Day 9

Question 1

• Recall the process of multiplication.*
• Note that with signed integers, we use sign extension and the last number becomes its 2’s complement and than being added to the accumulator*

Answer 1

Question 2

Recall signed multiplication.

Compute it yourself

Answer 2

Question 3

Compute -5/8 * 6/8

Answer 3

Question 4

• Recall the method for integer division*
• Now slove 45/6*

Answer 4

Question 5

Answer 5

Question 6

Describe the exponent of 32 bit machine

Answer 6

8 bits are reserved for the exponent.

23 bits are reserved for the mantisa.

1 bit is reserved for sign.

because we’re using 2’s complement we have a range of -127<=x<=128.

We add 127 to x to have positive numbers.

Zero is taken by x = -127, and mantisa all zeros.

x = 128 implies infinity, so..

-126<=x<=127 -> (+127) -> 1<=x<=254

That is, exponent can shift from -126 to 127.

If the number is normalized, it starts with 1, thus the smallest mantisa is 1.0, and largest is 2-2^-23.

If the number is not normalized, it starts with 0, thus the smallest mantisa is 2^-23 and largest is 1-2^-23

Question 7

• With addition, you want the two numbers to have the same exponent - 1/2*
• With substraction, you want the two number to have the same exponent -1/2*
• 1) the greater one of the two*
• 2) the smaller one of the two*

Answer 7

1

2

Question 8

Explain associative cache

Answer 8

Address: 13|3 =>BlockNumber|LineInBlock

Size of block = size of line in cache.

Tag memory = an array in the cache in which block numbers are saved

Valid bits = 1 if block is present in the cache.

Time of search O(n), where n is the number of blocks in the cache.

Question 9

Write about cache:

• Average read time
• Average write time
• Average memory-access time

Answer 9

Question 10

• With miss-hit there are two approches.*
• Name them and explain.*

Answer 10

Question 11

• With miss-hit there are two write approaches*
• Name them and explain*

Answer 11

Question 12

• With hit there are two write approaches*
• Name them and explain*

Answer 12

Question 13

What are the advantages of mapping system for both I/O and memory?

Answer 13

• Data transfer to/from CPU is the same for both I/O and memory.
• Number of board connections is smaller

Question 14

• What is the size of the I/O space?
• How a device is identified?
• How many device can be connected?
• How can we know whether address refers to I/O or Memory?
• Where are the control bits are stored?

Answer 14

• 2^10 addresses
• by two addresses(registers)
  
  • data register
  • status register
• 20 MSB’s in the address tell us if it belongs to memory or I/O
• Control bits going from CPU to I/O are stored in the command register
• Control bits going from I/O to CPU are stored in the status register.