Which would be the most efficient way to store files on your computer?

________ a database rearranges data and objects in a database to make its size smaller

hodyreva [135]

<span>An attribute in a relation of a database that serves as the primary key of another relation in the same database is called a

</span>

8 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

You are a network administrator for a large bagel manufacturer that has 32 bakeries located throughout the United States, United

Anni [7]

Answer:

See the components in explaination

Explanation:

In order to make it as IPv6, few key components should be supported, those components are given below:

The infrastructure must support the enhanced protocol StateLess Address Auto-Configuration (SLAAC).

Static addressing with DHCPv6, dynamic addressing with DHCPv6 and SLAAC are the methods used to configure the IPv6. The network administrator should able to understand and implement the IPv6 through the DHCPv6.

Other than the implementation, working of IPv4 and IPv6 are same. Therefore, the administrator need not to learn new information for its working.

As the IPv6 address length is 128-bit and purpose is for everything on line to have an IP address. It must allow the internet to expand faster devices to get internet access quickly.

The DHCPv6 is not supported by all windows. Therefore, network administrator should check the corresponding Operating system (OS) would support the DHCPv6 for IPv6.

The network administrator must have good knowledge and skills on the IPv6.

The above mentioned key components should be verified by the network administrator in order to support for IPv6 project with DHCPv6.

4 0

2 years ago

In step 4 of the CSMA/CA protocol, a station that successfully transmits a frame begins the CSMA/CA protocol for a second frame

DerKrebs [107]

Answer:

There could be a collision if a hidden node problem occurs.

Explanation:

CSMA/CA(carrier sense multiple access/ collision avoidance) is a multiple access method in wireless networking, that allows multiple node to transmit. Collision avoidance of this method is based on preventing signal loss or downtime as a result of collision of transmitting multi signals.

If a node at step 4(transmit frame) sends the first frame, the node still needs to send a RTS(request to send) and to receive a Clear to send (CTS) from the WAP, the is to mitigate the issue of hidden node problem as all frame are treated as unit, so other listening nodes, not detected would seek to connect and transmit as well.

5 0

2 years ago

Write a function addOddMinusEven that takes two integers indicating the starting point and the end point. Then calculate the sum

snow_lady [41]

Answer:g

public static int addOddMinusEven(int start, int end){

int odd =0;

int even = 0;

for(int i =start; i<end; i++){

if(i%2==0){

even = even+i;

}

else{

odd = odd+i;

}

return odd-even;

}

Explanation:

Using Java programming language:

The method addOddMinusEven() is created to accept two parameters of ints start and end
Using a for loop statement we iterate from start to end but not including end
Using a modulos operator we check for even and odds
The method then returns odd-even
See below a complete method with a call to the method addOddMinusEven()

public class num13 {

public static void main(String[] args) {

int start = 2;

int stop = 10;

System.out.println(addOddMinusEven(start,stop));

}

public static int addOddMinusEven(int start, int end){

int odd =0;

int even = 0;

for(int i =start; i<end; i++){

if(i%2==0){

even = even+i;

}

else{

odd = odd+i;

}

return odd-even;

}

8 0

2 years ago