Ask question

Ask question

All categories

2 years ago

15

This exercise shows why each pivot (in eli1nination by pivoting) must be in a different row. (a) In Example 7, make the third pi

vot on entry (3, 3) instead of on entry (3, 2). Can you still read off the solution

Computers and Technology

1 answer:

Darya [45]2 years ago

8 0

Answer:gghggggggggggggg

Explanation:

You might be interested in

Information is stored on a compact disk in a coded pattern of tiny pits arranged in a track that spirals outward toward the rim

Maksim231197 [3]

Answer:

(i) The rotation speed must stay the same.

(ii) The rotation speed must increase.

(iii) The rotation speed must decrease.

Explanation:

According to Equation

10.10, the angular speed must therefore vary as the laser–lens system moves

radially along the disc. In a typical CD player, the constant speed of the surface at

the point of the laser–lens system is 1.3 m/s.

5 0

2 years ago

________ is the process of converting a poorly-structured table into two or more well-structured tables. optimization normalizat

bulgar [2K]

Answer:

Normalization

Explanation:

Converting a poorly-structured table and optimizing its database structure in order to reduce redundancy in relations can be referred to as database normalization. If the data is not redundant means that data inconsistencies and errors like deletion and insertion will get reduced or eliminated significantly.

4 0

2 years ago

What are the differences between a policy, a standard, and a practice? What are the three types of security policies? Where woul

hjlf

Answer:

The difference between a policy, a standard and a practice is as follow:

Policy: It can be defined as the written instructions that describe proper behavior.
Standard: It can be defined as the detailed statement of what must be done to comply with policy.
Practice: It can be defined as the examples of actions that would comply with policy.

The three types of security policies are:

Enterprise Information Sec. Policy (EISP) : High level policy that sets the strategic direction, scope, and tone for the organization's security efforts. Use: It is used to support the mission, vision and direction of the organization and sets the strategic direction, scope and tone for all security efforts

Issue Specific Sec. Policy (ISSP) : An organizational policy that provides detailed, targeted guidance to instruct all members of the organization in the use of a resource, such as one of its processes or technologies. Use: It is used to support routine operations and instructs employees on the proper use of these technologies and processes

System Specific Sec. Policy (SysSP): Organizational policies that often function as standards or procedures to be used wen configuring or maintaining systems. SysSPs can be separated into two general groups-managerial guidance and technical specifications- but may be written as a single unified document. Use: It is used as a standard when configuring or maintaining systems.
ISSP policy would be needed to guide the use of the web, email and use of personal use of office equipment.

3 0

2 years ago

Write pseudocode for a program which will ask a user to enter two numbers a and b. It then asks what operation is to be performe

sleet_krkn [62]

get int input for a

get int input for b

get string input for operator

if a is not int or b is not int throw exception and print error

if operator is not * / // or % throw exception and print error

if operator is * do multiplication of a and b and make answer c

else if operator is / do division of a and b and make answer c

else if operator is // do floor division of a and b and make answer c

else if operator is % do floor modulo of a and b and make answer c

print c

5 0

2 years ago

A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, an

yulyashka [42]

The question is incomplete. It can be found in search engines. However, kindly find the complete question below:

Question

Cites as a pitfall the utilization of a subset of the performance equation as a performance metric. To illustrate this, consider the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requires the execution of 1.0E9 instructions. 1. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2. 2. Another fallacy is to consider that the processor executing the largest number of instructions will need a larger CPU time. Considering that processor P1 is executing a sequence of 1.0E9 instructions and that the CPI of processors P1 and P2 do not change, determine the number of instructions that P2 can execute in the same time that P1 needs to execute 1.0E9 instructions. 3. A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, and consider that the processor with the largest MIPS has the largest performance. Check if this is true for P1 and P2. 4. Another common performance figure is MFLOPS (millions of floating-point operations per second), defined as MFLOPS = No. FP operations / (execution time x 1E6) but this figure has the same problems as MIPS. Assume that 40% of the instructions executed on both P1 and P2 are floating-point instructions. Find the MFLOPS figures for the programs.

Answer:

(1) We will use the formula:

CPU time = number of instructions x CPI / Clock rate

So, using the 1 Ghz = 10⁹ Hz, we get that

CPU time₁ = 5 x 10⁹ x 0.9 / 4 Gh

= 4.5 x 10⁹ / 4 x 10⁹Hz = 1.125 s

and,

CPU time₂ = 1 x 10⁹ x 0.75 / 3 Ghz

= 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

So, P2 is actually a lot faster than P1 since CPU₂ is less than CPU₁

(2)

Find the CPU time of P1 using (*)

CPU time₁ = 10⁹ x 0.9 / 4 Ghz

= 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

So, we need to find the number of instructions₂ such that CPU time₂ = 0.225 s. This means that using (*) along with clock rate₂ = 3 Ghz and CPI₂ = 0.75

Therefore, numbers of instruction₂ x 0.75 / 3 Ghz = 0.225 s

Hence, numbers of instructions₂ = 0.225 x 3 x 10⁹ / 0.75 = 9 x 10⁸

So, P1 can process more instructions than P2 in the same period of time.

(3)

We recall that:

MIPS = Clock rate / CPI X 10⁶

So, MIPS₁ = 4GHZ / 0.9 X 10⁶ = 4 X 10⁹HZ / 0.9 X 10⁶ = 4444

MIPS₂ = 3GHZ / 0.75 X 10⁶ = 3 x 10⁹ / 0.75 X 10⁶ = 4000

So, P1 has the bigger MIPS

(4)

We now recall that:

MFLOPS = FLOPS Instructions / time x 10⁶

= 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Therefore,

MFLOPS₁ = 1777.6

MFLOPS₂ = 1600

Again, P1 has the bigger MFLOPS

3 0

2 years ago