1 answer:
#include<iostream>
#include <fstream>
#include <vector>
#include <string>
#include "StatePair.h"
using namespace std;
int main() {
ifstream inFS;
int zip;
int population;
string abbrev;
string state;
unsigned int i;
vector<StatePair <int, string>> zipCodeState;
vector<StatePair<string, string>> abbrevState;
vector<StatePair<string, int>> statePopulation;
inFS.open("zip_code_state.txt");
if (!inFS.is_open()) {
cout << "Could not open file zip_code_state.txt." << endl;
return 1;
}
while (!inFS.eof()) {
StatePair <int, string> temp;
inFS >> zip;
if (!inFS.fail()) {
temp.SetKey(zip);
}
inFS >> abbrev;
if (!inFS.fail()) {
temp.SetValue(abbrev);
}
zipCodeState.push_back(temp);
}
inFS.close();
inFS.open("abbreviation_state.txt");
if (!inFS.is_open()) {
cout << "Could not open file abbreviation_state.txt." << endl;
return 1;
}
while (!inFS.eof()) {
StatePair <string, string> temp;
inFS >> abbrev;
if (!inFS.fail()) {
temp.SetKey(abbrev);
}
getline(inFS, state);
getline(inFS, state);
state = state.substr(0, state.size()-1);
if (!inFS.fail()) {
temp.SetValue(state);
}
abbrevState.push_back(temp);
}
inFS.close();
inFS.open("state_population.txt");
if (!inFS.is_open()) {
cout << "Could not open file state_population.txt." << endl;
return 1;
}
while (!inFS.eof()) {
StatePair <string, int> temp;
getline(inFS, state);
state = state.substr(0, state.size()-1);
if (!inFS.fail()) {
temp.SetKey(state);
}
inFS >> population;
if (!inFS.fail()) {
temp.SetValue(population);
}
getline(inFS, state);
statePopulation.push_back(temp);
}
inFS.close();
cin >> zip;
for (i = 0; i < zipCodeState.size(); ++i) {
}
for (i = 0; i < abbrevState.size(); ++i) {
}
for (i = 0; i < statePopulation.size(); ++i) {
}
}
Statepair.h
#ifndef STATEPAIR
#define STATEPAIR
#include <iostream>
using namespace std;
template<typename T1, typename T2>
class StatePair {
private:
T1 key;
T2 value;
public:
StatePair()
{}
void SetKey(T1 key)
{
this->key = key;
}
void SetValue(T2 value)
{
this->value = value;
}
T1 GetKey() { return key;}
T2 GetValue() { return value;}
void PrintInfo()
{
cout<<key<<" : "<<value<<endl;
}
};
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Answer:
a.) k² - 3k + 3
b.) 1/(1 - k)²
c.)
Explanation:
a.) A packet can make 1,2 or 3 hops
probability of 1 hop = k ...(1)
probability of 2 hops = k(1-k) ...(2)
probability of 3 hops = (1-k)²...(3)
Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)
= (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)
= k + 2k - 2k² + 3(1 + k² - 2k)
∴mean number of hops = k² - 3k + 3
b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3
if k = 0 then number of hops is 3
if k = 1 then number of hops is (1 - 3 + 3) = 1
multiple transmissions can be needed if K is between 0 and 1
The probability of successful transmissions through the entire path is (1 - k)²
for one transmission, the probility of success is (1 - k)²
for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)
for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on
∴ for transmitting a single packet, it makes:
∞ n-1
T = ∑ n(1 - k)²(1 - (1 - k)²)
n-1
= 1/(1 - k)²
c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)
from (a) above, mean number of hops when transmitting a packet = k² - 3k + 3
from (b) above, mean number of transmissions by a packet = 1/(1 - k)²
substituting: mean number of required packet =