To figure the costs of each combonations work all cases to show the cost. There

Question

To figure the costs of each combonations work all cases to show the cost. There should be 22 cases in total.

Each computer cost $6 dollars and you have the option for 3 computers.

Security software is $4 dollars per computer and will reduce threat by 75%

A0 --> Email Server -- $10 | A1 --> Payroll -- $15 |    A2 --> File Server -- $5

Probability of being hacked: 1 computer = 5% | 2 computers = 10% | 3 computers = 15%

These are the cases to figure out cost:

NO SECURITY SECURITY Partial Security PC1 PC2 PC3 PC1 PC2 PC3 PC1 PC2 PC3 1 PC 1,2,3 1 PC 1,2,3 3 PCs 1S 2U 3U 2 PCs 1,2 3 2 PCs 1,2 3 3 PCs 1S 2S 3U 2 PCs 1,3 2 2 PCs 1,3 2 3 PCs 1S 2U 3S 2 PCs 2,3 1 2 PCs 2,3 1 3 PCs 1U 2S 3U 3 PCs 1 2 3 3 PCs 1 2 3 3 PCs 1U 2S 3S 3 PCs 1U 2U 3S 2 PCs 1S 2S 3U 2 PCs 1S 3S 2U 2 PCs 1S 2U 3U 2 PCs 1U 2U 3S 2 PCs 1U 3U 2S 2 PCs 1U 2S 3S

Explanation / Answer

#include "des56.h"
#include
#include
#include

//Use these constants to specify ENCRYPT or DECRYPT in the fencrypt function call. (The middle parameter)
#define ENCRYPT 0
#define DECRYPT 1

/*This function takes the XOR of block A and block B and stores the result in block "result"*/
/*Note that it is possible that "result" is the same as "A" (or "B").*/
/*For example:
XORblock(block1,IV,block1);
Would XOR the first block with the IV and put the result in block1*/
void XORblock(char A[8],char B[8],char result[8])
{
int i = 0;
for(;i<8;i++)
result[i] = A[i]^B[i];
}

/*This function takes a block as input and prints the corresponding hex to the terminal*/
void printhex(unsigned char block[8])
{
int i=0;
for(;i<8;i++)
printf(" 0x%x ",block[i]);
puts("");
}

void cbc_enc_three_blocks(char block1[8],char block2[8], char block3[8], char IV[8],keysched *ks)
{
// You write this code

}

void cbc_dec_three_blocks(char block1[8],char block2[8], char block3[8], char IV[8],keysched *ks)
{
// You write this code
}

int main()
{

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