When making a presentation to an audience using powerpoint, you use notes page view.
a. true
b. false

One subtask in the game is to roll the dice. explain why is roll the dice an abstraction.

Consider a slotted aloha system, where the time slot equals the fixed duration of each packet. assume that there are 4 stations a, b,c, d sharing the medium. (a) stations a, b,c, d receive one packet each from higher layers at times 1.3, 1.5, 2.6,5.7 respectively. show which transmissions take place when, according to the slottedaloha protocol; describe all transmissions until all four packets have been successful. when needed, each station has access to the following sequence of random number, provided by a random number generator and drawn uniformly between 0 and 1: (1) station a draws numbers: 0.31, 0.27, 0.78, 0.9, 0.9, 0.11, 0. (2) station b draws numbers: 0.45, 0.28, 0.11, 0.83, 0.37, 0.22, 0. (3)station c draws numbers: 0.1, 0.2, 0.3, 0.4, 0. (4) station d draws numbers: 0.36, 0.77, 0.9, 0.1, 0.1, 0.1, 0.1, 0. (b) in slotted aloha, a station transmits in each time slot with a given probability. what probabilities would you assign to each of the four stations so as to: (i) maximize the efficiency of the protocol? (ii) maximize fairness among the four stations? (c) will the efficiency increase or decrease if we modify slotted aloha as follows: (i) get rid of slots and allow stations to transmit immediately? (ii) implement carrier sensing? (iii) implement collision detection? (iv) implement collision avoidance?

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i. e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }