A computer employs RAM chips of 1024 x 8 and ROM chips of 2048 x 4. The computer system needs 2K bytes of RAM, and 2K bytes of ROM and an interface unit with 256 registers each. A memory-mapped I/O configuration is used. The two higher -order bits of the address bus are assigned 00 for RAM, 01 for ROM, and 10 for interface.

a)How many RAM and ROM chips are needed? Draw a memory-address map for the system and Give the address range in hexadecimal for RAM, ROM

Can someone help me out , i solved the (a) part as: The following ROM and RAM Will be used 8 RAM chips and 4 ROM chips are required Explanation: RAM chip size = 256 multiply 8

Required memory size = 2000 bytes = 2 multiply 1024 bytes = 2 multiply 1024 multiply 8 bits

Total number of RAM chip required –

(2 multiply 1024 multiply 8)/(256 multiply 8) = 8 chips

ROM chip size = 1024 multiply 8

Required memory size = 4k bytes = 4 multiply 1024 bytes = 4 multiply 1024 multiply 8 bits

Total number of ROM chips required –

(4 multiply 1024 multiply 8)/(1024 multiply8) = 4 chips

But i am clueless as how to draw the Memory address map