Operating System Test 1

Concept:

Swap space is a space on a hard disk used as virtual memory extension of RAM.

Explanation:

Swap space is used in various ways by different operating systems, depending on the memory management algorithms in use. Systems that implement swapping may use swap space to hold an entire process image, including the code and data segments.

Swap space can reside in one of two places: it can be carved out of the normal file system or it can be in a separate disk partition.

Swap space is used to save the process data. Operating system that offer virtual memory commonly use the mapping interface for kernel services. For instance, to execute program, operating system maps the executable into memory and then transfers control to the entry address of the executable. The mapping interface is commonly used for kernel access to swap space on disk.

I. True: The many-to-one model maps many user-level threads to one kernel thread. Thread management is done by the thread library in user space, so it is efficient. However, the entire process will block if a thread makes a blocking system call. Also, because only one thread can access the kernel at a time, multiple threads are unable to run in parallel on multicore systems.

II. False: The one-to-one model maps each user thread to a kernel thread. It provides more concurrency than the many-to-one model by allowing another thread to run when a thread makes a blocking system call.

III. False: In one-to-one model creating a user thread requires creating the corresponding kernel thread. Because the overhead of creating kernel threads can burden the performance of an application, most implementations of this model restrict the number of threads supported by the system. 

Data:

Disk size = 750 GB

Data block size = 25000 B

Overhead per entry = 10 B

Formula: 

Total number of data disk block = \(\frac{Disk\; size}{Data\; block\; size}\)

Total space occupied by overhead = over head per entry × number of entries

Maximum file size = total file system size – space consumed by overhead

Calculation:

Total number of data disk block = \(\frac{750 \; GB}{25000\; B}\) = 0.03 x 2³⁰

Space occupied by overhead = 0.03 x 2³⁰ x 10 = 0.3 x 2³⁰ 

Maximum file size = 750 x 230 – 0.3 x 230

= 749.7 x 2¹0 MB = 767692.8 MB

Data:

In index node of a Unix- style file system,

Direct Block pointer = 12

Single Indirect Block pointer = 1

Double Indirect Block pointer = 1

Triple Indirect Block pointer = 1

Disk block address = disk block entries size = 32 bit = 4 B

Disk Block size = 1 KB

Calculation:

Number of entries in a block = \(\frac{{{\rm{Disk\;block\;size}}}}{{{\rm{Disk\;block\;address\;size}}}} = \frac{{1KB}}{{4\;B}} = 256 = {2^8}\)

File size = (12 direct + 1 single indirect + 1 double indirect + 1 triple-indirect) × 256 B

File size = (12 × 1 + 1 × 2⁸ + 1 × 2⁸ × 2⁸ + 1 × 2⁸ × 2⁸ × 2⁸)× 1 KB

File size = (0.0000 + 0.00006 + 0.06250 + 16) GB

File size = 16.06256 16 GB ≈ 16 GB

Value of k inside a while loop	Condition	If condition	fork calling
1	1%5 = 1	False	called
2	2%5 = 2	False	called
3	3%5 = 3	False	called
4	4%5 = 4	False	called
5	5%5 = 0	True(skip)	not called
6	6%5 = 1	False	called
7	7%5 = 2	False	called
8	8%5 = 3	False	called
9	9%5 = 4	False	called
10	10%5 = 0	True(skip)	not Called
11	11%5 = 1	False	called
	while(11) loop terminated	Doesn't executed	Doesn't executed

 

From the table, it is clear that fork is called 9 times

Formula:

If n times fork is called the number of processes created = 2n

Calculation:

Since n = 9

number of processes created = 29 = 512

number of child processes created = 29 - 1 = 511