Exercise 2: Find an upper bound and a lower bound for each of the function f(x). 1) f(x)=x²+2x+1. 2) f(x)=3x²+2x²+x 3) f(n)=(6n+4)(1+ign)

Computer Networking: A Top-Down Approach (7th Edition)
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Hello, can you please help me do exercise 2 (with subparts 1, 2 and 3) please

Exercise 1:
Consider two algorithms A and B that solve the same class of problems.
The time complexity of A is 5000n. The time complexity for B is ceiling(1.1") or 1.17
Find the time required to execute each for n=10, n=100, n= 1000, n=1000000
Tabulate your results? which algorithm has better performance for n>1000
Exercise 2:
Find an upper bound and a lower bound for each of the function f(x).
f(x)=x²+2x+1.
1)
2) f(x)= 3x²+2x² + x
3) f(n)=(6n+4)(1+lg n)
Exercise 3:
Consider the algorithm below:
1. ien;
2. while (i z1) {
3. x=x+1
4. i=i/2}
Trace the algorithm, and fill the table that indicates the number of times x=x+1 for the following
values of n
n
8
16
32
i (range)
(8,4,2,1)
#times x=x+1 executed
4
Find a theta notation in terms of n for the number of times the statement x-x+1 is executed based on
generalizing the results above for n=2". Note that if n=2", then k=log2 (n).
Transcribed Image Text:Exercise 1: Consider two algorithms A and B that solve the same class of problems. The time complexity of A is 5000n. The time complexity for B is ceiling(1.1") or 1.17 Find the time required to execute each for n=10, n=100, n= 1000, n=1000000 Tabulate your results? which algorithm has better performance for n>1000 Exercise 2: Find an upper bound and a lower bound for each of the function f(x). f(x)=x²+2x+1. 1) 2) f(x)= 3x²+2x² + x 3) f(n)=(6n+4)(1+lg n) Exercise 3: Consider the algorithm below: 1. ien; 2. while (i z1) { 3. x=x+1 4. i=i/2} Trace the algorithm, and fill the table that indicates the number of times x=x+1 for the following values of n n 8 16 32 i (range) (8,4,2,1) #times x=x+1 executed 4 Find a theta notation in terms of n for the number of times the statement x-x+1 is executed based on generalizing the results above for n=2". Note that if n=2", then k=log2 (n).
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