Question

Consider the organization of address bus and data bus in the following two ways • The...

Consider the organization of address bus and data bus in the following two ways

• The address bus operates in parallel with the data bus

• The address bus is multiplexed with the data bus

i) Compare and contrast the two modes of operations. You should briefly explain their advantages and disadvantages.

ii) With the aid of a diagram, explain what is burst mode and how burst mode can improve the efficiency of address buses.

b) A machine is running with a clock frequency of 2MHz.

i) An Assembly program contains 100 instructions. 30% of the instructions require 4 clock cycles, 50% instructions require 5 clock cycles and the remaining instructions require 2 clock cycles for execution. What is the total time required to execute the program?

ii) A system has a level 1 cache and a level 2 cache. The hit rate of the level 1 cache is 60% and the hit rate of the level 2 cache is 50%. An access to level 1 cache requires 1 cycle, an access to level 2 cache requires 4 cycles, and an access to main memory requires 10 cycles. What is the average access time?

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Answer #1

Hello,
          Please find the answer to part b attached as under. Please give a thumbs up rating if you find the answer useful! Have a rocking day ahead!

b

i. 30 instructions require 4 clock cycles = 120 clock cycles. 50 instructions take 5 clock cycles = 250 clock cycles. 20 instructions require 2 clock cycles = 40 clock cycles. Total number of clock cycles = 120 + 250 + 40 = 410 clock cycle. One clock cycle takes 1/2M seconds = 0.5 microseconds. Thus, total time for execution = 0.5 micro * 410 = 0.205 milliseconds.

ii.

Average access time = [H1*T1]+[(1-H1)*H2*T2]+[(1-H1)(1-H2)*Hm*Tm]

where

H1 = Hit rate of level 1 cache = 0.6
T1 = Access time for level 1 cache = 1 cycle
H2 = Hit rate of level 2 cache = 0.5
T2 = Access time for level 2 cache = 4 cycles
Hm = Hit rate of Main Memory = 1
Tm = Access time for Main Memory = 10 cycles

Thus, average access time = [0.6]+[0.8]+[2] = 3.4 cycles = 3.4 * 0.5 microseconds = 1.7 microseconds

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