Analysis of potential method with example

Stack operations
Push operation: Time complexity to push an item into a stack is O(1)

Pop operation: Time complexity to pop an item from a stack is O(1)

Multipop operation: Time complexity to pop k item from a stack is min(stack size, k).

Algorithm:

Multipop (S, k)
While not Stack_Empty (S) and k ! = 0
    do Pop (S)
    k=k – 1

Let us define the potential of a stack as the number of elements in the stack.
So, Φ(D₀) = 0 and Φ(D_i) ≥ 0.

Amortized cost of Push: According to potential function.

Amortised cost(C_i) = Actual cost(c_i ) + change in potential
C_i = c_i +Φ(D_i)-Φ(D_i-1) = 1+n-(n-1) = 2

So, C_i is also O(1)

Amortized cost of Pop: According to potential function:

Amortised cost(Ci) = Actual cost(ci ) + change in potential
Ci = ci +Φ(D_i-1)-Φ(D_i) = 1+n-1-(n) = 0

So, C_i is also O(1)

Amortized cost of MultiPop: According to potential function:

Amortised cost(Ci) = Actual cost(ci ) + change in potential
Ci = ci +Φ(D_i)-Φ(D_i-1) = k + s – k – s =0

So, C_i is also O(1)

Now, we have the final result as shown below in the table

So, the amortized cost of each operation is O(1) and the total cost of each operation is O(N). 
So time complexity of N operations is O(N).
Therefore, it is appropriate to quantify the change in potential as positive for low-cost operations and negative for high-cost operations.