What is the waiting time of each process for each of the scheduling a.pdf
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What is the waiting time of each process for each of the scheduling algorithms? Solution First-Come, First-Served (FCFS) Scheduling Process Burst Time P1 24 P2 3 P3 3 Suppose that the processes arrive in the order: P1 , P2 , P3 The Gantt Chart for the schedule is: P1 P2 P3 0 24 27 30 Waiting time for P1 = 0; P2 = 24; P3 = 27 SJF Process Arrival Time Burst Time P1 0.0 7 P2 2.0 4 P3 4.0 1 P4 5.0 4 SJF (non-preemptive) P1 P3 P2 P4 0 7 8 12 16 Average waiting time = [0 +(8-2)+(7-4) +(12-5)] /4 =4 Example of Preemptive SJF Proces Arrival Time Burst Time P1 0.0 7 P2 2.0 4 P3 4.0 1 P4 5.0 4 SJF (preemptive) P1 P2 P3 P2 P4 P1 0 2 4 5 7 11 16 Average waiting time = (9 + 1 + 0 +2)/4 =3 Round Robin (RR) Each process gets a small unit of CPU time (time quantum), usually 10-100 milliseconds. After this time has elapsed, the process is preempted and added to the end of the ready queue. If there are n processes in the ready queue and the time quantum is q, then each process gets 1/n of the CPU time in chunks of at most q time units at once. No process waits more than (n-1)q time units. Performance 1. q large _ FIFO 2. q small _ q must be large with respect to context switch, otherwise overhead is too high. Example of RR with Time Quantum = 4 Process Burst Time P1 24 P2 3 P3 3 The Gantt chart is: P1 P2 P3 P1 P1 P1 P1 P1 0 4 7 10 14 18 22 26 30 Average waiting time = [(30-24)+4+7]/3 = 17/3 =5.66 P1 P2 P3.
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![What is the waiting time of each process for each of the scheduling algorithms?
Solution
First-Come, First-Served (FCFS) Scheduling
Process Burst Time
P1 24
P2 3
P3 3
Suppose that the processes arrive in the order: P1 , P2 , P3
The Gantt Chart for the schedule is:
P1
P2
P3
0 24 27 30
Waiting time for P1 = 0; P2 = 24; P3 = 27
SJF
Process Arrival Time Burst Time
P1 0.0 7
P2 2.0 4
P3 4.0 1
P4 5.0 4
SJF (non-preemptive)
P1
P3
P2
P4
0 7 8 12 16
Average waiting time = [0 +(8-2)+(7-4) +(12-5)] /4 =4
Example of Preemptive SJF
Proces Arrival Time Burst Time
P1 0.0 7
P2 2.0 4](https://image.slidesharecdn.com/whatisthewaitingtimeofeachprocessforeachoftheschedulinga-230706072622-88004c89/85/What-is-the-waiting-time-of-each-process-for-each-of-the-scheduling-a-pdf-1-320.jpg)
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molecules (semi quantitative real-time PCR).
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fluorescent dyes that intercalate with any double-stranded DNA, and (2) sequence-specific DNA
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Please explain In a Drosophila dihybrid cross, a sepia male was mate.pdf
Please explain In a Drosophila dihybrid cross, a sepia male was mated to a apterous female.
Their phenotypically wild F_1 female produced the following offspring when crossed to
phenotypically wild F_1 males: Please set up a null hypothesis and use a Chi-square analysis to
see if the observation follows the Mendelian phenotypic ratio.
Solution
Answer:
Chi-Square value= 1.33
The degrees of freedom = 4-1 = 3
The probability level for df 3
Chi-Square value, 1.33 lies below 0.5 (probability), so we fail to reject null
hypothesis.PhenotypeObserved(O)Expected (E)O-E(O-E)2(O-E)2/EWild
type45045000.000.00Sepia16015010100.000.67Apterous140150-
10100.000.67sepia/aprerous505000.000.008008001.33.
Ganges River question The Ganges River, shown in Figure 4.57, has an.pdf
Ganges River question The Ganges River, shown in Figure 4.57, has an annual average flow rate
of 12,105 m3/s. The average temperature of the Ganges River was reported to be 24.68C. The
Ganges has become contaminated with raw sewage and industrial waste, so that the dissolved
oxygen in the river upstream of sewage discharges is 4.6 mg/L. Suppose that the raw sewage that
is untreated flows into the Ganges at a rate of 460 m3/s, the temperature is 258C, and the
dissolved oxygen in the sewage stream is completely depleted (i.e., DO 0). How would you
develop a plan for improving water quality in the Ganges River? 90\" 90 5 STAN 00 kn CHINA
BHU NDIA G.aga. Bana Bay of Bengal 20 Figure The Ganges River Basin Source: Based on
Rahaman, M.M. (2006) The Ganges water conflict: A comparative analysis of the 1977
Agreement and the 1996 Treaty
Solution
Ganges, the holiest river of India is 2500 kms long binding 5 different states of India and has a
huge cultural and spiritual signifincance. The Ganga River Basin is home to more than 600
million Indians which is the highest number of people supported by any river in the world.
However, this huge population has made the Ganges one of the most polluted rivers in the world.
Despite this iconic status and religious heritage, the Ganges today is facing formidable pollution
pressures and associated threats to its biodiversity and environmental sustainability. Increased
dumping of human waste, as well as industrial waste, is noted as the most common form of
pollution of the river. The studies indicate that a large proportion of pollution load in the river
came from the municipal wastewater generated in twenty-five Class I towns located on the banks
of the Ganges, each with a population exceeding 100,000. It constituted around 75% of the
pollution from all point-sources. Remaining 25% of the pollution from point-sources was mainly
due to untreated industrial effluent. Therefore, emphasis is to be given on interception and
diversion of wastewater and its treatment in Sewage Treatment Plants, before discharging into
river.
The data given are -
Annual average flow rate = Qr = 12105 m3/sec
DO in the river upstream of the sewage discharge = DOr = 4.6 mg/l
Flow rate of raw sewage flowing into the river = 460 m3/sec
DO of the mix is given by ( Qr x DOr + Qsewage x DOsewage ) / (Qr + Qsewage) which
according to the question gets depleted completely and one can observe the amount of sewage
that is being dump in the river every day.
Improving the water quality in Ganges can be initiated by preparing a plan, that guides
investment and choices. An obvious focus to be given on the critically polluted stretch and the
plan should prioritize the pollution hot spots and the investments with the greatest impact.Scarce
resources must be allocated to investments with the highest returns. The paucity of credible and
reliable water quality data on the Ganges must be rectified. The global experience shows that we
need good data, including.
Locomotion is important for many Bacteria Which of the following are .pdf
Locomotion is important for many Bacteria Which of the following are common forms of
motility? Transient Gliding Swimming Random walk Lateral Bacterial cells frequently
produce inclusions in the cytoplasm or periplasm. What are functions that these serve?
Exporting nutrients to the environment Nutrient reserve Producing membrane bound organelles
survival during desiccation or starvation signal transduction to neighboring cells Which of the
following cell wall polysaccharides are sensitive to lysozyme? Teichoic acid Chitin
Poptdoglycan Cellulose Pseudonmurein
Solution
Question 6) Gliding is the common form of the motility. This allows the bacteria to move over
the solid surface.It involves ejection of polysaccharide slime from exit points at both the ends of
the body.
Question 7) Nutrient reserve (Inclusion bodies are the reserve material for the later use, they
have glycogen which is a polymer that is carbohydrate reserve for carbohydrate and energy)
Question 8) Peptidoglycan (catalyze hydrolysis of 1,4-beta-linkages between N-acetylmuramic
acid and N-acetyl-D-glucosamine residues in a peptidoglycan in cell wall of the bacterial cell).
List the three subpathways of cellular respiration and the results o.pdf
List the three subpathways of cellular respiration and the results of each.
Solution
The 3 subpathways of cellular respiration are as follows:
1. Glycolysis followed by link reaction: Take place in the cytoplasm and results in one glucose
molecule being converted into two pyruvate molecules with a net yield of 2 ATP molecules. Two
NAD+ which act as hydrogen carriers are converted into NADH + H+
The link Reaction: Pyruvate from glycolysis is absorbed by the mitochondion. Pyruvate
undergoes oxidative decarboxylation, whereby hydrogen is removed to form NADH + H+, and
carbon dioxide is removed. The end result is an acetyl group, which is accepted by coenzyme A
to form Acetyl CoA.
2. Krebs cycle: Acetyl group is transferred from Acetyl CoA to oxaloacetate at the beginning of
the cycle to form a 6-carbon compound. The compound undergoes oxidative decarboxylation,
which is distributed through 6 reactions : 2 reactions result in CO2 and the remaining four result
in H+. Of the H+ produced, that from three of the four reactions goes to form NADH + H+ and
the remaining forms FADH2. ATP is also produced in one of these reactions and is termed a
substrate-level phosphorylation.
3. ETC (Electron Transport chain): is a series of electron carriers located in the membrane of the
mitochondia. NADH + H+ and FADH2 give their electrons to these carriers in the membrane,
which then \"carry\" the electron from one carrier to another. The elctron is losing enery as it is
passed from one carrier to another. While this is taking place, the concentration of H+ outside
the mitochondia is growing larger and larger and so must be fixed. The energy released as
electron pass along the transport chain is used to pump H+ accross the membrane of the
mitochondia. The flow of the H+ ions through an enzyme known as ATP synthase drives the
enzyme to produce ATP. At the end of ETC, the electrons are given to oxygen, which also
accepts excress hydrogens, to form water..
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- 1. What is the waiting time of each process for each of the scheduling algorithms? Solution First-Come, First-Served (FCFS) Scheduling Process Burst Time P1 24 P2 3 P3 3 Suppose that the processes arrive in the order: P1 , P2 , P3 The Gantt Chart for the schedule is: P1 P2 P3 0 24 27 30 Waiting time for P1 = 0; P2 = 24; P3 = 27 SJF Process Arrival Time Burst Time P1 0.0 7 P2 2.0 4 P3 4.0 1 P4 5.0 4 SJF (non-preemptive) P1 P3 P2 P4 0 7 8 12 16 Average waiting time = [0 +(8-2)+(7-4) +(12-5)] /4 =4 Example of Preemptive SJF Proces Arrival Time Burst Time P1 0.0 7 P2 2.0 4
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