Question

1. What is the largest power of two to which the following numbers are aligned?

a. 11010101000002

b. 0xA3318

c. 0x2F379C

d. 0x7043000

2.  How many address lines would be needed for each of the following amounts of memory space?

a. 819,200

b. 100 MB

3. How many 4K blocks of physical memory would be required after the program malloc’s 2 MB of virtual memory?

Answer 1

1. Alignment to the power of two is the largest power of 2 less than the given number

a. Assuming 11010101000002 is 1101010100000 to the base of 2

1101010100000 is aligned to 2¹².

b. 0xA3318

0xA3318 = (10100011001100011000)₂ is aligned to 2¹⁹.

c. 0x2F379C

0x2F379C = (1011110011011110011100)₂ is aligned to 2²¹.

d. 0x7043000

0x7043000 = (111000001000011000000000000)₂ is aligned to 2²⁶.

2. For 2ⁿ bits, we need n address lines

a. 819200 Assuming this is Bytes

819200 Bytes = 6553600 bits = 0x640000 bits < 2²³ and > 2²². So we needs 23 address lines

b. 100MB

100MB = 100 x 8 Mbits = 100 x 8 x 1024 kbits = 100 x 8 x 1024 x 1024 bits = 100 x 2²³ bits < 2³⁰ and > 2²⁹. So we need 30 address lines

3. If memory is aligned to 4KB blocks, 2MB memory would require (2x1024KB) / (4KB) = 512 blocks

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