Tutorial Work

1

a. There are $4\times 10^6$ memory locations, therefore maximum value is $0b1111010000100011111111$, hence $22$ bits. b. There are $8\times 10^6$ memory locations, therefore maximum value is $0b11110100001000111111111$, hence $23$ bits.

This was using powers of $10$ instead of powers of $2$.

Given $4\text{M}\times 16$ memory:

1. If it is word-addressable, $2^2\cdot 2^{20}=2^{22}$. The exponent is number of bits, hence $22$ bits.
2. If it is byte-addressable, $2^2\cdot 2^{20}\cdot 2^1=2^{23}$. Hence $23$ bits.

2

a. There are $2\times 10^6$ memory locations, therefore maximum value is $0b111101000010001111111$, hence $21$ bits. b. There are $8\times 2\times 10^6$ memory locations, therefore maximum value is $0b111101000010001111111111$, hence $24$ bits.

Given $2\text{M}\times 32$ memory:

1. If it is word-addressable, $2^1\cdot 2^{20}\times 2^1=2^{22}$. Hence, $22$ bits.
2. If it is nybble-addressable, $2^1\times 2^{20}\times 2^3=2^{24}$. Hence, $24$ bits.

3

a. Lowest address is $0$, largest address is $2^{20}-1$. b. Lowest address is $0$, largest address is $2^{19}-1$.

4

a. $275=0b100010011$, hence requires $9$ bits. b. $16-9=7$, $2^7=128$ words.

5

a. Load 5 is $0b0001000000000101$ or $0\text{x}1005$. b. Add 5 is $0b0011000000000101$ or $0\text{x}3005$. c. Halt is $0b0111000000000000$ or $0\text{x}7000$.

6

$$\begin{array}{rl}MAR& \leftarrow X\\ MBR& \leftarrow AC\\ M[MAR]& \leftarrow MBR\end{array}$$

7

MARIE can handle 16-bit data, so the AC must be 16-bits wide. MARIE’s memory is limited to 4096 address locations, MAR only needs to be 12 bits wide to hold largest address.