computer-Networks
| Question 184 |
Two hosts are connected via a packet switch with 107 bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 10000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is _______.
| (1575) | |
| (1576) | |
| (1577) | |
| (1578) |
Question 184 Explanation:
Given that there is a single switch between the source and destination.
Bandwidth of the each link is 107 bits/sec.
Given that propagation delay between host to switch and switch to host is same i.e. 20 microseconds.
Given that 35 microseconds of buffering time is required by the switch.
Total data we need to send is 10000 bits.
Given that packet size is 5000 bits.
Number of packets we need to send is 10000/5000 = 2 packets.
Total time for the first packet to reach from source to switch = Transmission time + propagation time
Transmission time=(5000 bits)/(107 bits/sec) = 500 microseconds
Total time between source to switch is = 500 + 20 = 520 microseconds
Time required for the packet to reach from switch to destination is
= Buffer time+Transmission time + propagation time = 35 + 500 + 20 = 555 microseconds.
Total time required for the first packet to reach to destination from the source = 520 + 555 =1075 microseconds.
While transferring first packet from switch to destination source starts sending of it’s second packet to switch, that means from 1055 microsecond on wards transmission of 2nd packet is started by the source.
By the time first packet is reached to the destination 2nd packet is completely available at the switch.
Time required for the 2nd packet to reach to the destination from the switch is
= transmission time + propagation time = 500+20 = 520 microseconds
Therefore total time required for the two packets to reach to destination from the source
= 1055+ 520 = 1575 microseconds.
Bandwidth of the each link is 107 bits/sec.
Given that propagation delay between host to switch and switch to host is same i.e. 20 microseconds.
Given that 35 microseconds of buffering time is required by the switch.
Total data we need to send is 10000 bits.
Given that packet size is 5000 bits.
Number of packets we need to send is 10000/5000 = 2 packets.
Total time for the first packet to reach from source to switch = Transmission time + propagation time
Transmission time=(5000 bits)/(107 bits/sec) = 500 microseconds
Total time between source to switch is = 500 + 20 = 520 microseconds
Time required for the packet to reach from switch to destination is
= Buffer time+Transmission time + propagation time = 35 + 500 + 20 = 555 microseconds.
Total time required for the first packet to reach to destination from the source = 520 + 555 =1075 microseconds.
While transferring first packet from switch to destination source starts sending of it’s second packet to switch, that means from 1055 microsecond on wards transmission of 2nd packet is started by the source.
By the time first packet is reached to the destination 2nd packet is completely available at the switch.
Time required for the 2nd packet to reach to the destination from the switch is
= transmission time + propagation time = 500+20 = 520 microseconds
Therefore total time required for the two packets to reach to destination from the source
= 1055+ 520 = 1575 microseconds.
Correct Answer: A
Question 184 Explanation:
Given that there is a single switch between the source and destination.
Bandwidth of the each link is 107 bits/sec.
Given that propagation delay between host to switch and switch to host is same i.e. 20 microseconds.
Given that 35 microseconds of buffering time is required by the switch.
Total data we need to send is 10000 bits.
Given that packet size is 5000 bits.
Number of packets we need to send is 10000/5000 = 2 packets.
Total time for the first packet to reach from source to switch = Transmission time + propagation time
Transmission time=(5000 bits)/(107 bits/sec) = 500 microseconds
Total time between source to switch is = 500 + 20 = 520 microseconds
Time required for the packet to reach from switch to destination is
= Buffer time+Transmission time + propagation time = 35 + 500 + 20 = 555 microseconds.
Total time required for the first packet to reach to destination from the source = 520 + 555 =1075 microseconds.
While transferring first packet from switch to destination source starts sending of it’s second packet to switch, that means from 1055 microsecond on wards transmission of 2nd packet is started by the source.
By the time first packet is reached to the destination 2nd packet is completely available at the switch.
Time required for the 2nd packet to reach to the destination from the switch is
= transmission time + propagation time = 500+20 = 520 microseconds
Therefore total time required for the two packets to reach to destination from the source
= 1055+ 520 = 1575 microseconds.
Bandwidth of the each link is 107 bits/sec.
Given that propagation delay between host to switch and switch to host is same i.e. 20 microseconds.
Given that 35 microseconds of buffering time is required by the switch.
Total data we need to send is 10000 bits.
Given that packet size is 5000 bits.
Number of packets we need to send is 10000/5000 = 2 packets.
Total time for the first packet to reach from source to switch = Transmission time + propagation time
Transmission time=(5000 bits)/(107 bits/sec) = 500 microseconds
Total time between source to switch is = 500 + 20 = 520 microseconds
Time required for the packet to reach from switch to destination is
= Buffer time+Transmission time + propagation time = 35 + 500 + 20 = 555 microseconds.
Total time required for the first packet to reach to destination from the source = 520 + 555 =1075 microseconds.
While transferring first packet from switch to destination source starts sending of it’s second packet to switch, that means from 1055 microsecond on wards transmission of 2nd packet is started by the source.
By the time first packet is reached to the destination 2nd packet is completely available at the switch.
Time required for the 2nd packet to reach to the destination from the switch is
= transmission time + propagation time = 500+20 = 520 microseconds
Therefore total time required for the two packets to reach to destination from the source
= 1055+ 520 = 1575 microseconds.