Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1 Gbps. What is the queueing delay of packet number 48 ?(in milliseconds, rounded to one decimal place, e. g. 0.01234 seconds would be entered as "12.3")

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of n-th packet is :  (n - 1) × L/R

where L :  packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate =  2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

answer: 500 wph.

explanation:

the equation above shows how temperature f, measured in degrees fahrenheit, relates to a temperature c, measured in degrees celsius. based on the equation, which of the following must be true?

a temperature increase of 1 degree fahrenheit is equivalent to a temperature increase of

5

9

degree celsius.

a temperature increase of 1 degree celsius is equivalent to a temperature increase of 1.8 degrees fahrenheit.

a temperature increase of

5

9

degree fahrenheit is equivalent to a temperature increase of 1 degree celsius.

a) i only

b) ii only

c) iii only

d) i and ii only