Friday, July 13, 2007

operating system

cont.#4.Given the following information;
Job list:
job number memory allocation
J1---740 k
J2---500k
J----700k
Memory List:
Memory Block Size
Block1------610k(low-order memory)
Block2------850k
Block3------700k(high-order memory)
Best-fit
In the Job List they have a Job1, Job2, Job3,..In memory list also have block1 to 3.The job1 has a memory requested is 740k so that the job 1 will go to block 2 while the size is 850k. In job 2 the memory requested is 500k and that's why they go to block 1 have a size of 610k. While the job3 has memory requested of 700k will go to block3 that have 700k.
First-fit
the job1 and the job2 they not allocate in block1 and 2.while the job3 is fit to the bock3
5.Given the following information:
Job List
job number memory requested
Job1---- 700k
job2----500k
Job3---740k
Memory List:
Memory Block Size
block1---610k (low-order memory)
block2---850k
block3---700k(high-order memory)
Best-fit:
job1 will go to block3. the job2 also go to block1. and the last job 3 goes to block2.
First-Fit:
6.
a.) Another variation of the free list algorithm is useful for very transient allocations of memory specially when all allocation are about the same size.
b.)It is compelled by there size and memory requested
7. Worst-fit the memory manager placed a process in the largest block of unallocated memory available. the placement will create the largest hold after the allocation, thus increasing the possibility that compare to best fit, another process can use the remaining space.
8. yes, the memory manager will know to adjust them by the value stored in relocation register all of the other values ( data values) one not marked and won't be changed after relocation other's numbers in the program, those indicating instruction, register, or constants use in the intruction, are also left alone

Wednesday, July 11, 2007

OPERATING SYSTEM

Exercises:

research topic:

Unit name - absolute size - exponential form
byte - 1 byte -2
kilobyte -1,024 byte -2^10
megabyte -1,048,576 bytes -2^20
gigabyte -1,073,741 bytes -2^30
terabyte -1,099,511,627,776 -2^40
petabyte -1,125,899,906,824,624 -2^50

Exercises:
A.Explain he following:
Multiprogramming. Why i s it used? A multiprogramming is a technique used to utilize maximum CPU time by running multiple programs simultaneously. The execution begins with the first program and continuous till an instruction waiting for a peripheral is reached, the context of this program is stored, and the second program is memory is given chance to run. The process continued until all program finished running. Multiprogramming has no guarantee that a program will run is timely manner.
B.Internal Fagmentation. How does it occur? The internal fragmentation occurs it when a fixed
partition is partially used by program, the remaining space within the partition is unavailable to any other job and that's the time internal fragmentation occur when there is another job followed on the space. So that it will not wasted.
C.Compaction: Why is it need? Compaction is very needed because it is the process of collecting fragments of available memory space into contiguous in block by moving programs and data in a
computer's memory disks, or known as garbage collection.
E.Relocation: How often should it performed? It depend on the process of address refferences in program.
#2. Describe the Major Disadvantages for each of the four memory allocation schemes presented in the chapter. The disadvantage of this memory allocation its an overhead process, so that while compaction is being done everything else must wait.
#3.Describe the Major Advantages for each of the memory alloatiuon schemes presented in the chapter. They could be divided into segments of variable sizes of equal size. Each page, or segment, could be stored wherever there was an empty block best enough to hold it.

Monday, July 9, 2007

Architecture

Processor
Intel® Pentium® Processor Extreme Edition

Intel® Pentium® processor Extreme Edition
Processor NumberΔ 840, 955, 965
Architecture 90 nm and 65 nm technology
L2 Cache 2x1MB and 2x2MB
L3 Cache NA
Clock Speed 3.20, 3.46 GHz and 3.73 GHz
Front Side Bus 800 MHz and 1066 MHz
Chipset
Intel® 975X Express chipset
Intel® 955X Express Chipset
Socket LGA775
Motherboard
Intel® Desktop Board D955XBK
Intel® Desktop Board D975XBX

Architecture

Processor
Intel® Pentium® Processor Extreme Edition

Intel® Pentium® processor Extreme Edition
Processor NumberΔ 840, 955, 965
Architecture 90 nm and 65 nm technology
L2 Cache 2x1MB and 2x2MB
L3 Cache NA
Clock Speed 3.20, 3.46 GHz and 3.73 GHz
Front Side Bus 800 MHz and 1066 MHz
Chipset
Intel® 975X Express chipset
Intel® 955X Express Chipset
Socket LGA775
Motherboard
Intel® Desktop Board D955XBK
Intel® Desktop Board D975XBX

ARCHITECTURE

Chipset Motherboard Manufacturer Model Socket Form Factor Graphics Brand Compare
View Detail AMD 690G Asus V3-M2A690G AM2 Barebone ATI Radeon X1250
View Detail AMD 690G Asus P1-M2A690G AM2 Barebone ATI Radeon X1250
View Detail AMD 690G Asus P2-M2A690G AM2 Barebone ATI Radeon X1250

JONATHAN CALIMPITAN
MERCEDITA DUSARAN
MYLA BAIR
JENEFE SORIANO
ROMEO ZAMBRANO
MELBA DUA

ARCHITECTURE

Processor Number L2 Cache Clock Speed Front Side Bus Speed Dual-Core
Intel® Virtualization Technology± Intel® 64Φ Execute Disable Bit°
65-nm Technology
540 1MB 1.86 GHz 533 MHz

440 512KB 2.00 GHz 800 MHz
430 512KB 1.80 GHz 800 MHz
420 512KB 1.60 GHz 800 MHz
± Intel® Virtualization Technology (Intel® VT) requires a computer system with an enabled Intel® processor, BIOS, virtual machine monitor (VMM) and, for some uses, certain platform software enabled for it. Functionality, performance or other benefits will vary depending on hardware and software configurations and may require a BIOS update. Software applications may not be compatible with all operating systems. Please check with your application vendor.

Wednesday, July 4, 2007

Founders of Intel
Gordon Moore
Bob Noyce

Stanley Mazor
Federico Faggin,
Robert Noyce
Stanley Mazor
Ted Hoff