to find the memory from address lines. you just have to make address liness power of 2, as shown below for 12 address lines 212 bits. 212 = 22 * 210 = 4 Kb
that means we can address 4 Kb from 12 address lines.........
2^12 = 4096
4096= 1024*4=(2^10)*4= 4 kbytes
18 address lines is 218 or 262,144 different addresses.
it maps the memory of 4096 byte i.e 2^12=4096 byte
The number of address lines needed to access N-KB is given by log2N Then the number of address lines needed to access 256KB of main memory will be log2256000=18 address lines.
You need 20 bits of address bus to address 1 Mb of memory.
18
Memory Lane - 1947 1948-02-18 was released on: USA: 18 February 1948
Memory Lane - 1947 1948-03-18 was released on: USA: 18 March 1948
Memory Lane - 1947 1948-05-18 was released on: USA: 18 May 1948
Memory Lane - 1947 1948-06-18 was released on: USA: 18 June 1948
Memory Lane - 1947 1948-08-18 was released on: USA: 18 August 1948
Memory Lane - 1947 1948-10-18 was released on: USA: 18 October 1948
Memory Lane - 1947 1948-09-18 was released on: USA: 18 September 1948
Memory Lane - 1947 1948-11-18 was released on: USA: 18 November 1948
Memory Lane - 1947 1948-12-18 was released on: USA: 18 December 1948