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TiliK225 [7]
3 years ago
9

We have noted that there can be no input queuing if the switching fabric is n times faster than the input line rates, assuming n

input lines all have the same rate. Explain why this is the case.
Computers and Technology
1 answer:
Ostrovityanka [42]3 years ago
7 0

Consider the given data:

Assume packet length=n

Maximum queuing delay= (n–1)D

All packets are of the same length, n packets arrive at the same time to the n input ports, and all n packets want to be forwarded to different output ports.

a)  The maximum delay for a packet for the memory = (n-1)D

b)  The maximum delay for a packet for the bus = (n-1)D

c)  The maximum delay for a packet for the crossbar switching fabrics=  0

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Please help me with question 1!
BaLLatris [955]

12,831 is the answer but if you care for the work then here it is:

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3 years ago
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The equation of certain traveling waves is y(x.t) = 0.0450 sin(25.12x - 37.68t-0.523) where x and y are in
NNADVOKAT [17]

Answer:

A. 0.0450

B. 4

C. 0.25

D. 37.68

E. 6Hz

F. -0.523

G. 1.5m/s

H. vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

I. -1.67m/s.

Explanation:

Given the equation:

y(x,t) = 0.0450 sin(25.12x - 37.68t-0.523)

Standard wave equation:

y(x, t)=Asin(kx−ωt+ϕ)

a.) Amplitude = 0.0450

b.) Wave number = 1/ λ

λ=2π/k

From the equation k = 25.12

Wavelength(λ ) = 2π/25.12 = 0.25

Wave number (1/0.25) = 4

c.) Wavelength(λ ) = 2π/25.12 = 0.25

d.) Angular frequency(ω)

ωt = 37.68t

ω = 37.68

E.) Frequency (f)

ω = 2πf

f = ω/2π

f = 37.68/6.28

f = 6Hz

f.) Phase angle(ϕ) = -0.523

g.) Wave propagation speed :

ω/k=37.68/25.12=1.5m/s

h.) vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

(i) vy(3.5m, 21s) = 0.045(-37.68) cos (25.12*3.5-37.68*21-0.523) = -1.67m/s.

5 0
4 years ago
A collision between gas atoms and electrons raises the energy levels of oxygen and nitrogen in the _________.
Gala2k [10]
Ionosphere i hope this helps
4 0
4 years ago
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Write a generator function named count_seq that doesn't take any parameters and generates a sequence that starts like this: 2, 1
-BARSIC- [3]

Answer:

#required generator function

def count_seq():

   #starting with number 2. using string instead of integers for easy manipulation

   n='2'

   #looping indefinitely

   while True:

       #yielding the integer value of current n

       yield int(n)

       #initializing an empty string

       next_value=''

       #looping until n is an empty string

       while len(n)>0:

           #extracting first digit (as char)

           first=n[0]

           #consecutive count of this digit

           count=0

           #looping as long as n is non empty and first digit of n is same as first

           while len(n)>0 and n[0]==first:

               #incrementing count

               count+=1

               #removing first digit from n

               n=n[1:]

           #now appending count and first digit to next_value

           next_value+='{}{}'.format(count,first)

       #replacing n with next_value

       n=next_value

#testing, remove if you don't need this

if __name__ == '__main__':

   #creating a generator from count_seq()

   gen=count_seq()

   #looping for 10 times, printing next value

   for i in range(10):

       print(next(gen))

Explanation:

6 0
3 years ago
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