Find the efficiency of the ring where data rate of the link is 4 mbps, number of stations are 20, separated by 100 meters and bit delay in each station is 2.5 bits. Assume early token reinsertion with packet size of 1000 bits and transmission speed is 2 x 108 m/sec.
Answers
Answered by
7
Let a be the uniform acceleration.
v² - u² = 2 a S => S = v² / 2 a as u = 0.
final velocity = v = √(2 a S)
time taken to reach this velocity : t1 = (v - u ) / a = v / a = √(2S/a)
Now the particle moves with this velocity for distance 2 S.
so time taken t2 = 2 S / v = 2 S / √(2aS) = √(2 S / a)
Now the particle moves with a uniform deceleration - a3 for a distance 3 S.
time taken t3 to stop.
2 * a3 * 3 S = v² - u² here v = 0 and u = √(2 a S)
=> a3 = a / 3
=> t3 = - u /a3 = 3 √( 2 S / a)
total time duration = t = t1 + t2 + t3 = 5 * √(2 S / a)
total distance travelled : S + 2S + 3S =6 S
=> average speed = 3 / 5 * √(2 a S)
maximum speed = √(2 a S)
Ratio = 3/5
v² - u² = 2 a S => S = v² / 2 a as u = 0.
final velocity = v = √(2 a S)
time taken to reach this velocity : t1 = (v - u ) / a = v / a = √(2S/a)
Now the particle moves with this velocity for distance 2 S.
so time taken t2 = 2 S / v = 2 S / √(2aS) = √(2 S / a)
Now the particle moves with a uniform deceleration - a3 for a distance 3 S.
time taken t3 to stop.
2 * a3 * 3 S = v² - u² here v = 0 and u = √(2 a S)
=> a3 = a / 3
=> t3 = - u /a3 = 3 √( 2 S / a)
total time duration = t = t1 + t2 + t3 = 5 * √(2 S / a)
total distance travelled : S + 2S + 3S =6 S
=> average speed = 3 / 5 * √(2 a S)
maximum speed = √(2 a S)
Ratio = 3/5
Similar questions