CSc 360 Assignment #3


Category: You will Instantly receive a download link for .zip solution file upon Payment


5/5 - (2 votes)

There are two parts to this assignment, a) and b). You should try them in the order listed below. There is no
C coding in this assignment, solutions are done entirely in Java. For this reason, only rose.exe is provided
for you (no C code).
a) Disk Scheduling Algorithms [25%]
In this part of the assignment we are going to practice disk scheduling by implementing two scheduling
algorithms. This will let you quantitatively observe how much of an impact even the slightest intelligent
scheduling can have.
One of the crudest disk scheduling algorithms is the First In, First Out (FIFO) disk scheduling algorithm.
In FIFO, requests are retrieved from the disk in the order they arrive. There is no regard for the disk head or
other queued requests. As you might guess, this can lead to inefficient results.
The FIFO algorithm is implemented in and will help you:
1. See how to work with the DiskScheduler parent class which will ultimately help with the other
scheduling algorithms.
2. As a baseline to demonstrate and compare the effectiveness of your scheduling algorithms.
The first algorithm you will impliment is the Shortest Seek Time First (SSTF) algorithm. Before sending
the next request, SSTF reviews all queued requests and their distance to the disk head’s current position. It
then sends the queued request with the minimum distance from the disk head. An empty class file is provided for you to fill out.
The second scheduling algorithm you will impliment is the Elevator algorithm, or more specifically the
simpler Circular Elevator algorithm (C-SCAN). The elevator algorithm resembles how a real life
elevator operates. The elevator in ECS will take a direction (up or down) and then service floor requests
along the way as it heads in that direction. In the Elevator algorithm, the disk head moves in one direction
and services the closest read and write requests as it progresses in a direction. When the disk head will
8/17/13 2/6
move to the end of the disk, even if there are no more requests in that direction. When it reaches the
end it switches its direction and continues the same logic, but in the opposite direction. In C-SCAN, when
the disk head reaches the end it jumps all the way back to the beginning (ignoring any requests on the way)
and continues in the same direction. CSCAN only needs to maintain one direction. An empty class file is provided for you to fill out.
To implement these algorithms, you need to write the insert() and remove() functions.
insert() is called anytime a read() or write() is called, and should be added into your internal data
remove() is called after current request has been completed, and the current request should be
removed from your internal structure.
One thing that is very important with these two functions is that the parent DiskScheduler assumes that after
either of these functions are called, current will be set to something. current should only ever be set to null
if there are no requests AND current should only be set in insert() if it is null. Picking a new current should
be done in remove() otherwise.
To help you get a feel for the code, I recommend reviewing files in this order (and running fcfs.bin):
1. lib-source/ The basic data structure that represents a request.
2. This is what will be used in the tests and will be making the requests.
3. The test case.
4. lib-source/ Focus on the read(), write() and run() methods.
5. The FCFS implementation. The //showQ(); commented out lines might be useful
for you.
The test case files and are provided for you.
b) File System Reading [45%]
A disk scheduler allows us to effectively read/write to a disk concurrently, but what can we do with this? A
major aspect of operating systems is the file system — the organization/structure of how persistent data are
retrieved and stored. The more common file system standards in use today are NTFS (Windows) and ext4
(Linux). These systems support a whole host of features like journaling, encryption, recovery, etc. To
understand the basics of file systems we will be implementing our own VERY crude file system that
resembles ext and works with inodes.
Before we begin, it is important to hammer in that a file system is just a structure imposed on a storage
device by the operating system. A device like your HDD, USB flash stick, MP3 player, etc. are all just
blank memory. It is the operating system (or device creator in many cases) that formats the memory into a
file system like NTFS or ext. Some device manufactures even use their own custom internal file systems.
File systems tend be structured into various structures and data blocks. In practise, these blocks usually
range in size, typically from 512 bytes to 32 kB. To keep things as simple and convenient as possible, our
blocks will be 4 bytes (i.e. a java int type or int32 in C). The total number of blocks in our file system is
190. Each block is pointed to by a pointer typically 4 to 8 bytes. For simplicity, our pointers will be 4 bytes
(int) as well.
8/17/13 3/6
In most file systems, the first blocks of the file system are dedicated to booting the operating system. This
aspect is so critical, there are usually duplicates of these boot blocks in case of corruption. Luckily, we do
not need to boot; thus, the first blocks of the file system will contain our “files”. Following the ext model, all
our files are described by inodes.
An inode is simply a data structure that contains metadata about the file AND pointers to blocks containing
the files data (or data blocks). The ext inode contains a lot of metadata like access times, access privileges,
system flags, etc. This inode also has 15-16 pointers. The first 12 pointers point directly to data blocks. The
13th pointer points to a data block that contains only pointers to other data blocks. It is called an indirect
data block. The 14th pointer is a double indirect block, which points to data block full of pointers to
indirect data blocks. I will let you guess what the 15th and 16th pointers are :).
To greatly simplify things, our inode will consist of only 12 pointers, all of which point directly to data
blocks (NO indirect data block pointers). In addition, the only metadata we will store is the file size (or
total number of data blocks used). There are a finite number of inodes intially formatted for a file system.
You can actually determine the inode id for files by using ls -i — try this out in cygwin some time. The
number of initialized inodes is usally quite vast, but for our file system we will only have 8 inodes.
In order for a file system to write files it needs to keep track of which data blocks are free to write to. One
mechanism to keep track of free data blocks is by using a bitmask table. That is we set aside several blocks
(int32) and have each bit in the block represent whether a block is free (0) or used (1). Our disk has 192
data blocks, so we only need 6 blocks for the bitmap (i.e. 1 block = 32 bits and 32 * 6 = 192).
8/17/13 4/6
After the bitmap blocks are the actual 192 data blocks. All together the disk is 302 in length. For a diagram
of how the disk is made up see the image below:
In this part of the assignment you will be implementing read_file, write_file and delete_file methods for
this file system. Completing these will also require working with the data block bitmask to get and free
available data blocks. All of the code (apart from test case fiddling) is done in code/b/
There are three very basic tests provided for you:, and
1. It is recommended you begin by implementing read_file() because it is just pointer following and
does not involve working with the bitmask. You can test your implementation on
2. Next, you might want to complete delete_file(). Deleting a file is pretty straightforward, but requires
you to complete free_blocks(). If you are new to bitwise/bitshift operations or could use a refresher,
you might find this useful: Bitwise and Bit Shift Operators. The test case exists to
help test delete_file().
8/17/13 5/6
3. Finish off with write_file(). Writing a file will require you to grab some free blocks, which will
require you to complete get_free_blocks(). This too will require some bitwise/bitshift operations. Use to test the basics of this.
Some misc notes:
Some of constants declared at the top of should be useful for you.
The and disk.load(10) functions will save the disk to and load the disk from a file
called “vdisk_010” respectively.
I included a file code/b/vdisk_001. Although no test is written for it, you may or may not find it useful
for writing a test case.
A directory called “soln” contains binaries for:
sstf.bin: A solution to DiskSSTF.
cscan.bin: A solution to DiskCSCAN.
TestRead.bin: Solution to the read_file() case.
TestDelete.bin: Solution to the delete_file() test case.
TestWrite.bin: Solution to the write_file() test case.
rose.exe: identical to the one in the rose folder.
Run each and see what they do.
The assignment is to be done in teams of two. Each member will receive the same mark. Each team member
needs to understand all aspects of the assignment — midterms will be based around understanding aspects of
the assignments. It is HIGHLY recommended that team members do the assignment together.
Your evaluation will be based mostly on the correctness of your solutions. Testing is one way to evaluate
correctness. Code inspection is another. Even if your code runs, there may still be errors which can only be
uncovered by inspection. Obscured programming tricks and habits will make programs harder to
understand. Keep everything simple and elegant!
Code [70%]
Documentation [30%]
Doxygen and coding style (The solution should not be overly convoluted) [10%]
Descriptions of tests performed [10%]
Answer to questions [10%]
8/17/13 6/6
Answer the questions in doxygen. If you do not answer the questions on the mainpage, please provide a
link from the mainpage to the answers for each question (helpful for the marker and you).
Please include all necessary makefiles in each directory to generate the executables and make sure that
“make clean; make” will rebuild your solutions to each part in each subdirectories.
Submit a zipped file with the name assignment3.tgz.
assign3 should keep the structure of
To do this go into cygwin and navigate to the path above assign2 and enter:
tar czvf assignment3.tgz assign3
Please submit this file (assignment3.tgz) electronically through Connex before midnight on the due date.
Only one team member needs to submit.