BCA 4th semester long question solution 2078
9. What is CPU Scheduling? Write does the criteria for CPU scheduling? Consider the following set of processes, with the length of the CPU burst given in millisecond , draw Gant chart illustrating their execution and calculate average waiting time using:
a) First come First serve
b) Shortest Job first
c) None-preemptive priority (smaller number implies higher priority)
d) Round robin (quantum=1)
Process | Burst Time | Priority |
P0 | 10 | 3 |
P1 | 1 | 1 |
P2 | 2 | 5 |
P3 | 1 | 4 |
P4 | 5 | 2 |
BCA 4th semester long question solution 2078
a) First Come First Serve
Solution
Process | Arrival Time (AT) | Burst Time (BT) | Completion time (CT) | Turned around Time (TAT) | Waiting Time (WT) |
P0 | 0 | 10 | 10 | 10 | 0 |
P1 | 0 | 1 | 11 | 11 | 10 |
P2 | 0 | 2 | 13 | 13 | 11 |
P3 | 0 | 1 | 14 | 14 | 13 |
P4 | 0 | 5 | 19 | 19 Total=67 | 14 Total= 48 |
Note:- CT=From Gantt chart
TAT= CT-AT
WT=TAT-BT
P0=This process come first so we start at first AT=0 here in BT= 10
P1= P0 को BT लाई P1मा जोडनु पार्ने हुन्छ । Eg, [ P0 Burst time (BT)=10+P1(1)=11]
Gantt chart
P0 | P1 | P2 | P3 | P4 |
Total Waiting Time= 0+10+11+13+14
=48
Average waiting time=Total waiting time/ process
48/6
=9.6
Total turn around Time= 10+11+13+14+19
=67
Average turn around time= Total turn around time/ Process
= 67/5
= 13.4
b) Short job first
Solution
Process | Burst Time (BT) | Priority | Completion time (CT) | Turned around Time (TAT) | Waiting Time (WT) |
P0 | 10 | 3 | 10 | 10 | 0 |
P1 | 1 | 1 | 11 | 11 | 10 |
P2 | 2 | 5 | 14 | 14 | 12 |
P3 | 1 | 4 | 12 | 12 | 11 |
P4 | 5 | 2 | 15 | 15 Total=62 | 10 Total= 43 |
P0 | P1 | P2 | P3 | P4 |
Total waiting Time=0+10+12+11+10
=43
Average waiting time= Total waiting time/process
=43/5
=8.6
Total turnaround time= 10+11+14+12+15
=62
Average turn around time= total trunaround time/Process
=62/5
=12.4
C) Non-preemptive priority (small number implies higher)
Process | Burst Time (BT) | Priority | Completion time (CT) | Turned around Time (TAT) | Waiting Time (WT) |
P0 | 10 | 3 | 10 | 10 | 0 |
P1 | 1 | 1 | 11 | 11 | 10 |
P2 | 2 | 5 | 19 | 19 | 17 |
P3 | 1 | 4 | 17 | 17 | 16 |
P4 | 5 | 2 | 16 | 16 Total=73 | 11 Total= 54 |
P0 | P1 | P2 | P3 | P4 |
Total waiting Time=0+10+17+16+11
=54
Average waiting time= Total waiting time/process
=54/5
=10.8
Total turnaround time= 10+11+19+17+16
=73
Total turn around time= total trunaround time/Process
=73/5
=14.6
d) Round Robin (quantum = 1)
Process | Burst Time (BT) | Completion time (CT) | Turned around Time (TAT) | Waiting Time (WT) |
P0 | 10 | 10 | 10 | 0 |
P1 | 1 | 11 | 11 | 10 |
P2 | 2 | 15 | 15 | 13 |
P3 | 1 | 13 | 13 | 12 |
P4 | 5 | 19 | 19 Total=68 | 15 Total= 50 |
P0 | P1 | P2 | P3 | P4 | P2 | P4 |
Total waiting Time=0+10+13+12+15
=50
Average waiting time= Total waiting time/process
=50/5
=10
Total turnaround time= 10+11+15+13+19
=68
Total turn around time= total trunaround time/Process
=68/5
=13.6
BCA 4th semester long question solution 2078
Bachelor in computer Applications Full Marks:60
Course Title: Numerical Method Pass Marks: 24
Code No: CACS252 Time: 3 hours
Semester: IV
Group A
i. Which of the following is numerical error
a. Missing Information
b. Data Error
c. Roundup Error
d. Conversion Error
ii. Which of the following method is used solve the system of linear equation?
a. Bisection Method
b. False Position Method
c. Taylor series Method
d. Matrix Inverse Method
iii. How many significant digits in the number 105.78904560?
a. 11
b. 3
c. 8
d. 10
iv. What will be the solution of equations x+y=3 and 2x+2y=6?
a. Unique solution
b. Infinite Solution
c. No Solution
d. III-condition
v. Which of the following formula is four-point formula to estimate integration?
a. Trapezoidal Rule
b. Simpson’s 1/3 Rule
c. Simpson’s 3/8 Rule
d. Boole’s Rule
vi. When a differential equation is considered as an ordinary differential equation?
a. If it has one independent variable
b. if it has one dependent variable
c. more than one dependent variable
d. more than one independent variables