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balu736 [363]
1 year ago
12

Suppose that the following processes arrive for execution at the times indicated. Each process will run for the amount of time l

isted. In answering the questions, use non-preemptive scheduling, and base all decisions on the information you have at the time the decision must be made.
Process Arrival Time Burst Time
P1 0.0 8
P2 0.4 4
P3 1.0 1
a) What is the average turnaround time for these processes with the FCFS scheduling algorithm?

b) What is the average turnaround time for these processes with the SJF scheduling algorithm?
Computers and Technology
1 answer:
Len [333]1 year ago
5 0

Answer:

a) 10.53

b) 9.53

Explanation:

a) Average Turnaround Time: ( (8-0)+(12-0.4)+(13-1.0) ) / 3 = 10.53

b) Average Turnaround Time: ( (8-0)+(13-0.4)+(9-1.0) ) / 3 = 9.53

You might be interested in
Which decimal value (base 10) is equal to the binary number 1012?
GaryK [48]

Answer:

The decimal value of 101₂² base 2 = 25 base 10

Explanation:

The number 101₂² in decimal value is found as follows;

101₂ × 101₂ = 101₂ + 0₂ + 10100₂

We note that in 101 + 10100 the resultant 2 in the hundred position will have to be converted to a zero while carrying over 1 to the thousand position to give;

101₂ + 0₂ + 10100₂ = 11001₂

Therefore;

101₂² =  11001₂

We now convert the result of the square of the base 2 number, 101², which is 11001₂ to base 10 as follows;

Therefore converting 11001₂ to base 10 gives;

11001₂= 1 × 2⁴ + 1 × 2³ + 0 × 2² + 0 × 2 ¹ + 1 × 2⁰

Which gives;

16 + 8 + 0 + 0 + 1 = 25₁₀.

7 0
1 year ago
Compare the elements of the basic Software Development Life Cycle with 2 other models. How are they similar? How are they differ
Artemon [7]

Answer:

Explanation:

One of the basic notions of the software development process is SDLC models which stands for Software Development Life Cycle models. SDLC – is a continuous process, which starts from the moment, when it’s made a decision to launch the project, and it ends at the moment of its full remove from the exploitation. There is no one single SDLC model. They are divided into main groups, each with its features and weaknesses. The most used, popular and important SDLC models are given below:

1. Waterfall model

2. Iterative model

3. Spiral model

4. V-shaped model

5. Agile model

Stage 1. Planning and requirement analysis

Each software development life cycle model starts with the analysis, in which the stakeholders of the process discuss the requirements for the final product.

Stage 2. Designing project architecture

At the second phase of the software development life cycle, the developers are actually designing the architecture. All the different technical questions that may appear on this stage are discussed by all the stakeholders, including the customer.  

Stage 3. Development and programming

After the requirements approved, the process goes to the next stage – actual development. Programmers start here with the source code writing while keeping in mind previously defined requirements. The programming by itself assumes four stages

• Algorithm development

• Source code writing

• Compilation

• Testing and debugging

Stage 4. Testing

The testing phase includes the debugging process. All the code flaws missed during the development are detected here, documented, and passed back to the developers to fix.  

Stage 5. Deployment

When the program is finalized and has no critical issues – it is time to launch it for the end users.  

SDLC MODELS

Waterfall – is a cascade SDLC model, in which development process looks like the flow, moving step by step through the phases of analysis, projecting, realization, testing, implementation, and support. This SDLC model includes gradual execution of every stage completely. This process is strictly documented and predefined with features expected to every phase of this software development life cycle model.

ADVANTAGES  

Simple to use and understand

DISADVANTAGES

The software is ready only after the last stage is over

ADVANTAGES

Management simplicity thanks to its rigidity: every phase has a defined result and process review

DISADVANTAGES

High risks and uncertainty

Iterative SDLC Model

The Iterative SDLC model does not need the full list of requirements before the project starts. The development process may start with the requirements to the functional part, which can be expanded later.  

ADVANTAGES                                        

Some functions can be quickly be developed at the beginning of the development lifecycle

DISADVANTAGES

Iterative model requires more resources than the waterfall model

The paralleled development can be applied Constant management is required

Spiral SDLC Model

Spiral model – is SDLC model, which combines architecture and prototyping by stages. It is a combination of the Iterative and Waterfall SDLC models with the significant accent on the risk analysis.

4 0
1 year ago
An attacker distributes hostile content on Internet-accessible Web sites that exploit unpatched and improperly secured client so
julia-pushkina [17]

Answer:

Since the attack impacts the client SANS Critical Security Controls. This act affects the Inventory of authorized and unauthorized software running on the victim's machine.

5 0
1 year ago
Read 2 more answers
A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, an
yulyashka [42]

The question is incomplete. It can be found in search engines. However, kindly find the complete question below:

Question

Cites as a pitfall the utilization of a subset of the performance equation as a performance metric. To illustrate this, consider the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requires the execution of 1.0E9 instructions. 1. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2. 2. Another fallacy is to consider that the processor executing the largest number of instructions will need a larger CPU time. Considering that processor P1 is executing a sequence of 1.0E9 instructions and that the CPI of processors P1 and P2 do not change, determine the number of instructions that P2 can execute in the same time that P1 needs to execute 1.0E9 instructions. 3. A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, and consider that the processor with the largest MIPS has the largest performance. Check if this is true for P1 and P2. 4. Another common performance figure is MFLOPS (millions of floating-point operations per second), defined as MFLOPS = No. FP operations / (execution time x 1E6) but this figure has the same problems as MIPS. Assume that 40% of the instructions executed on both P1 and P2 are floating-point instructions. Find the MFLOPS figures for the programs.

Answer:

(1) We will use the formula:

                                       CPU time = number of instructions x CPI / Clock rate

So, using the 1 Ghz = 10⁹ Hz, we get that

CPU time₁ = 5 x 10⁹ x 0.9 / 4 Gh

                    = 4.5 x 10⁹ / 4 x 10⁹Hz = 1.125 s

and,

CPU time₂ = 1 x  10⁹ x 0.75 / 3 Ghz

                  = 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

So, P2 is actually a lot faster than P1 since CPU₂ is less than CPU₁

(2)

     Find the CPU time of P1 using (*)

CPU time₁ = 10⁹ x 0.9 / 4 Ghz

                = 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

So, we need to find the number of instructions₂ such that  CPU time₂ = 0.225 s. This means that using (*) along with clock rate₂ = 3 Ghz and CPI₂ = 0.75

Therefore,   numbers of instruction₂ x 0.75 / 3 Ghz = 0.225 s

Hence, numbers of instructions₂ = 0.225 x 3 x  10⁹ / 0.75  = 9 x 10⁸

So, P1 can process more instructions than P2 in the same period of time.

(3)

We recall  that:

MIPS = Clock rate / CPI X 10⁶

  So, MIPS₁ = 4GHZ / 0.9 X 10⁶ = 4 X 10⁹HZ / 0.9 X 10⁶ = 4444

        MIPS₂ = 3GHZ / 0.75 X 10⁶ = 3 x 10⁹ / 0.75 X 10⁶ = 4000

So, P1 has the bigger MIPS

(4)

  We now recall that:

MFLOPS = FLOPS Instructions / time x 10⁶

              = 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Therefore,

                  MFLOPS₁ = 1777.6

                  MFLOPS₂ = 1600

Again, P1 has the bigger MFLOPS

3 0
1 year ago
Match the job skills with the correct job role.
Usimov [2.4K]

Answer:

B-1,A-4,C-2,D-3

Explanation:

5 0
1 year ago
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