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3 years ago

What is indirect program memory addressing? Explain the working of the following instructions?

Computers and Technology

1 answer:

Elanso [62]3 years ago

4 0

Indirect main memory addressing is described as the method in which the variable's address is stored in a mind register and also a command is utilized to point towards the memory registers which hold the register.
Instruction is used to direct towards a register that displays the results of the variables.
The register, in turn, refers to that variable's address in simplistic words. As just a result, it is indeed considered a passive program, memory address.

The instructions operate as follows:

JMP AX :

The JMP instruction is being used to execute an unconditional jump.
AX is indeed the label's name.
JMP AX codes are being used to transfer control of the flow program to the AX label.

JMP LIST:

The JMP command can be used to execute an unconditional jump, JMP LIST[DX] is the label's name.
The JMP LIST[DX] program is being used to transfer control of its flow programs to the specified section.
It is the segment to which the flow control is transmitted.

JMP NEAR PTR[DI+3]:

Unconditional jumps could be done with the JMP command as you'll see, the label is JMP NEAR PTR[DI+3].
It's being used to transmit flow control to a particular section, throughout this case [DI+3].
The close keyword indicates that perhaps the code segment would be in the line of code being nearby.

Learn more:

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3 years ago

Oxnard Casualty wants to ensure that their e-mail server has 99.98 percent reliability. They will use several independent server

sattari [20]

Answer:

The smallest number of servers required is 3 servers

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Explanation:

Given

$Reliability = 99.98\%$

$Individual Servers = 95\%$

Required

Minimum number of servers needed

Let p represent the probability that a server is reliable and the probability that it wont be reliable be represented with q

Such that

$p = 95\%\\$

It should be noted that probabilities always add up to 1;

So,

$p + q = 1$ '

Subtract p from both sides

$p - p + q = 1 - p$

$q = 1 - p$

Substitute $p = 95\%\\$

$q = 1 - 95\%$

Convert % to fraction

$q = 1 - \frac{95}{100}$

Convert fraction to decimal

$q = 1 - 0.95$

$q = 0.05$

------------------------------------------------------------------------------------------------

<em>To get an expression for one server</em>

The probabilities of 1 servers having 99.98% reliability is as follows;

$p = 99.98\%$

Recall that probabilities always add up to 1;

So,

$p + q = 1$

Subtract q from both sides

$p + q - q = 1 - q$

$p = 1 - q$

So,

$p = 1 - q = 99.98\%$

$1 - q = 99.98\%$

Let the number of servers be represented with n

The above expression becomes

$1 - q^n = 99.98\%$

Convert percent to fraction

$1 - q^n = \frac{9998}{10000}$

Convert fraction to decimal

$1 - q^n = 0.9998$

Add $q^n$ to both sides

$1 - q^n + q^n= 0.9998 + q^n$

$1 = 0.9998 + q^n$

Subtract 0.9998 from both sides

$1 - 0.9998 = 0.9998 - 0.9998 + q^n$

$1 - 0.9998 = q^n$

$0.0002 = q^n$

Recall that $q = 0.05$

So, the expression becomes

$0.0002 = 0.05^n$

Take Log of both sides

$Log(0.0002) = Log(0.05^n)$

From laws of logarithm $Loga^b = bLoga$

So,

$Log(0.0002) = Log(0.05^n)$ becomes

$Log(0.0002) = nLog(0.05)$

Divide both sides by $Log0.05$

$\frac{Log(0.0002)}{Log(0.05)} = \frac{nLog(0.05)}{Log(0.05)}$

$\frac{Log(0.0002)}{Log(0.05)} = n$

$n = \frac{Log(0.0002)}{Log(0.05)}$

$n = \frac{-3.69897000434}{-1.30102999566}$

$n = 2.84310893421$

$n = 3 (Approximated)$

<em>Hence, the smallest number of servers required is 3 servers</em>

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3 years ago