Aiza S. Cabrera January 9, 2010
BSIT-3
I. Concept Questions
1. Name the five key concepts about an operating system that you think a user needs to know and understands.
Answer:
1.1 User Command Interface
This is the interface from which the user issues commands to the operating system. Also called the “shell” (container inside which the entire user interface is presented), this is the visible interface with which users interact. For most users, the User Interface is the operating system.
1.2 Device Manager
This component monitors all devices, channels, and control units. The Device Manager must select the most efficient method for allocation of a system’s devices, printers, terminals, disk drives, and other hardware. This is based on a pre-determined scheduling policy, and the Device Manager makes allocations, starts operations, and ultimately deallocates devices.
1.3 Processor Manager
This component is responsible for allocating the central processing unit (CPU). The status of each process must be tracked, and the Processor Manager handles matters such as process prioritization and multithreading. The tasks of the Processor Manager can be divided into two main categories: accepting or rejecting incoming jobs (handled by the Job Scheduler) and determining which process is given access to the CPU and for how long (handled by the Process Scheduler).
1.4 Memory Manager
This component controls main memory. It evaluates the validity of each memory request, and allocates memory space (as needed and available). For multi-user systems, the Memory Manager maintains a log of what memory resources are in use by which users. When items stored in main memory are no longer needed, the Memory Manager handles memory deallocation.
1.5 File Manager
This component tracks every file in the system. These files include data files, assemblers, compilers, and applications. The File Manager can use predetermined access policies to enforce restrictions on file access. It also handles all other file permissions. The File Manager allocates file resources by opening a particular file and deallocates resources by closing the file.
3. Explain the following:
a. Internal Fragmentation. How does it occur?
Internal fragmentation occurs when storage is allocated without ever intending to use it. This space is wasted. While this seems foolish, it is often accepted in return for increased efficiency or simplicity. The term "internal" refers to the fact that the unusable storage is inside the allocated region but is not being used.
b. External Fragmentation. How does it occur?
External fragmentation is the phenomenon in which free storage becomes divided into many small pieces over time. It is a weakness of certain storage allocation algorithms, occurring when an application allocates and deallocates ("frees") regions of storage of varying sizes, and the allocation algorithm responds by leaving the allocated and deallocated regions interspersed. The result is that although free storage is available, it is effectively unusable because it is divided into pieces that are too small to satisfy the demands of the application. The term "external" refers to the fact that the unusable storage is outside the allocated regions.
c. Compaction. Why is it needed?
4. Cache Memory how it works?
Answer:
A CPU cache is a cache used by the central processing unit of a computer to reduce the average time to access memory. The cache is a smaller, faster memory which stores copies of the data from the most frequently used main memory locations. As long as most memory accesses are cached memory locations, the average latency of memory accesses will be closer to the cache latency than to the latency of main memory.
When the processor needs to read from or write to a location in main memory, it first checks whether a copy of that data is in the cache. If so, the processor immediately reads from or writes to the cache, which is much faster than reading from or writing to main memory.
5. Which is the fastest cache’s L1, L2, L3? Why?
Answer:
L1 cache is the fastest among the three because it has a capacity between 4 to 16 kilobytes and accessing speeds of 10 nanoseconds, while the L2 cache can reach sizes of 512 kilobytes and a speed of only 20 to 30 nanoseconds.
II. Memory Utilization Problem
1. Given the following information:
Table 1A
| Job Number | Memory Requested |
| J1 | 700KB |
| J2 | 500KB |
| J3 | 740KB |
| J4 | 850KB |
| J5 | 610KB |
a. Use the best-fit algorithm to allocate the memory blocks to the five arriving jobs.
| Memory Block | Memory size | Job number | Job size | Status | Internal Fragmentation |
| 1132 | 700 | J1 | 700KB | Busy | 0 |
| 1003 | 720 | J5 | 610KB | Busy | 110 |
| 1114 | 800 | J3 | 740KB | Busy | 60 |
| 2310 | 750 | - | - | Free | |
| 1755 | 610 | J2 | 500KB | Busy | 110 |
b. Use the first-fit algorithm to allocate the memory blocks to the five arriving jobs.
| Memory Block | Memory size | Job number | Job size | Status | Internal Fragmentation |
| 1132 | 700 | J1 | 700 | Busy | 0 |
| 1003 | 720 | J2 | 500 | Busy | 220 |
| 1114 | 800 | J3 | 740 | Busy | 60 |
| 2310 | 750 | J5 | 610 | Busy | 140 |
| 1755 | 610 | - | - | Free | |
c. Use the next-fit algorithm to allocate the memory blocks to the five arriving jobs.
| Memory Block | Memory size | Job number | Job size | Status | Internal Fragmentation |
| 1132 | 700 | - | - | Free | |
| 1003 | 720 | J5 | 610 | Busy | 110 |
| 1114 | 800 | J3 | 740 | Busy | 60 |
| 2310 | 750 | J1 | 700 | Busy | 50 |
| 1755 | 610 | J2 | 500 | Busy | |
d. Use the worst-fit algorithm to allocate the memory blocks to the five arriving jobs.
| Memory Block | Memory size | Job number | Job size | Status | Internal Fragmentation |
| 1132 | 700 | - | - | Free | |
| 1003 | 720 | J5 | 610 | Busy | 110 |
| 1114 | 800 | J1 | 700 | Busy | 100 |
| 2310 | 750 | J2 | 500 | Busy | 250 |
| 1755 | 610 | - | - | Free | |
2. Given the following information:
Table 2A
| Job Number | Memory Requested |
| J1 | 30KB |
| J2 | 50KB |
| J3 | 30KB |
| J4 | 25KB |
| J5 | 35KB |
Table 2B
| ORIGINAL STATE OF MAIN MEMORY |
| 100KB(P1) |
| 25KB(P2) |
| 25KB(P3) |
| 50KB(P4) |
| 30KB(P5) |
a. Create a memory layout for the fixed partition after job entry based on the given information (Table 2A and Table B)
J1 30KB
J2 50KB
J3 30KB
J4 25KB
J5 35KB
| ORIGINAL STATE |
| 100KB
|
25KB |
|
|
| 50KB
|
|
| Job After Entry |
|
|
|
|
|
| J2 50KB
|
| J3 30KB |
P1
P2
P3
P4
P5
b. Before Job 6 (30KB) and Job 7 (45KB) arrives, there are three jobs done already for processing which J2,J3,J4. Create an initial memory layout for the dynamic partition based on the given information (Table 2A)
J1 30KB
J2 50KB
J3 30KB
J4 25KB
J5 35KB
| ORIGINAL STATE |
| 100KB
|
25KB |
|
|
| 50KB
|
|
| Job After Entry |
|
|
|
|
|
|
|
| |
P1
P2
P3
P4
P5
J2,J3,J4 are already done
| OS |
|
|
|
|
|
| J7 45KB
|
| J6 30KB |
P1
P2
P3
P4
P5
J6 and 7 arrives
3. Illustrate and find the page number with the displacement of a given program line:
Job1 is 1600 lines long
PS=200 and LNTBL=542.
Answer:
LNTBL divided by PS and the result is the Page Number and the remainder is the displacement
542 ÷ 200=2 the remainder is 142
Therefore the displacement is 142
2. List three tangible (physical) resources of a computer system and explain how it works.
Answer:
2.1 IRQ
Short for Interrupt request, IRQ is a signal that has a direct line to the computer processor, allowing it to stop the processor momentarily and decide what to do next. Every IBM compatible computer has a maximum of 15 IRQs and are prioritized in the computer according to the importance of the device. See IRQ Listing for a list of IRQs,which may be available or are currently used.
2.2 I/O
Input Output (I/O) represents the locations in memory that are designated by use of various devices to exchange information amongst themselves and the rest of the PC. See IRQ Listing for a list of IRQs and I/O ranges.
2.3 DMA
DMA, or Direct Memory Access, are pathways provided by the hardware to allow the hardware direct access to the computer's memory. See DMA Listing for listing of DMA channels.
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