Lecture-6.1: Direct Mapping Math Problem Solution, Part-1 (COA)

Practice Problems Based on Direct Mapping

Question-01: Consider a direct mapped cache of size 16 KB with block size 256 bytes. The size of main memory is 128 KB. Find-

1)     Number of bits in tag

2)     Tag directory size

Solution:

Given-

Cache memory size = 16 KB

Block size = Frame size = Line size = 256 bytes

Main memory size = 128 KB

We consider that the memory is byte addressable.

Number of Bits in Physical Address-

We have, 

Size of main memory

= 128 KB

= 2¹⁷ bytes

Thus, Number of bits in physical address = 17 bits 

Number of Bits in Block Offset-

We have,

Block size

= 256 bytes

= 2⁸ bytes

Thus, Number of bits in block offset = 8 bits

Number of Bits in Line Number-

Total number of lines in cache

= Cache size / Line size

= 16 KB / 256 bytes

= 2¹⁴ bytes / 2⁸ bytes

= 2⁶ lines

Thus, Number of bits in line number = 6 bits

Number of Bits in Tag-

Number of bits in tag

= Number of bits in physical address – (Number of bits in line number + Number of bits in block offset)

= 17 bits – (6 bits + 8 bits)

= 17 bits – 14 bits

= 3 bits

Thus, Number of bits in tag = 3 bits

Tag Directory Size-

Tag directory size

= Number of tags × Tag size

= Number of lines in cache × Number of bits in tag

= 2⁶ x 3 bits

= 192 bits

= 24 bytes

Thus, size of tag directory = 24 bytes

Question-02: Consider a direct mapped cache of size 512 KB with block size 1 KB. There are 7 bits in the tag. Find-

1)     Size of main memory

2)     Tag directory size 

Solution:

Given-

Cache memory size = 512 KB

Block size = Frame size = Line size = 1 KB

Number of bits in tag = 7 bits

We consider that the memory is byte addressable.

Number of Bits in Block Offset-

We have,

Block size

= 1 KB

= 2¹⁰ bytes

Thus, Number of bits in block offset = 10 bits

Number of Bits in Line Number-

Total number of lines in cache

= Cache size / Line size

= 512 KB / 1 KB

= 29 lines

Thus, Number of bits in line number = 9 bits

Number of Bits in Physical Address-

Number of bits in physical address

= Number of bits in tag + Number of bits in line number + Number of bits in block offset

= 7 bits + 9 bits + 10 bits

= 26 bits

Thus, Number of bits in physical address = 26 bits

Size of Main Memory-

We have,

Number of bits in physical address = 26 bits

Thus, Size of main memory

= 2²⁶ bytes

= 64 MB

Tag Directory Size-

Tag directory size

= Number of tags x Tag size

= Number of lines in cache x Number of bits in tag

= 2⁹ x 7 bits

= 3584 bits

= 448 bytes

Thus, size of tag directory = 448 bytes