Encrypting your link and protect the link from viruses, malware, thief, etc! Made your link safe to visit.

Compute the physical address for the specified operand in each of the following instructions.The register contents and variable are as follows: (CS)=0A00H, (DS)=0B00H,

 

Question:

None Q: Compute the physical address for the specified operand in each of the following instructions.The register contents and variable are as follows: (CS)=0A00H, (DS)=0B00H, (SS)=0D00H,(SI)=OFF0H, (DI)=00BOH, (BP)=00EAH and (IP)=0000H, LIST=00F0H, AX=4020H, BX=2500H.1) Destination operand of the instructionMOV LIST [BP+DI] , AX2) Source operand of the instructionMOV CL, [BX+200H]3) Destination operand of the instructionMOV [DI+6400H] , DX4) Source operand of the instructionMOV AL, [BP+SI-400H]

Step 1

Given question has asked to compute the physical address for specific operand in given instruction
The register contents and variable are as follows:
(CS)=0A00H
(DS)=0B00H
(SS)=0D00H
(SI)=0FF0H
(DI)=00B0H
(BP)=00EAH
(IP)=0000H
LIST=00F0H
AX=4020H
BX=2500H

1) MOV LIST[BP+DI],AX

2) MOV CL,[BX+200]

3) MOV AL,[BP+SI-400]

Step 2

1) MOV LIST[BP+DI],AX

This is an example of Base indexed addressing mode-

Here, destination operand physical address is calculated as base register plus an index register.

Given instruction moves data from AX to address pointed by (BP+DI) in LIST segment.

(DI)=00B0H
(BP)=00EAH

LIST=00F0H

So, physical address of destination operand, PA= LIST*10H+BP+DI
                                                                                  =0F00H+00EAH+00B0H

                                                                                   =109A

 

Step 3

2. MOV CL,[BX+200]

This is an example of Base indexed relative addressing mode-

Here, source operand physical address is calculated as base register plus an index register.

Given instruction moves data from the address pointed by (BX+200) in Data segment(DS) to CL register.

BX=2500H

(DS)=0B00H

So, physical address of source operand, PA =DS*10H+BX+200

                                                                         =B000H+2500H+200

                                                                          =B200H

 

 

 

Step 4

3.MOV[DI+6400H],DX

This is an example of Base indexed addressing mode-

Here, destination operand physical address is calculated as displacement plus an index register.

Given instruction moves data from DX to address pointed by (DI+6400H) in data segment segment.

(DI)=00B0H
(DS)=0B00H

So, physical address of destination operand, PA= DS*10H+DI+6400H

                                                                                 =0B00H*10+00B0H+6400H

                                                                                 =B000H+00B0H+6400H

                                                                                 =114B0

 

 

 

 

Step 5

4.MOV AL,[BP+SI-400]

Here, source operand physical address is calculated as base register plus an index register and subtracting displacement.

Given instruction moves data from the address pointed by (BP+SI-400) in Data segment(DS) to AL register.

(BP)=00EAH

(SI)=0FF0H

(DS)=0B00H

So, physical address of source operand, PA =DS*10H+BP+SI-400

                                                                           =B000H+00EAH+0FF0H-400

                                                                           =BCDAH

ST

Search This Blog

Labels

Report Abuse

QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

  Question: Show transcribed image text Answer:
  Introduction To find incremental revenue and cost, we subtract Current revenue and cost from Projected revenue and cost. Answer Current Situation Projected Sales and Profit Incremental Revenues and Costs Total Revenue Php1,500,000 Php1,800,000 + Php3000,000 Variable Cost 575,000 425,000 - Php 332,000 Direct Fixed Costs 625,000 700,000 + Php 75,000 Indirect Fixed Costs 100,000 100,000 No change Profit Php 200,000 Php 575,000 + Php 375,000 Profit increases by Php 375,000, thus a new camera and coffee maker must be purchased.

Contributors