A computer system uses 32-bit memory addresses and is byte-addressable. What is the maximum addressable memory capacity of this system?
With 32-bit byte-addressable addresses, total addressable memory is $2^32$ bytes = 4 GB.
In a 32-bit big-endian machine, the 32-bit integer value 0x12345678 is stored at memory address 0x1000. What byte value is stored at memory address 0x1003?
In big-endian, MSB is stored at lowest address. 0x1000 = 0x12, 0x1001 = 0x34, 0x1002 = 0x56, 0x1003 = 0x78.
How many 32K x 1 RAM chips are required to construct a memory capacity of 256 Kbytes?
Total bits = 256 KB = 2048 Kbits. Chip capacity = 32 Kbits. Number of chips = 2048 / 32 = 64 chips.
A ROM is used to store the multiplication table of two 8-bit unsigned integers. What is the minimum size of the ROM required?
Inputs = 8 + 8 = 16 address bits (64K locations). Maximum product of two 8-bit numbers requires 16 bits. Thus size is 64 K x 16 bits.
Perform addition of two 8-bit 2's complement numbers: $A = 01001101 (+77)$ and $B = 11101001 (-23)$. What are the values of Carry (C) and Overflow (V) flags?
01001101 + 11101001 = (1)00110110 (+54). Carry out $C = 1$. Carry into MSB = 1, Carry out = 1 => Overflow $V = 1 XOR 1 = 0$.