Introduction:

Amortized analysis is technique to find

minimum time required to perform a order of

Data-structure operations when a set of inputs

effects the next order of input we look at amortized analysis. We assume that a

order of operation can cost higher than the total cost. Amortized analysis is

not applicable everywhere it is applied whenever the order in one problem

effecting the other order in that problem. For calculating the worst-case

running time of any kind of data-structure operation in the order, p; if the order

contains m operations, then mp is an upper bound for the total running time.

This is technically correct but it may come with a surprisingly output and give

a wrong or unexpected result. Now from above paragraph we concluded the main

ideas of our topic is: Amortized analysis applied when a order of data

structure is effecting next order. Amortized analysis is itself an upper bound

which results as an average performance for each operation in worst case time.

Amortized analysis is exercised with the whole cost of order of operations. It does

not deals

with the cost of a single operation in that order. Let’s

suppose that

the amortized cost of insertion into a splay tree with m items is O

(log m), so when I insert ’54’ into this tree, the cost will be O

(log m). In fact, inserting ’54’ might require O

(m) operations. It is only appropriate to say, when I insert

m items into a tree, the average time for

each operation will be O (log m).

History:

Amortization in finance

means to pay off a debt i.e. loans and

mortgage, by smaller payments made over a period of time.

A method aggregate analysis which is now known as

Amortized analysis is a technique which was introduced by Robert Tarjan.

According to him and as he wrote in 1985 in his paper that we can surprisingly

achieve upper bound and lower bound for many varieties of algorithm. This

technique is used by many researchers. Robert Tarjan reveals that by using this

technique we can obtain “self-adjusting” data structures that are efficient but

must have amortized complexity low.

Amortization plays a vital

role in the analysis of many other standard algorithms and data

structures, including

·

maximum flow (In optimization theory, maximum flow problems involve

finding a feasible flow through a single-source, single-sink flow

network that is maximum)

·

Fibonacci heaps In a field of computer science,is

a data structure for priority

queue operations, consisting of a collection

of heap-ordered trees. It has a

better amortized running time than

many other priority queue data structures including the binary heap and binomial heap. Michael L. Fredman and Robert

E. Tarjan developed Fibonacci heaps in 1984

and published them in a scientific journal in 1987. They named Fibonacci heaps

after the Fibonacci numbers, which are

used in their running time analysis.

·

Dynamic arrays In a field of computer science, is a array that is growable array, resizable

array, dynamic table, mutable array, or array list is a random

access, variable-size list data

structure that allows elements to be added or

removed. It is supplied with standard libraries in many modern mainstream

programming languages. Dynamic arrays overcome a limit of static arrays, which have a fixed capacity that needs to be specified

at allocation.

Comparison to other analysis techniques:

Worst-case analysis can give overly negative

(pessimistic) bounds for orders of

operations, because this type of analysis could not tackle

with the operations on same kind of data structure. Amortized analysis may result

into more than the

practical worst-case bound by observing such kind

of interactions into account. The bound that is given

by amortized analysis is, a worst-case

bound on the average time per

operation; a specific operation in a order of

data-structure may have more cost

than this bound, but the minimum cost on whole of

the operations in any valid order will always perform

within the bound. Amortized analysis is same as average case

analysis, however it is for average cost over a order of operations.

Methods

of Amortized analysis:

We are going to discuss three methods:

·

Aggregate method

·

Potential method

·

Accounting method

Aggregate

Method:

The aggregate method, where the all over running time for a order

of operations is analyzed.

Ø In aggregate analysis,

we will proof that for every m, a order of m operations takes

Worst-case time T (m) in

total.

Ø In the worst

case, the average cost, or amortized-cost, per operation is therefore T(m)/m.

Note that this amortized cost applies to each operation, even when there are

several types of operations in the order.

The other two methods are:

Ø The accounting

method

Ø The potential

method

Stack

operations:

In our first example we will

discuss to insert a new element we can do two operation push and pop let assume

the cost in O(1).

PUSH.S: x

pushes object x onto

stack S.

POP.S

pops

the top element/object of stack S and will return

back a popped object. Now make sure that you are not calling pop on an empty

stack.

Ø What will happen

if you call pop operation on empty stack?

Surely, it will give us an

error.

Since every operation that we can

implement on stack takes O (1) time. The total cost of a order of number of

PUSH and POP operations is therefore n, and the actual running time for number

of operations is Q (n).

Now let we add one more

functionality/operation in stack operation of MULTIPOP (S,k),

where S is stack and k is a number of objects. MULTIPOP (S,k) perform

will do some simple task it will removes the k top objects of

stack S, it will result as empty stack if the

objects are fewer than k. For sure, we will always suppose that k is

positive number negative can never ever be taken if this happens the MULTIPOP operation

will not change the stack and leaves the stack same as it was previous. In the

following pseudocode, the operation STACK-EMPTY returns TRUE if there are no

objects currently on the stack, and FALSE otherwise.

The top 4 objects

are popped when we call MULTIPOP(S,4). The next

operation is MULTIPOP(S,8) which empties the stack since there

were fewer than 8 objects remaining.

According to you what in the

running time of Multipop(S,k)? The time analysis is linear for numbers of POP operations

executed, hence we can assume that the time analysis of multipop for every PUSH

and POP operation is 1.

The while loop will run

till stack is not empty and k is not equal zero that only the when we call

multipop with higher number even than it is working for each iteration of while

loop pop operation will be called as you can see in line 2. Thus, the total cost

of MULTIPOP is Q(sk),now if we recall the amortized analysis we can see

clearly the function give us linear time as whole but a part of order is

expensive than whole. Let us consider a order of

numbers of PUSH, POP, and MULTIPOP operations on an empty stack. The worst-case

cost of a MULTIPOP operation in the order s O(n), since the stack size is

minimum upto n. The worst-case time of any stack operation is therefore O(n),

and hence a order of n operations costs O(n2), since we may have

O(n) MULTIPOP operations costing O(n) each. Although this analysis is correct,

the O(n2) result, which we obtained by considering the worst-case

cost of each operation on each functionality, is not tightly bound. Using

aggregate analysis, we can obtain a better upper bound that considers the

entire order of n operations. In fact, although a single MULTIPOP operation can

be expensive, any order of n PUSH, POP, and MULTIPOP operations on an initially

empty stack can cost at most O(n).We can pop each object from the stack at most

once for each time we have pushed it onto the stack. Therefore, the number of

times that POP can be called on a nonempty stack, including calls within

MULTIPOP, is at most the number of PUSH operations, which is at most n. For any

value of n, any order of n PUSH, POP, and MULTIPOP operations takes a total of

O (n) time. The average cost of an operation is O(n)/n= O(1). In aggregate

analysis, we assign the amortized cost of each operation to be the average

cost. In this example, therefore, all three stack operations have an amortized

cost of O(1).We emphasize again that although we have just shown that the

average cost, and hence the running time, of a stack operation is O(1). We

actually showed a worst-case bound

of O(n) on a order of n operations. Div this total cost by n yielded the

average cost per operation, or the amortized cost.

Incrementing a

binary counter:

The worst case running

time occurs when all i bits are flipped, so increment (A) has running time O(i).

In a order of n increment operations, few increments will cause that many bits

to flip. Indeed, bit 0 flips with every increment bit 1 flips with every 2nd

increment bit 2 flips with every 4th increment

Total number of bit

flips in n increment operations is

n + n/2 + n/4 + … +

n/2i 0, then ?i >

ci. Operation i stores work in the data structure for later use.

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