Speed of memory versus speed of CPU, Computer Engineering

Assignment Help:

In the past there was a large gap between speed of a memory andprocessor. So a subroutine execution for an instruction for illustration floating point addition may have to follow a lengthy instruction sequence. Question is; if we make it a machine instruction then just one instruction fetch will be needed and rest will be done with control unit sequence. So a 'higher level' instruction can be added to machines in an effort to improve performance.

Though this supposition is not very valid in present era where Main memory is supported with Cache technology. Cache memories have decreased the difference between memory and CPU speed and so an instruction execution by a subroutine step may not be that difficult.

Let's explain it with help of an illustration:

Suppose floating point operation ADD A, B needs the subsequent steps (presuming the machine doesn't have floating point registers) and registers being used for exponent are E1, E2, and EO (output); for mantissa M1, M2 and MO (output):

  • Load exponent of A in E1
  • Load mantissa of A in M1
  • Load exponent of B in E2
  • Load mantissa of B in M2
  • Compare E1 and E2

-  If E1 = E2 then MO ← M1 + M2 and EO ← E1

Normalise MO and adjust EO

  • Result will be contained in MO, E1

Else if E1< E2 then find the difference = E2 - E1

  • Shift Right M1, by difference
  • MO ← M1 + M2 and EO ← E2
  • Normalise MO and adjust EO
  • Result is contained in MO, EO

 Else E2 < E1, if so find the difference = E1 - E2

  • Shift Right M2 by difference above
  • MO ← M1 + M2 and EO ← E1
  • Normalise MO and adjust E1 into EO
  • Result is contained in MO, EO

 Store the above results in A

 Checks overflow underflow if any.

If all above steps are coded as one machine instruction then this simple instruction will need many instruction execution cycles. If this instruction is made as part of machine instruction set as: ADDF A,B (Add floating point numbers A and B and store results in A) then it would just be a single machine instruction. All above steps needed will then be coded with help of micro-operations in form of Control Unit Micro-Program. Soonly one instruction cycle (though a long one) may be required. This cycle will need only one instruction fetch. While in the program memory instructions will be fetched.

Though faster cache memory for data and Instruction stored in registers can create an almost similar instruction execution environment. Pipelining can further increase such speed. So creating an instruction as above may not result in faster execution.


Related Discussions:- Speed of memory versus speed of CPU

What do you mean by prototype, Q. What do you mean by Prototype? A prot...

Q. What do you mean by Prototype? A prototyping approach lay emphasis on the construction model of a system. Designing and building a scaled-down though functional version of a

Operating system.., what is network operating system design issues

what is network operating system design issues

Determine the nand gate, Find out the two inputs when the NAND gate output ...

Find out the two inputs when the NAND gate output will be low. Ans. The output of NAND gate will be low if the two inputs are 11. The Truth Table of NAND gate is shown

Define http, Http Hyper Text Transfer Protocol: The WWW protocol that...

Http Hyper Text Transfer Protocol: The WWW protocol that performs the request and regain functions of a server. Generally seen as the first part of a website address. It is t

Operation of micro controller, Consider the hardware design as shown. Withi...

Consider the hardware design as shown. Within the target system the EPROM would contain the hex data as shown below   Address  Assembly code   8000             86   8001

Determine frame time and propagation time in a lan, Maximum channel utiliza...

Maximum channel utilization in a LAN is defined by frame time (t f ) and propagation time (t p ). It is defined by (A) t p /t f (B) t f /t p  (C) 1 + (t f /t p )

Data validation and data transfer, Data Validation condition: The following...

Data Validation condition: The following condition stated below must be met for a data to be valid as is shown in the figure 3a below. 1. The data on the SDA line must remain stabl

C program, find area uder the curve y=f(x) between x=a and x=b Posted Date...

find area uder the curve y=f(x) between x=a and x=b Posted Date: 9/1/2012 1:50:09 AM

What is base register addressing, Q. What is Base Register Addressing ? ...

Q. What is Base Register Addressing ? An addressing technique in which content of an instruction specifies base register is added to address field or displacement field of the

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd