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Explain what might happen if two stations are accidentally assigned the same hardware address?

Firdavs [7]

Answer: If two different station are addressed with the same hardware address then there are chances of occurrence of the failure in the network at irregular intervals.The failure or error will occur because of the both the devices are seen as one by the network due to same address.

An intelligent network system id used ,it can identify the error can help in the prevention of the failure.Other option for configuring the situation is assigning the MAC(media access control)address to devices which are unique in nature thus, no device can have same address.

5 0

2 years ago

A _______ bulb contains a high-pressure gas. Oils from the hands can affect the expansion of the glass, which can shorten the li

shusha [124]

Answer:

I'd go with the 2nd one

If it heats up more & more, then it's lifespan will be shortened more & more. The more it heats up, the less durability it has.

6 0

2 years ago

11.19 LAB: Max magnitude Write a function max_magnitude() with two integer input parameters that returns the largest magnitude v

AlekseyPX

Answer:

# the user is prompt to enter first value

user_val1 = int(input("Enter first value: "))

# the user is prompt to enter second value

user_val2 = int(input("Enter second value: "))

# max_magnitude function is defined

# it takes two parameters

def max_magnitude(user_val1, user_val2):

# if statement to compare the values

# it compare the absolute value of each

if(abs(user_val2) > abs(user_val1)):

return user_val2

elif (abs(user_val1) > abs(user_val2)):

return user_val1

# max_magnitude is called with the

# inputted value as parameter

print(max_magnitude(user_val1, user_val2))

Explanation:

The code is written in Python and well commented. A sample image of program output is attached.

3 0

2 years ago

Suppose a computer has 16-bit instructions. The instruction set consists of 32 different operations. All instructions have an op

Kazeer [188]

Answer:

2^7= 128

Explanation:

An instruction format characterizes the diverse part of a guidance. The fundamental segments of an instruction are opcode and operands. Here are the various terms identified with guidance design: Instruction set size tells the absolute number of guidelines characterized in the processor. Opcode size is the quantity of bits involved by the opcode which is determined by taking log of guidance set size. Operand size is the quantity of bits involved by the operand. Guidance size is determined as total of bits involved by opcode and operands.

6 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