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zalisa [80]
2 years ago
11

Suppose a 4096 bit message is sent along a 10Mbit/sec transmission link. Suppose that the latency between the sender and the rec

eive is 4 milliseconds. What is message transmission time?
Computers and Technology
1 answer:
umka21 [38]2 years ago
8 0

Answer: 8.1 msec.

Explanation:

If we are told that we have a transmission link of sending 10 Mbits in one second, setting up a direct proportion, we can find out how much time is needed to send 4096 bits along the same channel, as follows:

10⁶ bits = 1 sec

4096 bits = x  ⇒ x = 4096 bits. 1 sec / 10⁶ bits = 4.1 msec.

As we have 4 msec of latency between sender and the receiver, we need to add these 4 msec to the transit time, so we have a total message transmission time of 8.1 msec.

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def getpointValue(subject):

   if subject >= 75:        point = 4; grade = 'A'

   elif subject >= 60:        point = 3; grade ='B'

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subjects= ["Accounting","Marketing","Economics","MIS"]

acct = int(input(subjects[0]+": "))

mkt = int(input(subjects[1]+": "))

eco = int(input(subjects[2]+": "))

mis = int(input(subjects[3]+": "))

acctgrade = getpointValue(acct)

print(subjects[0]+" Grade: "+str(acctgrade[1]))

mktgrade = getpointValue(mkt)

print(subjects[1]+" Grade: "+str(mktgrade[1]))

ecograde = getpointValue(eco)

print(subjects[2]+" Grade: "+str(ecograde[1]))

misgrade = getpointValue(mis)

print(subjects[3]+" Grade: "+str(misgrade[1]))

totalpoint = (acctgrade[0] + mktgrade[0] + ecograde[0] + misgrade[0]) * 3

gpa = totalpoint/12

print("GPA: "+str(gpa))

Explanation:

<em>The solution I provided uses a function to return the letter grade and the point of each subject</em>

<em>I've added the explanation as an attachment where I used comment to explain difficult lines</em>

<em />

Download txt
6 0
3 years ago
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