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3 years ago

____ is the process of drawing a series of increasingly detailed DFDs, until all functional primitives are identified.

Computers and Technology

1 answer:

cupoosta [38]3 years ago

8 0

Answer:

a. Leveling

Explanation:

Leveling is also called decomposing.
It is a method to depict more detail in lower-level Data Flow Diagrams
It is the technique to expand or decompose DFDs into more details or levels to extract more information and identify all functional primitives until no further detail is required.
An information system can be comprised of multiple processes. Leveling partitions each process into sub-processes which identifies more details about the process with each level.
0-level DFD: it starts with context diagram which depicts the whole system as a single process e.g. Grading system.
1-level DFD: the context diagram is decomposed in this level. This means a single process is expanded into multiple processes which extracts more details about a process of a system.
2- level DFD: this level decomposes the the process of level 1 going into further details.
This decomposition and expansion continues until no further details are required and all functional primitives are identified.

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Assume a TCP sender is continuously sending 1,090-byte segments. If a TCP receiver advertises a window size of 5,718 bytes, and

Arturiano [62]

Answer:

for the 5 segments, the utilization is 3.8%

Explanation:

Given the data in the question;

segment size = 1090 bytes

Receiver window size = 5,718 bytes

Link transmission rate or Bandwidth = 26 Mbps = 26 × 10⁶ bps

propagation delay = 22.1 ms

so,

Round trip time = 2 × propagation delay = 2 × 22.1 ms = 44.2 ms

we determine the total segments;

Total segments = Receiver window size / sender segment or segment size

we substitute

Total segments = 5718 bytes / 1090 bytes

Total segments = 5.24587 ≈ 5

Next is the throughput

Throughput = Segment / Round trip

Throughput = 1090 bytes / 44.2 ms

1byte = 8 bits and 1ms = 10⁻³ s

Throughput = ( 1090 × 8 )bits / ( 44.2 × 10⁻³ )s

Throughput = 8720 bits / ( 44.2 × 10⁻³ s )

Throughput = 197.285 × 10³ bps

Now Utilization will be;

Utilization = Throughput / Bandwidth

we substitute

Utilization = ( 197.285 × 10³ bps ) / ( 26 × 10⁶ bps )

Utilization = 0.0076

Utilization is percentage will be ( 0.0076 × 100)% = 0.76%

∴ Over all utilization for the 5 segments will be;

⇒ 5 × 0.76% = 3.8%

Therefore, for the 5 segments, the utilization is 3.8%

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3 years ago

To what extent do you think st. cyprian was effective in his efforts to bring inherited christian teachings to bear on the unpre

Pepsi [2]

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3 years ago

What are your thoughts on the last nintendo direct

allsm [11]

Answer:

I don't know? How do you feel about them?

Explanation:

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2 years ago

Read 2 more answers

What special signal is issued periodically from an Access Point and contains the network transmission rate and service set ident

olya-2409 [2.1K]

Answer:

beacon frame

Explanation:

Beacon frame is a management frame In computer networks, known to be in IEEE 802.11 based WLANs. These frames are transmitted periodically and they contain all the information a station will require before it can rightly transmit a frame.

When it comes to announcing the presence of devices in a wireless computer network (WLAN), Beacon frames are used, and they can also be used in the synchronization of the devices and services

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2 years ago

Write a function that receives two numbers, m and n and calculates and displays the sum of the integers from m to n. For example

Evgesh-ka [11]

Answer:

The program to this question as follows:

Program:

//header file

#include <stdio.h> //defining header file

void inc(int m,int n) //defining method inc

{

int sum, i; //declaring variable

sum=m; //holding value of m variable

for(i=++m;i<=n;i++) //loop for calculate number between given range

{

sum=sum+i; //adding value

}

printf("sum of the integer is : %d",sum);//print value

}

int main() //defining method

{

int m,n; //defining integer variable

printf("Enter m value: "); //message

scanf("%d",&m); //input value by user in variable

printf("Enter n value: "); //message

scanf("%d",&n);//input value by user in variable

inc(m,n); //calling method

return 0;

}

Output:

Enter m value: 3

Enter n value: 7

sum of the integer is :25

Explanation:

In the above code, an "inc" function is declared, that accepts integer parameters that are "m and n", inside the method two integer variable "sum and i" is declared, which is used in the loop.

The loop uses the user parameter to count value and uses the sum variable to calculate there sum.
In the main method, two integer variable n and m are declared, which take value from the user end, and at the last, we call the inc method, that prints sum value.

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2 years ago