Solutions to End of Chapter Problems
Problem 1:
Because data can only be stored in 2 states (on/off, open/closed, etc.).
Problem 2:
Main memory is very expensive compared to secondary storage.
Problem 3:
Secondary memory is very slow compared to main memory.
Problem 4:
(a)
(b)
(c)
Problem 5:
False. Running programs are stored in main memory (i.e., Random Access Memory).
Problem 6:
Secondary memory (the hard drive).
Problem 7:
The operating system.
Problem 8:
(a)
(b)
(c)
Problem 9.
Writing the value of the position below each bit we have
Secondary memory (the hard drive)
Main memory (i.e., Random Access Memory)
Secondary memory (the hard drive)
9216 bits / 8 bits per byte = 1152 bytes
1152 bytes / 4 bytes per word = 288 words
1152 bytes / 1024 bytes per KB = 1.125 KB
1 0 0 0 0 0
32 16 8 4 2 1
Thus 1000002  3210
Problem 10.
101012  2110 .
Problem 11:
0
27
1
26
1
25
1
24
0
23
1
22
0
21
1
20
0(128) + 1(64) + 1(32) + 1(16) + 0(8) + 1(4) + 0(2) + 1(1) = 64 + 32 + 16 + 4 + 1 = 11710
Problem 12:
00101110 in base 2 is equivalent to 46 in base 10.
Problem 13.
7810  1001110 2
Problem 14:
4310  101011 2
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Problem 15:
28 = 256 and 27 = 128 , so there is a 1 in the 27 position.
223 – 128 = 95. Since 26 = 64, there is a 1 in the 26 position.
95 – 64 = 31. Since 25 = 32 and 24 = 16, there is a 1 in the 24 position.
31 – 16 = 15. Since 23 = 8, there is a 1 in the 23 position.
15 – 8 = 7. Since 22 = 4, there is a 1 in the 22 position.
7 – 4 = 3. Since 21 = 2, there is a 1 in the 21 position.
3 – 2 = 1. Since 20 = 1, there is a 1 in the 20 position.
1 1 0 1 1 1 1 1
27 26 25 24 23 22 21 20
= 110111112
Problem 16.
0x27 is equivalent to 39 in base-10.
Problem 17:
0xBEE is equivalent to 3054 in base-10.
Problem 18.
0x100 is 100000000 in binary.
Problem 19:
0xF1E is equivalent to 111100011110 in binary.
Problem 20:
0xB9B
Problem 21.
Since the first hexadecimal digit is C, the first four bits are 1100. Thus, the fourth bit is a zero.
Note how more difficult this question would have been if the address was provided in base-10
instead of in base-16.
Problem 22:
The answer is 0x0000006B. The picture is:
Address
06B
06C
06D
06E
06F
070
071
072
073
074
What is stored
C is stored in these two bytes
B is stored in
these four bytes
A is stored in
these four bytes
Problem 23:
The letter k is stored as ASCII 0x6B which in binary is 01101011 .
Problem 24:
The letter g is stored as ASCII 0x67.
Problem 25:
0x138 + 0x6 = 0x13E.
Problem 26.
0x8D4.
Problem 27:
0x113D
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Problem 28:
(a)
0804839d
0804839e
0804839f
080483a0
080483a1
080483a2
080483a3
080483a4
(b)
The four preceding addresses are (from lowest to highest):
08048398
08048399
0804839a
0804839b
(c) In ASCII, 0x55 = U, 0x53 =S, 0x4E = N, 0x41 = A, 0x21 = ! Thus: USNA!
Problem 29.
16 into 730 gives a quotient of 45 with a remainder of A
16 into 45 gives a quotient of 2 with a remainder of D
16 into 2 gives a quotient of 0 with a remainder of 2
Thus, 73010  0x2DA . Since we are asked to use the number of hex digits appropriate for the
x86 architecture, we must use 8 hex digits. Thus, the final answer is 0x000002DA
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