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Unix 1. what do you mean by File Management Techniques? 2. Tell me about Memory Protection? 3. I need some examples of deadlocks 4. There are three common storage placement policy in a variable partition allocation, which ine can be considered as the best? and why?

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Posted on Dec 03, 2009

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Examples of operating system

Examples of operating system are:
1. BSD, 2. Linux 3. Mac OS X 4. Microsoft Windows 5. UNIX

Jul 02, 2011 | Computers & Internet


How to divide bandwidth in squid

What is Bandwidth throttling :

Consider we are having 512Kbps Internet connection line. And the system administrator wants to divide the whole bandwidth into two sections. Read More One for the normal users and the other for special users who need high speed internet connection. So that all the normal users can use 256 Kbps and the special users can use the remaining 256 Kbps. So in this case you have to make a bandwidth throttling to differentiate the whole single pipe line into two sections. This can be achieved using squid proxy server.

Squid main configuration file ----- /etc/squid/squid.conf
Squid log file ----- /var/log/squid/access.log
Cache log file ----- /var/log/squid/cache.log

Bandwidth throttling in squid is done using Delay Pools. Delay pools uses bucketing system.

Throttle Calculations
512Kbps(Kilo bits per second) -----> 64Kbytes -----> 64000bytes
256Kbps -----> 32Kbytes -----> 32000bytes
128Kbps -----> 16Kbytes -----> 16000bytes
64Kbps -----> 8Kbytes -----> 8000bytes

Points to remember in Delay Pools:
There are basically three things to note, they are - delay pools, delay class, delay parameters.

Delay pool --- Defines how many pools we want to use
Delay Class ---- Defines type of the pool you are going to use.
Delay Parameter � allots the restrictions and fill rate/maximum bucket size.

As I have said previously in this documentation delay pools uses bucketing system.
Now there are three types of buckets

Class 1 pool: A single aggregate bucket, shared by all users
Class 2 pool: One aggregate bucket, 256 individual buckets
Class 3 pool: One aggregate bucket, 256 network buckets, 65,536 individual buckets

If you still have any problem in the above syntax (ie) aggregate, network and individual buckets to understand, then here is a simple syntax/example for all these

For Class 1 delay pool
delay_parameters 1 32000/32000

For Class 2 delay pool
delay_parameters 1 48000/48000 48000/48000

For Class 3 delay pool
delay_parameters 2 32000/32000 8000/8000 16384/16384

Example Setup:
One 512Kbps pipe line. We want to distribute the whole pipe line into 2. One for normal users and other for special users as follows
Normal users --- 128 Kbps
Special Users --- 384 Kbps

Configuration Setup File /etc/squid/squid.conf:
Before getting into the delay pools setup first create the acl(Access Control List). Through acl you can define rules according to your requirements. Find the Access Control section in the squid.conf file.
Here is an example:

acl superusers src
acl mynetwork src

Here in the above example only three users are special users who need 256Kbps bandwidth which are listed in the specialusers label and the whole network including he special users are labeled as mynetwork.

After creating the users you have to allow them to access the internet. Below line specifies for allowing the mentioned labeled users.

http_access allow superusers
http_access allow mynetwork

Now comes the funny part Delay Pools. Here We will deal with a basic example for delay pools.

The example is as follows: We are having 512Kbps pipe line connection. We want to divide it into segments one for the special users, web servers and the other for the normal users in the organization. The special users and the web servers are given 384Kbps speed and the remaining 128Kbps for the normal users.

Delay Pools for super users:
Check for the delay pools section in the squid.conf file.
Start the configuration for delay pools as follows
########## Delay Pools############

delay_pools 2

As described above we have to create 2 delay pools, one for each delay class.
######### Defining Delay pool 1 in class 2 #########

delay_class 1 2
delay_parameters 1 48000/48000 48000/48000
delay_access 1 allow superusers

The first line specifies Delay Class which defines delay pool 1 for the delay class 2.

Why we are using delay class 2 here?

The first part on the second line ie �1� defines the pool One(1)

The second part on second line defines the aggregate 48000/48000(restore/max). where restore is the number of bytes (not bits - modem and network speeds are usually quoted in bits) per second placed into the bucket, and maximum is the maximum number of bytes which can be in the bucket at any time.

The third part on second line is individual buckets again 48000/48000(restore/max). All the special users and the web servers should obtain the same speed of 384 Kbps.

The third line allows the super users to fall in that bucket.

######### Defining Delay pool 2 in class 3 #########

Delay Pools for normal users:

delay_class 2 3
delay_parameters 2 32000/32000 8000/8000 16384/16384
delay_access 2 allow !superusers

The first line specifies Delay Class which defines delay pool 2 for the delay class 3.

The first part on the second line ie �2� defines the pool Two(2) of class three(3)

The second part on second line defines the aggregate 32000/32000(restore/max) as aggregate (ie for whole).
Note : - If you use -1/-1. �-1� indicates �unlimited�.

The third part on second line is network bucket. We have defined here 8000/8000 .

The fourth part on second line is for individual.

The third line specifies to throttle all the users except the super users.

How to check:
First set proxy settings in your browser as follows:
Go to any client machine and open an IE(Internet Explorer). Select "Tools" menu then select "Internet Options" then select "Connections" tab. After that select the "LAN Settings" button, you will get a new pop-up window there you select "Use proxy server for your LAN" check box and finally give your squid proxy servers ip address and the port on which the squid proxy server is configured. Now press "OK" and again press "OK". Cloase the window and now open a new window.
Second Check by downloading any large file

Start to download any lage file not less than 10MB size. Because when ever you download any small file it used to download it quickly. When you start downloading large file at first if you note some starting bytes of the file gets downloaded quickly and later it gets slowed down.

Keep a note on the download pop-up box which shows the speed limits in "kbps". When we start downloading note on the pop-up box the speed limits in kbps shows high value and then slowly gets decreased. This means when you start downloading the bucket gets filled and slowly it decreases and becomes stable to the specified limit in Kbps according to the configuration. Remember again as i told restore value is used to set the download speed, and the max value lets you set the size at which the files are to be slowed down from.

on Apr 24, 2010 | fedora Linux Operating System

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I cant got the wave gts 5253 themes

You can download themes from , for example.
1. Download a theme (an .smt file) to your computer, note the location of the file so you'll find it later 2. Connect your phone to your computer via USB cable, and select "Mass Storage" mode on the phone 3. Transfer the .smt theme file to your phone's memory, into "Themes" directory 4. Disconnect the USB cable 5. Go to Settings -> Display and Light -> Theme 6. Select the new theme you want to install

Mar 16, 2011 | Computers & Internet

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I have a trial version of excel 7 on my computer; it has expired what doI need to buy to enable wha have?

process allocation Max
p1 0 0 1 2 0 0 1 2
p2 2 0 0 0 2 7 5 0
p3 0 0 3 4 6 6 5 6
p4 2 3 5 4 4 3 5 6
p5 0 3 3 2 0 6 5 6

total recourses in a system (r1,r2,r3,r4)=(6,7,12,12)
(i) Calculate available resource martrix & need matrix for state of
(ii) determine saftey if Iexits for the system applying banker's
(iii) p3 request for(0,1,0,0) . does this request lead to a deadlock?

Feb 01, 2010 | Microsoft Windows Vista Home Premium with...

2 Answers

Can not reinstall os on mt acer 246lc it bleeps 4 times and swithes off

Without knowing what motherboard is in the unit - it will be difficult to determine what the 4 beeps means.

Generally, 4 beeps means issues with :

- memory
- video card
- keyboard
- motherboard failure

Things to try -

- make sure there is no keys stuck down
- remove the battery and AC adapter, remove and re-install the memory sticks (reseating the memory)

If this does not work, your problem may be more serious. Let us know the results of the above suggestions.

Nov 17, 2009 | Acer Computers & Internet

2 Answers

How to be solve banker algorithm?example

I m providing you this from my college notes

Banker's Algorithm

* multiple instances of resource types IMPLIES cannot use resource-allocation graph

* banks do not allocate cash unless they can satisfy customer needs when a new process enters the system

* declare in advance maximum need for each resource type

* cannot exceed the total resources of that type

* later, processes make actual request for some resources

* if the the allocation leaves system in safe state grant the resources

* otherwise, suspend process until other processes release enough resources

Banker: Data Structures define MAXN 10 /* maximum number of processes */
#define MAXM 10 /* maximum number of resource types */
int Available[MAXM]; /* Available[j] = current # of unused resource j */
int Max[MAXN][MAXM]; /* Max[i][j] = max demand of i for resource j */
int Allocation[MAXN][MAXM]; /* Allocation[i][j] = i's current allocation of j*/
int Need[MAXN][MAXM]; /* Need[i][j] = i's potential for more j */
/* Need[i][j] = Max[i][j] - Allocation[i][j] */


X <= Y iff X[i] <= Y[i] for all i

(0,3,2,1) is less than (1,7,3,2)

(1,7,3,2) is NOT less than (0,8,2,1)

Each row of Allocation and Need are vectors: Allocation_i and Need_i

Banker: Example


10 5 7

Later Snapshot:

Max - Allocation = Need Available
P0 7 5 3 0 1 0 7 4 3 3 3 2
P1 3 2 2 2 0 0 1 2 2
P2 9 0 2 3 0 2 6 0 0
P3 2 2 2 2 1 1 0 1 1
P4 4 3 3 0 0 2 4 3 1

Banker: Safety Algorithm

* consider some sequence of processes

* if the first process has Need less than Available

* it can run until done

* then release all of its allocated resources

* allocation is increased for next process

* if the second process has Need less than Available

* ...

* then all of the processes will be able to run eventually

* IMPLIES system is in a safe state

Banker: Safety Algorithm

STEP 1: initialize
Work := Available;
for i = 1,2,...,n
Finish[i] = false
STEP 2: find i such that both
a. Finish[i] is false
b. Need_i <= Work
if no such i, goto STEP 4
Work := Work + Allocation_i
Finish[i] = true
goto STEP 2
if Finish[i] = true for all i, system is in safe state

Banker: Safety Example

Using the previous example, P1,P3,P4,P2,P0 satisfies criteria.

Max - Allocation = Need <= Work Available
P1 3 2 2 2 0 0 1 2 2 3 3 2 3 3 2
P3 2 2 2 2 1 1 0 1 1 5 3 2
P4 4 3 3 0 0 2 4 3 1 7 4 3
P2 9 0 2 3 0 2 6 0 0 7 4 5
P0 7 5 3 0 1 0 7 4 3 10 4 7
10 5 7<<< initial system

Banker: Resource-Request Algorithm

STEP 0: P_i makes Request_i for resources, say (1,0,2)
STEP 1: if Request_i <= Need_i
goto STEP 2
else ERROR
STEP 2: if Request_i <= Available
goto STEP 3
else suspend P_i
STEP 3: pretend to allocate requested resources
Available := Available - Request_i
Allocation_i := Allocation_i + Request_i;
Need_i := Need_i - Request_i
STEP 4: if pretend state is SAFE
then do a real allocation and P_i proceeds
restore the original state and suspend P_i

Banker: Resource-Request Algorithm [129]

Say P1 requests (1,0,2)

Compare to Need_1: (1,0,2) <= (1,2,2)

Compare to Available: (1,0,2) <= (3 3 2)

Pretend to allocate resources:

Max - Allocation = Need Available
P0 7 5 3 0 1 0 7 4 3 2 3 0<<<
P1 3 2 2 3 0 2<<< 0 2 0<<<
P2 9 0 2 3 0 2 6 0 0
P3 2 2 2 2 1 1 0 1 1
P4 4 3 3 0 0 2 4 3 1

Is this safe? Yes: P1, P3, P4, P0, P2

Can P4 get (3,3,0)? No, (3,3,0) > (2,3,0) Available

Can P0 get (0,2,0)? (0,2,0) < (2,3,0) Available

Pretend: Available goes to (2,1,0)

Thanks And Regards

May 13, 2009 | Microsoft Windows XP Professional

1 Answer

What are the differences between windows vista and unix


Hmmm... This is a BIG question, and it has a HUGE answer.
Have you ever seen Unix and Vista? I think you would immediately see the difference between the two.
1. Vista has a Graphic User interface, Unix is a command line interface.
2. Totally different systems (Filesystem, application, Kernel... whatever)
3. Age - Unix is ooold, and Vista is new (Unix dates back from the 1980's)

Well i can go on like this for days ...Google it more.

Darko, Serbia

Mar 19, 2009 | Microsoft Computers & Internet

1 Answer


Improving UNIX server performance NOTES.INI settings Most NOTES.INI settings that affect IBM® Lotus® Domino® server performance apply to all UNIX® platforms.
NSF_Buffer_Pool_Size_MB Many machines that run UNIX have very large amounts of physical RAM. Use the parameters NSF_Buffer_Pool_Size_MB or PercentSysAvailable Resources to control how much memory Domino is allowed to use. Each Domino instance on a UNIX machine can reference a maximum of 4GB of RAM.
Disk and memory requirements When a UNIX system runs Domino server software, the server must have enough disk space for program and data files and enough memory to handle swapping and the number of processes. You can also change several system parameters to improve server performance.
Disk I/O tuning Maintaining multiple file systems for operating system files, swap space, transaction logs, and data improves overall server performance.
Use RAID 0+1 hardware for the disk drives on which the data files reside. Use multiple smaller disk drives instead of a few large disk drives for Domino data. Domino does not perform simple predictable sequential reads; therefore, disable Read Ahead Cache and enable Write Cache.
Keeping swap space on its own separate striped volumes improves server performance at high loads. The transaction logs should always be on the most reliable and highest performing disk subsystem available to the system hosting the Domino server. Transaction logging must be on its own disk drives for improved server restart time, reliability, and availability. The logged transactions are written to disk as fast serial writes to a sequential file that is of configurable size in 4K blocks.
Console and database logging To improve server performance, limit the amount of information that is logged to the log file (LOG.NSF) and the console.

Feb 21, 2009 | Computers & Internet

1 Answer


Fragmentation occurs when the operating system cannot or will not allocate enough contiguous space to store a complete file as a unit, but instead puts parts of it in gaps between other files (usually those gaps exist because they formerly held a file that the operating system has subsequently deleted or because the operating system allocated excess space for the file in the first place). Larger files and greater numbers of files also contribute to fragmentation and consequent performance loss. Defragmentation attempts to alleviate these problems.
Example Consider the following scenario, as shown by the image on the right:
410px-file_system_fragmentation.svg.png An otherwise blank disk has 5 files, A, B, C, D and E each using 10 blocks of space (for this section, a block is an allocation unit of that system, it could be 1K, 100K or 1 megabyte and is not any specific size). On a blank disk, all of these files will be allocated one after the other. (Example (1) on the image.) If file B is deleted, there are two options, leave the space for B empty and use it again later, or compress all the files after B so that the empty space follows it. This could be time consuming if there were hundreds or thousands of files which needed to be moved, so in general the empty space is simply left there, marked in a table as available for later use, then used again as needed.[1] (Example (2) on the image.) Now, if a new file, F, is allocated 7 blocks of space, it can be placed into the first 7 blocks of the space formerly holding the file B and the 3 blocks following it will remain available. (Example (3) on the image.) If another new file, G is added, and needs only three blocks, it could then occupy the space after F and before C. (Example (4) on the image). Now, if subsequently F needs to be expanded, since the space immediately following it is no longer available, there are two options: (1) add a new block somewhere else and indicate that F has a second extent, or (2) move the file F to someplace else where it can be created as one contiguous file of the new, larger size. The latter operation may not be possible as the file may be larger than any one contiguous space available, or the file conceivably could be so large the operation would take an undesirably long period of time, thus the usual practice is simply to create an extent somewhere else and chain the new extent onto the old one. (Example (5) on the image.) Repeat this practice hundreds or thousands of times and eventually the file system has many free segments in many places and many files may be spread over many extents. If, as a result of free space fragmentation, a newly created file (or a file which has been extended) has to be placed in a large number of extents, access time for that file (or for all files) may become excessively long.
The process of creating new files, and of deleting and expanding existing files, may sometimes be colloquially referred to as churn, and can occur at both the level of the general root file system and in subdirectories. Fragmentation not only occurs at the level of individual files, but also when different files in a directory (and maybe its subdirectories), that are often read in a sequence, start to "drift apart" as a result of "churn".
A defragmentation program must move files around within the free space available to undo fragmentation. This is a memory intensive operation and cannot be performed on a file system with no free space. The reorganization involved in defragmentation does not change logical location of the files (defined as their location within the directory structure).

Jul 20, 2008 | Microsoft Windows Server Standard 2003 for...

3 Answers

Getting "Not enough memory available to complete this operation" error message using Sound Recorder

This problem occurs because of a design limitation in Sound Recorder and not because of any problem with the underlying multimedia. That means that the program is unable to access more than 2 gigs of memory by design.

Apr 08, 2008 | Microsoft Windows XP Professional With...

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