Question

How to find address lines for given words/bytes in memory?

Answer 1

Address format: Tag = 20 bits; Line = 6 bits; Word = 6 bits. Number of addressable units = 232 bytes; number of blocks in main memory = 226; Number of lines in cache = 26 = 64; size of tag = 20 bits. b. Address format: Tag = 26 bits; Word = 6 bits.

Answer 2

hey mate

question:

How can I determine how many address and data lines memory has and the memory size?

How can I determine how many address and data lines memory has and the memory size?1024*8 means that you have:

How can I determine how many address and data lines memory has and the memory size?1024*8 means that you have:1024 locations

How can I determine how many address and data lines memory has and the memory size?1024*8 means that you have:1024 locations8 bits per location

Address Lines:

Address Lines:Assuming that number of address lines (address bits) is n, how can we find n? If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 4 locations (0, 1, 2, and 3). As you can see, number of addressable locations = 2^n.

Address Lines:Assuming that number of address lines (address bits) is n, how can we find n? If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 4 locations (0, 1, 2, and 3). As you can see, number of addressable locations = 2^n.Given that number of addressable locations = 1024, then 1024=2^n

Address Lines:Assuming that number of address lines (address bits) is n, how can we find n? If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 4 locations (0, 1, 2, and 3). As you can see, number of addressable locations = 2^n.Given that number of addressable locations = 1024, then 1024=2^nThis means that n=log(1024) to the base 2.

Address Lines:Assuming that number of address lines (address bits) is n, how can we find n? If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 4 locations (0, 1, 2, and 3). As you can see, number of addressable locations = 2^n.Given that number of addressable locations = 1024, then 1024=2^nThis means that n=log(1024) to the base 2.Thus, n=10.

Address Lines:Assuming that number of address lines (address bits) is n, how can we find n? If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 4 locations (0, 1, 2, and 3). As you can see, number of addressable locations = 2^n.Given that number of addressable locations = 1024, then 1024=2^nThis means that n=log(1024) to the base 2.Thus, n=10.

Address Lines:Assuming that number of address lines (address bits) is n, how can we find n? If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 4 locations (0, 1, 2, and 3). As you can see, number of addressable locations = 2^n.Given that number of addressable locations = 1024, then 1024=2^nThis means that n=log(1024) to the base 2.Thus, n=10. Data Lines:

Address Lines:Assuming that number of address lines (address bits) is n, how can we find n? If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 4 locations (0, 1, 2, and 3). As you can see, number of addressable locations = 2^n.Given that number of addressable locations = 1024, then 1024=2^nThis means that n=log(1024) to the base 2.Thus, n=10. Data Lines:You have 8 bits for every location, therefore your memory needs a data bus with 8 lines. Every time you read a location (by loading its address on the address bus), the 8 bits that are stored at that location are loaded (by the memory chip) on the 8-line data bus.

Address Lines:Assuming that number of address lines (address bits) is n, how can we find n? If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 4 locations (0, 1, 2, and 3). As you can see, number of addressable locations = 2^n.Given that number of addressable locations = 1024, then 1024=2^nThis means that n=log(1024) to the base 2.Thus, n=10. Data Lines:You have 8 bits for every location, therefore your memory needs a data bus with 8 lines. Every time you read a location (by loading its address on the address bus), the 8 bits that are stored at that location are loaded (by the memory chip) on the 8-line data bus.Memory size:

Address Lines:Assuming that number of address lines (address bits) is n, how can we find n? If n=1, you can only address 2 locations (0 and 1). If n=2, you can address 4 locations (0, 1, 2, and 3). As you can see, number of addressable locations = 2^n.Given that number of addressable locations = 1024, then 1024=2^nThis means that n=log(1024) to the base 2.Thus, n=10. Data Lines:You have 8 bits for every location, therefore your memory needs a data bus with 8 lines. Every time you read a location (by loading its address on the address bus), the 8 bits that are stored at that location are loaded (by the memory chip) on the 8-line data bus.Memory size:As obvious, your memory has 1024*8 bits (8192 bits). Or simply, 1024 bytes

i hopes its helps.

please follow me!! ✌️✌️

@abhi.