SAD interview experience

To have a documentary about our SAD project we are in demand to have an interview to the company that we have chosen, a company which we were going to implement a system. My experience in our SAD interview is not so difficult  because the owner of the company is not that strict regarding to the matter that we were discussing. The owner is the one that we have interviewed he is nice , he's easy to talk to. He answers our questions precisely and directly.We ask him about company's history, mission and vision.We also ask him if what are those things that causes his company's problems and he said that one of his company's problem is their payroll system. They are using the manual system for their payroll. Manual system that causes time consuming and sometimes it may lead to a wrong calculations of their employees salaries. Which we think their company must be implemented with an automatic or an advance technology for their payroll system.

CS323: Case Study 4

READY to WAITING and WAITING to RUNNING.

Process is a program in execution. It contains the program code and its current activity. Amultitasking operating system may just switch between processes to give the appearance of many processes executing concurrently or simultaneously, though in fact only one process can be executing at any one time on a single-core CPU(unless using multi-threading or other similar technology). A process can be in one of many possible states:

New (Hold)

                  Where the process is being created but has not been admitted to the pool of executable processes by the operating system.

Ready

                  The process is waiting to be assigned to a processor.

Running

                  The instructions are being executed.

Waiting

                  The process is waiting for some event to occur.

Terminated (Finished)

                  The process has finished execution.

Process State Transition Diagram

In this process state diagram, I will explain why there is no transition; from the READY to WAITING and from the WAITING to RUNNING.

From the READY state to the WAITING state, there is no transition because a job in the WAITING state is waiting for peripheral device response which must be received before the CPU can effectively be used again.  A process in the READY queue is ready in all aspects to make effective use of the CPU.  If a job in the READY state cannot proceed because a required device fails, it should be sent back to the HOLD state, not the WAITING state. The WAITING state is a service from the RUNNING state that the OS is not ready to perform and an access to a resource is not yet available. It initiates the I/O and must wait for the result where the waiting is for a process to provide the input. There is only a transition from READY to RUNNING which is the OS chooses one of the processes in the READY state and assigns CPU to it.

From the WAITING state to the RUNNING state, the Process Scheduler selects processes from the READY state for the CPU.  By passing the READY queue would make process management impossible. The RUNNING state have only three transition: from RUNNING state to FINISHED; from RUNNING state to READY state; and from RUNNING state to WAITING state. From RUNNING state to FINISHED, the process is terminated by the OS if it has completed or aborted. From RUNNING state to READY state, the most common for this transition are: the running process has expired his time slot; the running process gets interrupted because a higher priority process is in the READY state. From RUNNING state to WAITING state, a process is put to this state if it requests for some thing for which it must wait.  There is only a transition of WAITING to READY where a process from a WAITING state is moved to a READY state when the event for which it has been waiting occurs.

Therefore, if there is a transition from the READY to WAITING and from the WAITING to RUNNING the process in the state diagram would no longer work and the process would be complicated.

CS323: Case Study 3

Load the following jobs into memory using fixed partition following a certain memory allocation method (a. best-fit, b. first-fit, c. worst-fit).

Memory Block	Size
Block 1	50K
Block 2	200K
Block 3	70K
Block 4	115K
Block 5	15K

a. Job1 (100k) f. Job6 (6k)

turnaround: 3 turnaround: 1

b. Job2 (10k) g. Job7 (25k)

turnaround: 1 turnaround: 1

c. Job3 (35k) h. Job8 (55k)

turnaround: 2 turnaround: 2

d. Job4 (15k) i. Job9 (88k)

turnaround: 1 turnaround: 3

e. Job5 (23k) j. Job10 (100k)

turnaround: 2 turnaround: 3

*turnaround – how long it will stay in the memory.

a.)BEST-FIT

>> We will first allocate job 1. Since job 1 has 100 k in size, in searching for its location the only possible block that can accommodate job 1 are blocks 2 and 4. In best fit method, the smallest difference between the block size and job size will be the block that the job will be allocating. So in job 1, block 4 has a smaller wasted memory than block 2 which means job 1 will be allocated in block 4. Job 2 with 10 k size can be accommodated by the five blocks and the smallest wasted memory among all the five blocks is block 5 with a memory waste of 5 k. Block 1 will accommodate job 3 since it has the smallest memory waste. Job 4 will be in block 3 since block 5 has already job 2 in it. Job 5 will occupy block 2 even though it will have a memory waste of 177 k. Though it’s a very large memory waste, job 5 has no choice because block 2 can accommodate its memory. Those first five jobs can be seen in column a. In column b, job 2 and job 4 has already finished its job since they both have a turnaround of 1. Since job 2 and 4 left the memory, the jobs on waiting list can now be accommodated. Job 6 has 6 k as we can see in table 1, block 5 has the smallest memory waste and since job 2 is already finish, job 6 now will occupy block 5.  Now Job 7 is left with block 3 because block 1 which has only 25 k memory waste is still occupied by job 3. In column c, job 3 and job 5 with a turnaround of 2 is already finished. Job 8 will be occupying block 3 because it ill only give 15 k of memory waste. Though job 9’s smallest memory waste is block 4, block 4 is still busy with job 1 which means job 9 will be accommodated by block 2. In column d, job 1 is already finished with its turnaround of three. Job 10 can now be accommodated by block 4. On the next column e it shows that job 8 with turnaround of 2 is already finished and job 10 which will be its 2^nd turnaround and job 9 on its 3^rd turnaround. Since in column e job 9 is in its 3^rd turnaround, job 10 will be left in this column still continuing its job until this column. And in by column g all jobs are already finished. 

b.)FIRST-FIT

>>In first fit method, the first block that can accommodate the size of the job will be the block that will accept the job. No matter how large or small the memory waste will be. In this table in column a, job 1 with 100 k size will be allocated by block 2 with a size of 200 k. Job 2 with 10 k can now be accommodate by block 1 with 50 k. unlike with the first job that will need a 100 k. next will be the job 3. Job 3 size is 35k, when it goes to the first free block which is block 3, it will be enough to accommodate that job because block 3 has 70 k which is greater than the size of job 3. Job 4 with 15 k will now be allocated in block 4 with 115 k in size. Though there is an exact size in the block 5, it is still not allowable since we are using first fit. The remaining block now is block 5 with 15 k only. Job 5 needs 23 k so this means job 5 will be put in waiting list. Job 6 with 6 k of memory will be accommodated by block 5. In column b, job 2, 4 and 6 is now finished with its turnaround of 1. Job 5 now one the waiting list will now occupy block 1 and job 7 will be occupying block 4. Block 5 is left unoccupied because the last 3 jobs on the waiting list have a greater memory size than block 5. In column c, job 3 with a turnaround of 2 and job 7 with turnaround of 1 will finished. Making way for job 8 to occupy block 3 and job 9 to occupy block 4. On the next column job 1 is finished, this will allow job 10 to enter in the memory. The next column is just the same with jobs finished at the end of their turnaround. 

C.)WORST-FIT

>>Worst fit is the completely the opposite of best fit because in worst fit the block with largest memory that is wasted will be the location of that certain job. Job 1 is occupying block 2 with 100 k of memory waste. This goes with the jobs 2, 3 and 4. Job 5 can only occupy the first four blocks so the next job which is job six that can be accommodated by block 5 will now enter in the memory, leaving job 5 in the waiting list.  In column b jobs 2,4 and 6 are already finished with turnaround of 1. Job 5 can now occupy block 4 since it has the huge size of wasted memory while job 7 will occupy block 1 that is only available with its size that a block can accommodate. In the next column which is c, job 7 and job 3 is already finished. Job 8 will now enter in the memory though it is the smallest memory waste, job 8 should enter the memory because there is an available space for it. Column d shows job 1 and job 5 is finished. This will allow job 9 and job 10 to enter and execute. Since job 9 and job 10 has the same turnaround they will end up together. 

Computer Architecture Case Study 2

In a multiprogramming and time-sharing environment, several users share the system simultaneously. This situation can result in various security problems. Name at least two of these problems. Can we ensure the same degree of security in a time-share machine as we have in a dedicated machine? 

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   Stealing or copying of other files

1. 
   

        This usually happen in multiprogramming and time-sharing environment, in most cases the user’s files (the original owner of the files) was being over written by other users. This dilemma will lead to stealing or copying of files which greatly affects many users and because of that it create a negative impact in all other users(who have their original files taken by others) to engage in a multiprogramming and time-sharing environment. They believe that once your file where found by other users they will directly copy it and even without your permission that is why most of the users were very vigilant when it comes to this particular machine. And also the ability of the virus to enter in a specific files which causes the files to be damage. In multiprogramming and time-sharing environment it usually happen because of the number of users which is actually connected to each other in which the virus easily transfer and spread from user to another, and if take for granted it may cause a lot more problems. But virus is easily detected by anti-computer virus so there is a small possibility that a virus can enter on your computer.

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  Using system resources (CPU, disk space) without proper accounting

     In a multiprogramming and time-sharing environment basically when the system resources such as CPU and disk space are not properly accounted it causes the printer to mix output by sending data while some other user's file is printing. This thing may result to scrambling of mix output which cases the users to have an unfamiliar data or output and most probably the user will find it difficult in compiling his/her mixed outputs again and that will lead him/her hard time in working with it.

Can we ensure the same degree of security in a time-share machine as we have in a dedicated machine?

Definitely no, why do I say so? Firstly, because in a time-sharing machine as what I have explained earlier it is not safe to use all files that you've worked for is not permanently secured and there is a tendency that it might be copied by other users when found.

            Secondly, it is virus prone. In most cases virus-spread is vulnerable most especially when there are lots of users on that time share machine. It may cause further damage on the files and other important programs if you don’t take it seriously. While in a dedicated machine not all files but most files were secured from being copied by other users and because it is dedicated machine it is not virus prone so there is no need to worry with regards to virus.

            And lastly, when we talk about the degree of security dedicated machine is more reliable because not all users can enter on it in return they don’t have the chance to change any files so it is applicable to anyone. Even though time-sharing machine is very helpful because you work with other users or other people but when it comes to the degree of security dedicated machines is much more reliable.

Computer Architecture Case Study 1

Explain why we need to study Computer Architecture? What is its significance with your field as an IT or CS student? Differentiate Computer Architecture and Computer Organization?

Ø  As an IT student we need to study Computer Architecture so that we can easily familiarize the whole view or design of a certain computer. When you already study computer architecture, you will also learn the importance and characteristic of computer which is visible to us. We will also know the features of a computer if we study its architecture.

Ø  The significance of computer architecture for me as an IT student, I easily know where to plug and unplug a certain peripherals such as keyboards, mice, monitors pc speakers, printers, scanners, web cams, etc.    

 Ø  Computer Architecture & Computer Organization

Computer Architecture shows the whole view of the design of a computer and tells it’s each importance and characteristic while Computer Organization let us know how computer architectures’ features are being implemented such as control signals, interfaces, memory technology, etc. It is also the capabilities of a certain functio units.