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What is the predicted value of the fifth CPU burst (in µsec) if exponential averaging is used in the case of the shortest job first scheduling algorithm? The length of CPU bursts µsec is {t1, t2, t3, t4} = {4, 5, 8, 7}, smoothening factor is 0.6 and predicted value of the first CPU burst is 8 µsec.
Correct answer is between '6.9,7.0'. Can you explain this answer?
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What is the predicted value of the fifth CPU burst (in µsec) if ...
Exponential Averaging Overview
Exponential averaging is a method used to predict future CPU bursts based on past observations. It helps in estimating the next CPU burst time using a weighted average of previous bursts.
Formula for Exponential Averaging
The formula for predicting the next CPU burst (B_n) is as follows:
B_n = α * t_n + (1 - α) * B_(n-1)
Where:
- α = smoothening factor (0.6 in this case)
- t_n = actual burst time of the nth CPU burst
- B_(n-1) = predicted burst time from the previous burst
Given Values
- Previous CPU bursts: {t1, t2, t3, t4} = {4, 5, 8, 7} microseconds
- Smoothening factor (α) = 0.6
- Initial predicted value (B_1) = 8 microseconds
Prediction Steps
1. Predicting the 2nd Burst (B_2):
- B_2 = 0.6 * t1 + 0.4 * B_1
- B_2 = 0.6 * 4 + 0.4 * 8 = 6.4 microseconds
2. Predicting the 3rd Burst (B_3):
- B_3 = 0.6 * t2 + 0.4 * B_2
- B_3 = 0.6 * 5 + 0.4 * 6.4 = 5.84 microseconds
3. Predicting the 4th Burst (B_4):
- B_4 = 0.6 * t3 + 0.4 * B_3
- B_4 = 0.6 * 8 + 0.4 * 5.84 = 7.334 microseconds
4. Predicting the 5th Burst (B_5):
- B_5 = 0.6 * t4 + 0.4 * B_4
- B_5 = 0.6 * 7 + 0.4 * 7.334 = 6.934 microseconds
Conclusion
The predicted value of the fifth CPU burst using exponential averaging is approximately 6.934 microseconds, which falls between 6.9 and 7.0 microseconds, confirming the provided answer.
Exponential averaging is a method used to predict future CPU bursts based on past observations. It helps in estimating the next CPU burst time using a weighted average of previous bursts.
Formula for Exponential Averaging
The formula for predicting the next CPU burst (B_n) is as follows:
B_n = α * t_n + (1 - α) * B_(n-1)
Where:
- α = smoothening factor (0.6 in this case)
- t_n = actual burst time of the nth CPU burst
- B_(n-1) = predicted burst time from the previous burst
Given Values
- Previous CPU bursts: {t1, t2, t3, t4} = {4, 5, 8, 7} microseconds
- Smoothening factor (α) = 0.6
- Initial predicted value (B_1) = 8 microseconds
Prediction Steps
1. Predicting the 2nd Burst (B_2):
- B_2 = 0.6 * t1 + 0.4 * B_1
- B_2 = 0.6 * 4 + 0.4 * 8 = 6.4 microseconds
2. Predicting the 3rd Burst (B_3):
- B_3 = 0.6 * t2 + 0.4 * B_2
- B_3 = 0.6 * 5 + 0.4 * 6.4 = 5.84 microseconds
3. Predicting the 4th Burst (B_4):
- B_4 = 0.6 * t3 + 0.4 * B_3
- B_4 = 0.6 * 8 + 0.4 * 5.84 = 7.334 microseconds
4. Predicting the 5th Burst (B_5):
- B_5 = 0.6 * t4 + 0.4 * B_4
- B_5 = 0.6 * 7 + 0.4 * 7.334 = 6.934 microseconds
Conclusion
The predicted value of the fifth CPU burst using exponential averaging is approximately 6.934 microseconds, which falls between 6.9 and 7.0 microseconds, confirming the provided answer.
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What is the predicted value of the fifth CPU burst (in µsec) if exponential averaging is used in the case of the shortest job first scheduling algorithm? The length of CPU burstsµsec is {t1, t2, t3, t4} = {4, 5, 8, 7}, smoothening factor is 0.6 and predicted value of the first CPU burst is 8 µsec.Correct answer is between '6.9,7.0'. Can you explain this answer?
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What is the predicted value of the fifth CPU burst (in µsec) if exponential averaging is used in the case of the shortest job first scheduling algorithm? The length of CPU burstsµsec is {t1, t2, t3, t4} = {4, 5, 8, 7}, smoothening factor is 0.6 and predicted value of the first CPU burst is 8 µsec.Correct answer is between '6.9,7.0'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about What is the predicted value of the fifth CPU burst (in µsec) if exponential averaging is used in the case of the shortest job first scheduling algorithm? The length of CPU burstsµsec is {t1, t2, t3, t4} = {4, 5, 8, 7}, smoothening factor is 0.6 and predicted value of the first CPU burst is 8 µsec.Correct answer is between '6.9,7.0'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the predicted value of the fifth CPU burst (in µsec) if exponential averaging is used in the case of the shortest job first scheduling algorithm? The length of CPU burstsµsec is {t1, t2, t3, t4} = {4, 5, 8, 7}, smoothening factor is 0.6 and predicted value of the first CPU burst is 8 µsec.Correct answer is between '6.9,7.0'. Can you explain this answer?.
What is the predicted value of the fifth CPU burst (in µsec) if exponential averaging is used in the case of the shortest job first scheduling algorithm? The length of CPU burstsµsec is {t1, t2, t3, t4} = {4, 5, 8, 7}, smoothening factor is 0.6 and predicted value of the first CPU burst is 8 µsec.Correct answer is between '6.9,7.0'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about What is the predicted value of the fifth CPU burst (in µsec) if exponential averaging is used in the case of the shortest job first scheduling algorithm? The length of CPU burstsµsec is {t1, t2, t3, t4} = {4, 5, 8, 7}, smoothening factor is 0.6 and predicted value of the first CPU burst is 8 µsec.Correct answer is between '6.9,7.0'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the predicted value of the fifth CPU burst (in µsec) if exponential averaging is used in the case of the shortest job first scheduling algorithm? The length of CPU burstsµsec is {t1, t2, t3, t4} = {4, 5, 8, 7}, smoothening factor is 0.6 and predicted value of the first CPU burst is 8 µsec.Correct answer is between '6.9,7.0'. Can you explain this answer?.
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