3. A, (first choice)
4. C (Third choice from top)
Answer:
O(N!), O(2N), O(N2), O(N), O(logN)
Explanation:
N! grows faster than any exponential functions, leave alone polynomials and logarithm. so O( N! ) would be slowest.
2^N would be bigger than N². Any exponential functions are slower than polynomial. So O( 2^N ) is next slowest.
Rest of them should be easier.
N² is slower than N and N is slower than logN as you can check in a graphing calculator.
NOTE: It is just nitpick but big-Oh is not necessary about speed / running time ( many programmers treat it like that anyway ) but rather how the time taken for an algorithm increase as the size of the input increases. Subtle difference.
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
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