Engineering High School

Suppose we have four jobs in a computer system, in the order JOB1, JOB2, JOB3 and JOB4. JOB1 requires 8 s of CPU time and 8 s of I/O time; JOB2 requires 4 s of CPU time and 14 s of disk time; JOB3 requires 6 s of CPU time; and, JOB4 requires 4 s of CPU time and 16 s of printer time. Define the following quantities for system utilization: • Turnaround time = actual time to complete a job • Throughput = average number of jobs completed per time period T • Processor utilization = percentage of time that the processor is active (not waiting) Compute these quantities (with illustrations if needed) in each of the following systems: a. A uniprogramming system, whereby each job executes to completion before the next job can start its execution. b. A multiprogramming system that follows a simple round-robin scheduling. Each process gets 2 s of CPU time turn-wise in a circular manner.

Answers

Answer 1

Answer:

Explanation:

a. Uniprogramming System:

Turnaround Time: For JOB1, the turnaround time is 16 s (8 s CPU time + 8 s I/O time). For JOB2, the turnaround time is 18 s (4 s CPU time + 14 s disk time). For JOB3, the turnaround time is 6 s (6 s CPU time). For JOB4, the turnaround time is 20 s (4 s CPU time + 16 s printer time).

Throughput: In a uniprogramming system, the throughput is equal to 1 job per time period, as only one job can be processed at a time.

Processor Utilization: For JOB1, the processor utilization is 8/16 = 50%. For JOB2, the processor utilization is 4/18 = 22.2%. For JOB3, the processor utilization is 6/6 = 100%. For JOB4, the processor utilization is 4/20 = 20%.

b. Multiprogramming System:

Turnaround Time: For JOB1, the turnaround time is 24 s (8 s CPU time + 8 s I/O time + 8 s waiting). For JOB2, the turnaround time is 24 s (4 s CPU time + 14 s disk time + 6 s waiting). For JOB3, the turnaround time is 16 s (6 s CPU time + 10 s waiting). For JOB4, the turnaround time is 32 s (4 s CPU time + 16 s printer time + 12 s waiting).

Throughput: In a multiprogramming system, the throughput is equal to 4 jobs per time period, as all four jobs can be processed concurrently.

Processor Utilization: For JOB1, the processor utilization is 8/24 = 33.3%. For JOB2, the processor utilization is 4/24 = 16.7%. For JOB3, the processor utilization is 6/16 = 37.5%. For JOB4, the processor utilization is 4/32 = 12.5%.

Answer 2

Here are the turnaround time, throughput, and processor utilization for the four jobs in a uniprogramming and round-robin system:

Uniprogramming

Turnaround time:

JOB1: 16 s

JOB2: 22 s

JOB3: 12 s

JOB4: 20 s

Throughput: 1 job/16 s

Processor utilization: 50%

Round-Robin

Turnaround time:

JOB1: 18 s

JOB2: 14 s

JOB3: 10 s

JOB4: 18 s

Throughput: 2 jobs/16 s

Processor utilization: 62.5%

The difference in turnaround time between the two systems is due to the fact that in a uniprogramming system, the CPU is idle while a job is waiting for I/O. In a round-robin system, the CPU is always running, even if it is only running a short burst of code for another job.

The difference in throughput between the two systems is also due to the fact that in a round-robin system, the CPU is always running, so more jobs can be completed in a given amount of time.

The processor utilization is higher in a round-robin system because the CPU is not idle as often. However, the turnaround time is also higher because the jobs have to share the CPU.

Read more about throughput here:

https://brainly.com/question/28390124

#SPJ2

Related Questions

Can someone convert this code into python? Please I need this code immediately.
int main(void)
{
double a[5],b[5],c=0;
int i;
for(i=0;i<5;i++)//입력
{
scanf("%lf %lf",&a[i],&b[i]);
}
for(i=0;i<5;i++)
{
c+=(b[i]-a[i]-1.0>0) ? ((b[i]-a[i]-1.0>4) ? 4.0 : b[i]-a[i]-1.0): 0;
}
c/=0.5;
c*=5000;
if(c>=150000)
c*=0.95;
else if(c<=50000)
c*=1.05;

printf("%.0f",c);
}

Answers

Here's the Python equivalent of the given code:

```
def main():
a = [0] * 5
b = [0] * 5
c = 0

for i in range(5): # Input
a[i], b[i] = map(float, input().split())

for i in range(5):
c += ((b[i] - a[i] - 1.0) > 0) * (4.0 if (b[i] - a[i] - 1.0) > 4 else (b[i] - a[i] - 1.0))

c /= 0.5
c *= 5000

if c >= 150000:
c *= 0.95
elif c <= 50000:
c *= 1.05

print("{:.0f}".format(c))

main()
```

Note that the `scanf` function has been replaced with Python's `input` function, and the `printf` function has been replaced with Python's `print` function with formatted output.

Construct a triangle , having its perimeter = 55 mm and the ratio of its angles 2:1:3 . I want the steps along with the drawing . this is ENGINEERING GRAPHICS grade 11 question .​

Answers

Answer:

  see attached

Explanation:

You want to construct a triangle with a perimeter of 55 mm and angles in the ratio 2:1:3.

Angles

The 2+1+3 = 6 ratio units represent 180°, the sum of angles in a triangle. Hence each unit in the ratio is 180°/6 = 30°.

The angles are 60° : 30° : 90°.

Perimeter

From your knowledge of "special triangles," you know the side lengths in a 30°-60°-90° triangle have the ratios 1 : √3 : 2.

That is, the perimeter of the triangle is (1+√3 +2) = 3+√3 times the length of the shortest side.

Side lengths

The shortest side of the triangle will have the length ...

  [tex]\dfrac{55\text{ mm}}{3+\sqrt{3}}=\dfrac{(3-\sqrt{3})(55\text{ mm})}{(3-\sqrt{3})(3+\sqrt{3})}= \dfrac{(165-55\sqrt{3})\text{ mm}}{3^2-(\sqrt{3})^2}\\\\\\=(27.5-\dfrac{55}{6}\sqrt{3})\text{ mm}\approx 11.6229\text{ mm}[/tex]

The longest side will be double this length, about 23.2457 mm. The side of intermediate length will be (55√3 -55)/2 ≈ 20.1314 mm.

Construction

If you're doing the drawing using only a compass and straightedge, you will start by constructing perpendicular lines (the right angle). Mark off 1 and two lengths of the shortest side on one of these perpendicular segments. Then draw the hypotenuse as 2 times the short length from the end of the short length. It will intersect the perpendicular segment at √3 times the short length.

The tricky part here is constructing a line segment of the desired irrational length. One way to do this is to construct a 30-60-90 triangle with a unit-length short side, then "unfold" it to a line segment with parts that are 1, √3, and 2 units long. A "proportioning" construction can divide the 55 mm length of the perimeter into these proportions, (10 mm might be a good "unit length" for this.)

__

Additional comment

The desired dimensions are sufficiently small that an accurate compass/straightedge construction will be difficult. It is unlikely your pencil line width is much less than about 0.2 mm, so already the error is about 2% in any length.

The attached drawing was made by telling the software what lengths we wanted in "exact" terms. Its calculation is generally good to 10 sf or better. To scale the drawing properly, you'd have to make the longest side about 23.25 mm, or 0.9152 inches.

Can somebody help me with these. with solution

1 Two balls are drawn in succession without replacement from a box containing 4 red balls and 3 black balls. Let Y be the random variable, where Y is the number of the red balls. What is the expected value of this random event?

2. Two balls are drawn in succession from a box without repetition containing 3 red 3 blue and four yellow marbles. What is the probability that they are of the same color?

Answers

Answer:

See below, please.

Explanation:

Part 1.

Let Y be the random variable representing the number of red balls drawn in succession. The possible outcomes are (R, R), (R, B), (B, R), and (B, B). We can calculate the probability of each outcome as follows

P(Y=0) = P(B, B) = (3/7) * (2/6) = 1/7

P(Y=1) = P(R, B) + P(B, R) = (4/7) * (3/6) + (3/7) * (4/6) = 12/42 + 12/42 = 24/42

P(Y=2) = P(R, R) = (4/7) * (3/6) = 2/7

The expected value of Y is given by:

E(Y) = Σ yi * P(Y=y)

= 0 * (1/7) + 1 * (24/42) + 2 * (2/7)

= 0 + 8/14 + 4/7

= 16/14

= 1.14

Therefore, the expected value of Y is 1.14.

Part 2.

The total number of ways to draw two marbles from the box is 10C2 = 45. The number of ways to draw two marbles of the same color is 3C2 + 4C2 = 3 + 6 = 9. Therefore, the probability of drawing two marbles of the same color is

P(same color) = 9/45 = 1/5

Therefore, the probability that the two balls are of the same color is 1/5.

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