uc3m_cs

Materials used for Computer Structure

Exercise 1

(statement) Convert the following 16-bit binary number to hexadecimal: 1101001011101011

(solution) 1101001011101010 -> 1101 0010 1110 1011 -> 0xD2EB

Exercise 2

(statement) Convert the following hexadecimal number to binary: F73AB592

(solution) F73AB592 -> 1111 0111 0011 1010 1011 0101 1001 0010

Exercise 3

(statement)

Consider a hypothetical computer with a word width of 24 bits with 55 registers that addresses the memory by bytes.
Answer the following questions:
  1. How many bits are used for the memory addresses?
  2. What is the size of the registers?
  3. How many bits are stored in each memory location?
  4. How many memory locations can be addressed? Express the result in KiB.
  5. How many bits are needed to identify the registers?

(solution)

Answers:
  1. 24 bits are used (the word-width of the computer if no other particular information is given).
  2. 24 bits (the word-width if no other particular information is given).
  3. In each memory position is stored a byte because the memory is addressed by bytes.
  4. You can address 2^24 memory locations. In each memory position is stored a byte.
    Then the memory size is: 2^24 bytes => 2^24 / 2^10 = 2^14 = 16*1024 KiB.
  5. As there are 55 registers, it is necessary to have round_ex(ln2(55)) = 6 bits.
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