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katovenus [111]
3 years ago
5

why is wrong timing, poor technology, bad implementation and politics in business affected the OSI model as a standard

Computers and Technology
1 answer:
salantis [7]3 years ago
5 0

Answer:

In the academic environment, TCP / IP was much preferred because the OSI model was thought to be inferior to the TCP/IP model

Explanation:

The following show how business affected the OSI model as a standard:

WRONG TIMING

When OSI was made, TCP / IP was already in use. Also nobody wanted to be the first to start using OSI.

POOR TECHNOLOGY

Both the presentation and session layer that was dedicated in OSI was very little.

The technology used was very difficult to understand.

BAD IMPLEMENTATION

The initial implementation of OSI model was very slow and the OSI layer 7 model had bad quality.

POLITICS

In the academic environment, TCP / IP was much preferred because the OSI model was thought to be inferior to the TCP/IP model. They where rumors that the OSI was for the European and US government.

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Using programming knowledge, it is possible to use programming logic, we can differentiate and find the terms

<h3>Writing the code and understanding the programming logic:</h3>

<em>.data</em>

<em>sourceword:</em>

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<em>.word 0xCF</em>

<em>operand:</em>

<em>.long 0xAA</em>

<em>.text</em>

<em>.globl _start</em>

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<em>movw sourceword, %ax # ax=0x00AB ( move the word value of sourceword =0x00AB to ax register.)</em>

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<em>xorw %cx, %cx # cx=0000 ( cx=cx XOR cx, that is clear cx register.)</em>

<em />

<em>andw wordmask, %ax # ax=0x008B ( ax=ax AND wordmask= 0x00AB AND 0x00CF= 0x008B)</em>

<em>orw wordmask, %bx # bx=0x00EF (bx=bx OR wordmask= 0x00AB OR 0x00CF= 0x00EF.)</em>

<em>xorw wordmask, %cx # cx=0x00CF (cx=cx XOR wordmask= 0x0000 XOR 0x00CF=0x00CF.)</em>

<em />

<em>movl operand, %eax # eax=0x000000AA. ( copy the 32-bit value of operand=0x000000AA to eax.)</em>

<em>movl %eax, %ebx # ebx=0x000000AA.( copy the value of eax to ebx.)</em>

<em>movl %eax, %ecx # ecx=0x000000AA. ( copy the value of eax to ecx.)</em>

<em />

<em>shll $3,%eax # eax=0x00000550. ( logical shift left of eax=0x000000AA by 3 bits.)</em>

<em />

<em># 0x000000AA=00000000000000000000000010101010</em>

<em />

<em># =>00000000000000000000010101010000=0x00000550.</em>

<em />

<em>rorl $4,%ebx # ebx=0xA000000A. (Rotate right ebx=0x000000AA by 4 bits.)</em>

<em />

<em># 0x000000AA=00000000000000000000000010101010</em>

<em />

<em>3 =>10100000000000000000000000001010= 0xA000000A.</em>

<em />

<em>sarl %ecx # here the count value to shift the bits not mentioned.</em>

<em />

<em># It is the arithmetic shift right instruction. the shifted left bits filled with MSB bit.</em>

See more about logic program at brainly.com/question/14970713

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