The memory units that follow are specified by the number of words times the number of bits per word. how many address lines and input/output data lines are needed in each case?
A- 2G = 2 2^30 = 2^31 words therefore you need 31 bits to specify a memory location( 31 address line). - If your word has 32 bits then you need 32 data lines. - The # of bytes is (2 2^30 * 2^5 )/ 8 = (2^36) / 8 = 2^36 / 2^3 = 2^33 bytes
