### Bin Packing Assignment algorithm for EDF Scheduling

1. This algorithm is for independent preemptible tasks.
2. All the processors are identical. The tasks requires no resources except the processor time
3. The period is equal to the relative deadline.
4.

The sum of utilisations of the tasks assigned to a processor is always less than or equal to 1, the task set is EDF scheduled in that processor.

So, the problme reduces to making task assignments with the property that the sum of the utilisations of the tasks assigned to a processor does not exceed to 1.

 T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 ei 5 7 3 1 10 16 1 3 9 17 21 Pi 10 21 22 24 30 40 50 55 70 90 95 u(i)=ei/Pi 0.50 0.33 0.14 0.04 0.33 0.40 0.02 0.05 0.13 0.19 0.22

Arrange the tasks in the order of their utilisation

So the order will be L={T1,T6,T2,T5,T11,T10,T3,T9,T8,T4,T7}

0.33

 Tasks u(i) -> utilisation Processor pi Assignment Vector(U) T1 0.50 p1 (0.50) T6 0.40 p1 (0.90) T2 0.33 p2 (0.90, 0.33) T5 p2 (0.90,0.66) T11 0.22 p2 (0.90,0.88) T10 0.19 p3 (0.90,0.88,0.19) T3 0.14 p3 (0.90,0.88,0.33) T9 0.13 p3 (0.90,0.88,0.46) T8 0.05 p1 (0.95,0.88,0.46) T4 0.04 p1 (0.99,0.88,0.46) T7 0.02 p2 (0.99,0.90,0.46)

Steps

5. When T1 is scheduled, there is only one processor with utilisation 0.5, so T1 is allocated to p1
6. When T6 comes, again p1 can accommodate as 0.5+0.4=0.9 which is again less than total utilisation factor 1.
7. When T2 comes, its utilisation is 0.33 which cant be added to p1 since it will cross 1, so a new processor is added and it is p2.
8. Like wise all other tasks are scheduled according to their utilisation factor.
9.

>The Final schedule is given below

p1=> T1, T6,T8,T4

p2=> T2,T5,T11,T7

p3 => T10,T3,T9

1. kanwarjit singh30 May 2010 at 21:56

xplain how processors are assigned tasks deeply...plz

### Installing TexLive 2019 in Ubuntu 18.04

Installation of TexLive 2019 in Linux (Ubuntu 18.04 LTS)
TeX (Tech)

Installation of TexLive 2019

Please watch the video for full installation

I used .iso file to download, the Total size is 3.3GB for Linux,

Extract the .iso file to a folder and open a terminal

\$] sudo ./install-tl
(it goes into a terminal mode, which is faster compared to the GUI Mode)

\$] sudo ./install-tl -gui
after the installation, set the PATH, MANPATH and INFOPATH as suggested by LATEX

export PATH=\$PATH:/usr/local/texlive/2019/bin/x86_64-linux
export MANPATH=/usr/local/texlive/2019/texmf-dist/doc/man
export INFOPATH=/usr/local/texlive/2019/texmf-dist/doc/info

put these lines in to the /home/pradeepkumar/.bashrc

We have installed TexLive 2019 and texstudio.

To install texstudio

\$] sudo apt install texstudio
The look and feel of TexStudio looks like this image.

texlive, it install everyt…

### Implementing a new system call in Kernel version 2.6.32

A system call is used by application or user programs to request service from the operating systems. Since the user programs does not have direct access to the kernel whereas the OS has the direct access. OS can access the hardware through system calls only.The following files has to be modified for implementing a system call/usr/src/linux-2.6.32.5/arch/x86/kernel/syscall_table_32.S/usr/src/linux-2.6.32.5/arch/x86/include/asm/unistd_32.h/usr/src/linux-2.6.32.5/include/linux/syscalls.h/usr/src/linux-2.6.32.5/MakefileNew set of files to be createdCreate a new directory newcall/ inside the path “/usr/src/linux-2.6.32.5/” Create new files Makefile, newcall.c and put them in the /usr/src/linux-2.6.32.5/newcall/ folder Create new user files (in any folder of Linux) to test the system call
testnewcall.c, testnewcall.h (created in /home/pradeepkumar) syscall_table_32.S Find the file /usr/src/linux-2.6.32.5/arch/x86/kernel/syscall_table_32.S and add the following line at the end
"…

### Electrical Machine Design (equations)

FactorsDC Machine Transformers Induction Machines Synchronous MachinesOutput EquationPa=CoD2Ln, where Pa=P/h for generators, Pa=P for motorsFor Single Phase
Q=2.22 f Bm Ai Kw Aw d10-3
For Three Phase
Q=3.33 f Bm Ai Kw Aw d 10-3Q=CoD2 L ns
KVA Input Q=
HP * 0.746 / Cos f * hQ=CoD2 L ns
KVA Input Q=
HP * 0.746 / Cos f * h
For Turbo alternators
Q=1.11Bavac KwsVa2 L 10-3/nsOutput CoefficientCo=Bav ac* 10-3where Bav-magnetic loading and ac - electric loadingDNACo=11 Kws Bav ac 10-3Co=11 Kws Bav ac 10-3 Choice of Magnetic LoadingFlux Density in Teeth Frequency of Flux Reversals Size of machineDNAMagnetizing current, Flux Density, Iron lossIron loss, Stability, Voltage Rating, Parallel Operation, Transient ShortCircuit current Choice of Electric LoadingTemperature rise,
speed of machine, Voltage, Armature reaction, CommutationDNAOverload Capacity, Copper losses, Temperature rise, Leakage ReactanceCopper loss, Synchronous reactance, Temperature rise, Stray Load losses,
Voltage rating Flux …