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# roster method set

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Active set

In optimization, a problem is defined using an objective function to minimize or maximize, and a set of constraints

g_1(x)\ge 0, \dots, g_k(x)\ge 0

that define the feasible region, that is, the set of all x to search for the optimal solution. Given a point x in the feasible region, a constraint

g_i(x) \ge 0

is called active at x if g_i(x)=0 and inactive at x if g_i(x)>0. Equality constraints are always active. The active set at x is made up of those constraints g_i(x) that are active at the current point .

The active set is particularly important in optimization theory as it determines which constraints will influence the final result of optimization. For example, in solving the linear programming problem, the active set gives the hyperplanes that intersect at the solution point. In quadratic programming, as the solution is not mandatorily on one of the edges of the bounding polygon, an estimation of the active set gives us a subset of inequalities to watch while searching the solution, which reduces the complexity of the search.

In general an active set algorithm has the following structure:

Find a feasible starting point
repeat until "optimal enough"
solve the equality problem defined by the active set (approximately)
compute the Lagrange multipliers of the active set
remove a subset of the constraints with negative Lagrange multipliers
search for infeasible constraints
end repeat

Substitution method

In optical fiber technology, the substitution method is a method of measuring the transmission loss of a fiber. It consists of:

1. using a stable optical source, at the wavelength of interest, to drive a mode scrambler, the output of which overfills (drives) a 1 to 2&nbsp;meter long reference fiber having physical and optical characteristics matching those of the fiber under test,
2. measuring the power level at the output of the reference fiber,
3. repeating the procedure, substituting the fiber under test for the reference fiber, and
4. subtracting the power level obtained at the output of the fiber under test from the power level obtained at the output of the reference fiber, to get the transmission loss of the fiber under test.

The substitution method has certain shortcomings with regard to its accuracy, but its simplicity makes it a popular field test method. It is conservative, in that if it were used to measure the individual losses of several long fibers, and the long fibers were concatenated, the total loss obtained (excluding splice losses) would be expected to be lower than the sum of the individual fiber losses.

Some modern optical power meters have the capability to set to zero the reference level measured at the output of the reference fiber, so that the transmission loss of the fiber under test may be read out directly.

Potential method

In computational complexity theory, the potential method is a method used to analyze the amortized time and space complexity of an algorithm. It can be thought of as a generalization of the accounting method and the debit method. It is useful in cases where it is hard to assign credit to specific elements of the structure.

## The method

The potential function \phi is set to be a non-negative-valued function from states of the data structure in question. If c_i represents the actual cost of the ith operation, which updates the data structure from state A_{i-1} to state A_i, then the effective cost d_i of this operation is defined to be d_i = c_i + \phi(A_i) - \phi(A_{i-1}) (i.e., the effective cost is the actual cost plus the difference in potential).

If \phi is chosen so that the starting state of the data structure has potential 0, then the sum of the effective costs d_1,\dots,d_n is greater than or equal to the sum of the actual costs c_1,\dots,c_n. So if an upper bound on the effective cost of each operation can be shown, this implies an upper bound on the total cost of n operations, which gives an upper bound on the amortized cost per operation.

## Sample applications

### Table expansion

Determining the amortized cost of table expansion can be solved using the potential method. Let the potential function be the MAX of (the number of occupied cells minus the number of vacant cells, or 0). A regular insert incurs O(1) cost and increases the potential by O(1); thus, the effective cost of a regular insert is O(1). An insert that causes reallocation incurs an actual cost of O(n) but decreases the potential by O(n) (i.e \phi(A_i) - \phi(A_{i-1}) = - O(n) ), giving an effective cost of O(1) for this operation. Since each type of operation has O(1) effective cost, this implies an amortized cost of O(1) as well.

Accounting method

In the field of analysis of algorithms in computer science, the accounting method is a method of amortized analysis based on accounting. The accounting method often gives a more intuitive account of the amortized cost of an operation than either aggregate analysis or the potential method. Note, however, that this does not guarantee such analysis will be immediately obvious; often, choosing the correct parameters for the accounting method requires as much knowledge of the problem and the complexity bounds one is attempting to prove as the other two methods.

The accounting method is most naturally suited for proving a O(1) bound on time. The method as explained here is for proving such a bound.

## The method

Preliminarily, we choose a set of elementary operations which will be used in the algorithm, and arbitrarily set their cost to 1. The fact that the costs of these operations may in reality differ presents no difficulty in principle. What is important, is that each elementary operation has a constant cost.

Each aggregate operation is assigned a "payment". The payment is intended to cover the cost of elementary operations needed to complete this particular operation, with some of the payment left over, placed in a pool to be used later.

The difficulty with problems that require amortized analysis is that, in general, some of the operations will require greater than constant cost. This means that no constant payment will be enough to cover the worst case cost of an operation, in and of itself. With proper selection of payment, however, this is no longer a difficulty; the expensive operations will only occur when there is sufficient payment in the pool to cover their costs.

## Examples

A few examples will help to illustrate the use of the accounting method.

### Table expansion

It is often necessary to create a table before it is known how much space is needed. One possible strategy is to double the size of the table when it is full. Here we will use the accounting method to show that the amortized cost of an insertion operation in such a table is O(1).

Before looking at the procedure in detail, we need some definitions. Let T be a table, E an element to insert, num(T) the number of elements in T, and size(T) the allocated size of T. We assume the existence of operations create_table(n), which creates an empty table of size n, for now assumed to be free, and elementary_insert(T,E), which inserts element E into a table T that already has space allocated, with a cost of 1.

The following pseudocode illustrates the table insertion procedure: function table_insert(T,E) if num(T) = size(T) U := create_table(2 &times; size(T)) for each F in T elementary_insert(U,F) T := U elementary_insert(T,E)

Without amortized analysis, the best bound we can show for n insert operations is O(n2) &mdash; this is due to the loop at line 4 that performs num(T) elementary insertions.

For analysis using the accounting method, we assign a payment of 3 to each table insertion. Although the reason for this is not clear now, it will become clear during the course of the analysis.

Assume that initially the table is empty with size(T) = m. The first m insertions therefore do not require reallocation and only have cost 1 (for the elementary insert). Therefore, when num(T) = m, the pool has (3 - 1)&times;m = 2m.

Inserting element m + 1 requires reallocation of the table. Creating the new table on line 3 is free (for now). The loop on line 4 requires m elementary insertions, for a cost of m. Including the insertion on the last line, the total cost for this operation is m + 1. After this operation, the pool therefore has 2m + 3 - (m + 1) = m + 2.

Next, we add another m - 1 elements to the table. At this point the pool has m + 2 + 2&times;(m - 1) = 3m. Inserting an additional element (that is, element 2m + 1) can be seen to have cost 2m + 1 and a payment of 3. After this operation, the pool has 3m + 3 - (2m + 1) = m + 2. Note that this is the same amount as after inserting element m + 1. In fact, we can show that this will be the case for any number of reallocations.

It can now be made clear why the payment for an insertion is 3. 1 goes to inserting the element the first time it is added to the table, 1 goes to moving it the next time the table is expanded, and 1 goes to moving one of the elements that was already in the table the next time the table is expanded.

We initially assumed that creating a table was free. In reality, creating a table of size n may be as expensive as O(n). Let us say that the cost of creating a table of size n is n. Does this new cost present a difficulty? Not really; it turns out we use the same method to show the amortized O(1) bounds. All we have to do is change the payment.

When a new table is created, there is an old table with m entries. The new table will be of size 2m. As long as the entries currently in the table have added enough to the pool to pay for creating the new table, we will be all right.

We cannot expect the first \frac{m}{2} entries to help pay for the new table. Those entries already paid for the current table. We must then rely on the last \frac{m}{2} entries to pay the cost 2m. This means we must add \frac{2m}{m/2} = 4 to the payment for each entry, for a total payment of 3 + 4 = 7.

Question:1. The set of counting numbers between -3 and 5. Roster and rule? 2. The set of divisors of 36 greater than -5 but less than or equal to 18. Roster and rule? Thank you for answering. :)

Answers:In roster form, the elements are listed out as such. In rule method, the rule for building the set is specified 1) roster form : {1,2,3,4} rule method { x : x belongs to natural numbers and -3 < x < 5 } 2) roster form : { 1,2,3,4,6,9,12,18 } rule method {x : x/6 belongs to N and -5 < x <= 18}

Question:In algebra, the definition of rule is a description identifying the members of a set. The definition of a roster is how you specify a set when listing all members of a set. I kind of get the gist of this, but do you think you could explain it in greater detail and/or give examples? I'm kind of confused on how to identify this (aka, if I take a quiz over it sometime. Which may be in the near future. Like tomorrow.) Help is appreciated!

Answers:In roster form, you actually list all the elements. A = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z} is in roster form. The rule for this set is that A is the letters of the alphabet.

Question:,,i need to know what is the meaning of set, methods of describing a set and kinds of set!!! tnx alot!!!

Answers:a collection of a distinct object considered as whole. describing a set. 3 methods. a. by word b. by listing or roster method c. by set-builder rotation or rule defining method. hope those describing methods help. i was looking for other methods. need more learning. haha!