Question

7. A computer has a memory space of 8 GB.

a) How many address lines are required to span this address space, assuming it is byte-addressed?

b) This computer has a block of 2 GB 32-bit-wide memory built using 512 MB static RAM chips that are each 8 bits wide. How many RAM chips are required to implement the memory?

Answer 1

Iga) Memory space of computer that is storage = 8 GB IMB IGB - 1024 MB &GB converted in bits - 1024 KB 8 X 1024 X 1024x1024X 8 bits 1 KB - 1024B => 8X1024 X 1024 X 102 4 Bytes Formula to calculate number address lunes (n) are 1 Byte= 8 Bit n = log (space) [ space should be in bytes n = (848) log 2 loa (8X1024X1024x1024 GB 3) (space in lites langa (2² x 20 x 20 x 24 (233) - И 33 log a 2 [ by using log formula b M log = 33 XI = c log ab & = 33 address lines loga

05 2GB 32 bit uude memory is expanty - 512 MB RAM chup & bits wide Now Total 2 GB X 32 memory Now Total Number memory RAM chips X capacity per RAM chips = Number RAM chips X eparty /shop = 268 x 32 chips X 512 MB X8 Number = 2GB X 32 Number of chips 2 X 1024 MB X 32 1GB = 1024 MB 512 MB X 8 2 X 2 X 4 16 2 Number RAM chips = 16