# And not notice any performance issues if the pc is

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and not notice any performance issues if the PC is being used for non-intenstive tasks. Purchasing the CPU running at 4 GHz would be a redundant purchase in this scenario and the slower clock rated CPU would be a more financially smart decision for the user. 6. Determine the minimum number of bits you would need for a memory address if memory could have up to 80 million bytes, with and without a calculator. Assume byte addressing, and fully explain how you arrive at your answer.
2 21 is roughly 2 20 * 2 = 2 million < 80 million which is not enough 2 22 is roughly 2 21 * 2 = 4 million < 80 million which is not enough 2 23 is roughly 2 22 * 2 = 8 million < 80 million which is not enough 2 24 is roughly 2 23 * 2 = 16 million < 80 million which is not enough 2 25 is roughly 2 24 * 2 = 32 million < 80 million which is not enough 2 26 is roughly 2 25 * 2 = 64 million < 80 million which is not enough 2 27 is roughly 2 26 * 2 = 128 million > 80 million which is enough Thus, number of bits required = 27 We know that 2^20 will give us 1,048,576 bytes (not enough). From here I can incrementally increase the exponent until I get a value that is higher than 80,000,000. As shown in the above calculations, 2^26 was insufficient (67,108,864) but 2^27 (134,217,728) was enough to cover the byte requirement.
With calculator: 80 Million bytes = 80000000 bytes We will use logarithms to solve this problem. Log 2 80000000 = Log 10 80000000/Log 10 2 = 7.903089986991943/0.301029996 = 26.2534966349 Checking the answer= 2 26.2534966349 = 79999998.374623 (calculator error) 2 26 = 67108864 ; 2 27 = 134217728 So, 27 bits are more than enough to represent 80 million bytes.
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