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Old 11-08-2008
partition question

Hi
I have an exam question which is below and I am looking for answer of this question:
I would appreciate your help.

* You are the systems administrator of a mail server that services approximately 3,000 users. 2,000
users access their email via a POP-3 service, while the remaining 1,000 users access their email via a
Unix mail reader. Recently users have complained about speed of disk access, so a new 10 gigabyte
disk has been purchased for the mail server to go with the other 10 Gb disk already in place.
How would you partition out /var with these two separate 10 gigabyte disks? Note, these disks will
be on the same SCSI controller, but will not overload the SCSI bus in any way.

Last edited by jim mcnamara; 11-08-2008 at 07:25 AM.. Reason: homework is not allowed
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struct::disjointset(n)						Tcl Data Structures					    struct::disjointset(n)

__________________________________________________________________________________________________________________________________________________

NAME
struct::disjointset - Disjoint set data structure SYNOPSIS
package require Tcl 8.4 package require struct::disjointset ?1.0? ::struct::disjointset disjointsetName disjointsetName option ?arg arg ...? disjointsetName add-partition elements disjointsetName partitions disjointsetName num-partitions disjointsetName equal a b disjointsetName merge a b disjointsetName find e disjointsetName destroy _________________________________________________________________ DESCRIPTION
This package provides disjoint sets. An alternative name for this kind of structure is merge-find. Normally when dealing with sets and their elements the question is "Is this element E contained in this set S?", with both E and S known. Here the question is "Which of several sets contains the element E?". I.e. while the element is known, the set is not, and we wish to find it quickly. It is not quite the inverse of the original question, but close. Another operation which is often wanted is that of quickly merging two sets into one, with the result still fast for finding elements. Hence the alternative term merge-find for this. Why now is this named a disjoint-set ? Because another way of describing the whole situation is that we have o a finite set S, containing o a number of elements E, split into o a set of partitions P. The latter term applies, because the intersection of each pair P, P' of partitions is empty, with the union of all partitions covering the whole set. o An alternative name for the partitions would be equvalence classes, and all elements in the same class are considered as equal. Here is a pictorial representation of the concepts listed above: +-----------------+ The outer lines are the boundaries of the set S. | / | The inner regions delineated by the skewed lines | * / * | are the partitions P. The *'s denote the elements | * / | E in the set, each in a single partition, their |* / | equivalence class. | / * | | / * / | | * / * / | | / / | | / / * | | / * | | / * | +-----------------+ For more information see http://en.wikipedia.org/wiki/Disjoint_set_data_structure. API
The package exports a single command, ::struct::disjointset. All functionality provided here can be reached through a subcommand of this command. ::struct::disjointset disjointsetName Creates a new disjoint set object with an associated global Tcl command whose name is disjointsetName. This command may be used to invoke various operations on the disjointset. It has the following general form: disjointsetName option ?arg arg ...? The option and the args determine the exact behavior of the command. The following commands are possible for disjointset objects: disjointsetName add-partition elements Creates a new partition in specified disjoint set, and fills it with the values found in the set of elements. The command maintains the integrity of the disjoint set, i.e. it verifies that none of the elements are already part of the disjoint set and throws an error otherwise. The result of the command is the empty string. disjointsetName partitions Returns the set of partitions the named disjoint set currently consists of. disjointsetName num-partitions Returns the number of partitions the named disjoint set currently consists of. disjointsetName equal a b Determines if the two elements a and b of the disjoint set belong to the same partition. The result of the method is a boolean value, True if the two elements are contained in the same partition, and False otherwise. An error will be thrown if either a or b are not elements of the disjoint set. disjointsetName merge a b Determines the partitions the elements a and b are contained in and merges them into a single partition. If the two elements were already contained in the same partition nothing will change. The result of the method is the empty string. disjointsetName find e Returns the partition of the disjoint set which contains the element e. disjointsetName destroy Destroys the disjoint set object and all associated memory. BUGS, IDEAS, FEEDBACK This document, and the package it describes, will undoubtedly contain bugs and other problems. Please report such in the category struct :: disjointset of the Tcllib SF Trackers [http://sourceforge.net/tracker/?group_id=12883]. Please also report any ideas for enhancements you may have for either package and/or documentation. KEYWORDS
disjoint set, equivalence class, find, merge find, partition, partitioned set, union CATEGORY
Data structures struct 1.0 struct::disjointset(n)