ITC222 Solutions to Exercises 4

The solutions are to the exercises specified in the section Data representation of the Study Notes. The exercises are from Chapter 4 of the textbook, A Programmer's View of Computer Architecture by Goodman & Miller and some solutions are derived from those given in The Instructor's Manual by Miller.

Some of the exercises include material of general interest that goes beyond the requirements of this subject. These are indicated by an asterisk.

Exercise 4.2

320₁₀ = 1 0100 0000₂
70₁₀ = 100 0110₂
128₁₀ = 1000 0000₂
31₁₀ = 1 1111₂
25₁₀ = 1 1001₂
84₁₀ = 101 0100₂

Exercise 4.3

10 1110₂ = 46₁₀
0101₂ = 5₁₀
1111₂ = 15₁₀
1 0111 1101₂ = 381₁₀
11 1111₂ = 63₁₀
1010 1010₂ = 170₁₀

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Exercise 4.4

Direct calculator conversion gives an error on most calculators here, so proceed by longhand calculation or as follows.

log₂ 4005 = 11.96..., so 12 bits are needed in an unsigned representation or 13 in a signed representation.

Exercise 4.5

The range of a 6-bit unsigned representation is 0 to 63.
The range of a 6-bit two's complement representation is -32 to 31.

Exercise 4.6

We need to know the representation being used (two's complement, unsigned, sign-magnitude).

Exercise 4.7

decimal	2's complement	sign-magnitude	1's complement
-68	1011 1100	1100 0100	1011 1011
108	0110 1100	0110 1100	0110 1100
-120	1000 1000	1111 1000	1000 0111
-85	1010 1011	1101 0101	1010 1010
85	0101 0101	0101 0101	0101 0101
101	0110 0101	0110 0101	0110 0101
38	0010 0110	0010 0110	0010 0110
127	0111 1111	0111 1111	0111 1111
125	0111 1101	0111 1101	0111 1101
0	0000 0000	0000 0000	0000 0000

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Exercise 4.8

1111 1111 1010 1010
0000 0000 0101 0000
0000 0000 1111 1111
0000 0000 0111 1101
1000 0000 0111 1111
1111 1111 1010 1010

Exercise 4.9

The algorithm for converting a 16-bit sign-magnitude integer to an 8-bit sign magnitude integer, including check for validity is:

Interchange bits 7 and 15 (numbering from 0 as the least significant bit);
If the most significant byte is 0 remove it, otherwise the number does't fit in 8 bits;

Exercise 4.10

SAL allows integers in the 2's complement representation which range from 0x80000000 (-2147483648) to 0x7fffffff (2147483647).

Exercise 4.11

The algorithm for the binary to ascii conversion used in the solution is rather inelegant. This is because we haven't yet found out how to process arrays in memory. Consequently the ascii digits must be generated in order from most significant to least significant. A much better algorithm can be used for generating the digits in order from least significant to most significant.

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Exercise 4.12

Single precision IEEE FPS representation for -83.7₁₀:

Negative, so S = 1
83₁₀ = 101 0011 by calculator conversion.
Converting the decimal fraction to a binary fraction must be done longhand, unless you have a calculator that is smarter than mine.
```
2*0.7   0.4   1
----------------
2*0.4   0.8   0
2*0.8   0.6   1
2*0.6   0.2   1
2*0.2   0.4   0
----------------
2*0.4   0.8   0
```
where the break indicates the repetitive block. Thus 0.7₁₀ = 0.1(0110)₂ where the brackets indicate the repetitive block.
83.7₁₀ = 101 0011.1(0110) = 1.0100111(0110) * 2⁶, so the binary fraction after normalisation is F = 0100111(0110)
The exponent is biased 127, E = 6 + 127 = 1000 0101 by calculator conversion.
Gathering all together S E F is 0xc2a76666

Exercise 4.14

Additive inverse simply means the negative. Taking this operation twice gives the original number.

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Author: Ken Lodge
Last Update: 21 August, 1997

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