8 Gi * 256 = (2**3)*(2**30)*(2**8) = 2**41 = (2**1)*(2**40) = 2 Ti 16 Mi / 64 = (2**4)*(2**20)/(2**6) = 2**18 = 256 ki log2(16 Gi) = log2((2**4)*(2**30)) = log2(2**34) = 34 An 16 GiB memory has a 32-bit word size. How many words are contained in this memory? 16 GiB / 32b = (16 Gi/32)*(B/b) = (16 Gi/32)*8 = ((2**4)*(2**30)/(2**5))*2**3 = 2**32 = 4 Gi A memory has 8 Mi words. Each word contains 16 bits. How many bytes are contained in this memory? (8 Mi)*16b/8b = (8 Mi)*2 = 16 Mi If a memory has 256 Gi addresses, how many bits are required to address the memory. log2(256 Gi) = log2(2**38) = 38