What will be the content of memory location 250 in decimal after exclusion of the following program (here instructions are separated by semicolons, and numbers are represented in decimal)?: LDI R3, 248; LD R1, (R3); DEC R1, R1 INC R3, R3; LD R2, (R3); ADD R2, R2, R1 ; INC R3, R3; ST (R3), R2. (do not write any blank space) Data address memory ... 248 25 249 31 250
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- Convert given code to LEGv8 code:int f, g, y //global 64-bit variablesint sum (int a, int b) { // at memory address X0+1000.return (a +b)} int main (void) // at memory address X0 + 800{f=2;g=3;y= sum (f, g);return y;}Convert this code, making valid assumptions about registers and register use. Notethat brackets and global variable declarations are not affecting the addresses of the instructionsin memory.Solve the 8085 Write a program to load twenty memory locations starting from 8005H, where each location's content should increases by 2 over the previous one, however, the first location should contain 04H, assuming that the programs start at memory location 9009H?Given the following information:Job List:Memory Block List:JobNumber MemoryRequested Memory Block Memory Block SizeJob A 57K Block 1 900KJob B 920K Block 2 910KJob C 50K Block 3 200KJob D 701K Block 4 300KUse the first-fit algorithm to indicate which memory blocks are allocated to each of the four arriving jobs.
- 2. 2) You are required to write an Assembly Language program segment to perform theoperation Ci = where Ai and Bi represents a set of 50 memory locations each storing avalue such that the A values are stored starting from memory location 100 while the B valuesare stored starting from memory location 200. The results are to be stored starting frommemory location 300.Let's say that p is a pointer to memory and the next four bytes in memory (in hex) beginning at p's address are: 89 AB CD EF. After the following code is run on a little-endian computer what are the values of x, y, and the four bytes in memory beginning at p's address. Answer for x and y in four digits of upper-case hexadecimal (eg, 0AC6). Answer for memory as eight digits of upper-case hexadecimal with a single space between each byte (eg, 89 AB CD EF). x: y: mem at p:Modify below program to include response timeProgram:ROUND ROBIN CPU SCHEDULING ALGORITHM: #include<stdio.h>#include<conio.h>using namespace std; int main(){int i,j,n,bu[10],wa[10],tat[10],t,ct[10],max; float awt=0,att=0,temp=0;printf("Enter the no of processes--"); scanf("%d",&n);for(i=0;i<n;i++){printf("\nEnter Burst Time for process %d--", i+1); scanf("%d",&bu[i]);ct[i]=bu[i];}printf("\nEnter the size of time slice--"); scanf("%d",&t);max=bu[0];for(i=1;i<n;i++)if(max<bu[i])max=bu[i];for(j=0;j<(max/t)+1;j++)for(i=0;i<n;i++)if(bu[i]!=0)if(bu[i]<=t){tat[i]=temp+bu[i];temp=temp+bu[i];bu[i]=0;}else{bu[i]=bu[i]-t;temp=temp+t;}for(i=0;i<n;i++){wa[i]=tat[i]-ct[i];att+=tat[i];awt+=wa[i];}printf("\nThe Average Turnaround time is--%f",att/n);printf("\nThe Average Waiting time is--%f ",awt/n);printf("\n\tPROCESS\t BURST TIME \t WAITING TIME\tTURNAROUND TIME\n");for(i=0;i<n;i++)printf("\t%d \t %d \t\t %d \t\t %d…
- Modify below program to include response time: Program: PRIORITY CPU SCHEDULING ALGORITHM:#include<stdio.h>main(){int p[20],bt[20],pri[20], wt[20],tat[20],i, k, n, temp;float wtavg, tatavg;clrscr();printf("Enter the number of processes --- ");scanf("%d",&n);for(i=0;i<n;i++){p[i] = i;printf("Enter the Burst Time & Priority of Process %d --- ",i);scanf("%d %d",&bt[i], &pri[i]);}for(i=0;i<n;i++)for(k=i+1;k<n;k++)if(pri[i] > pri[k]){temp=p[i];p[i]=p[k];p[k]=temp;temp=bt[i];bt[i]=bt[k];bt[k]=temp;temp=pri[i];pri[i]=pri[k];pri[k]=temp;}wtavg = wt[0] = 0;tatavg = tat[0] = bt[0];6for(i=1;i<n;i++){wt[i] = wt[i-1] + bt[i-1];tat[i] = tat[i-1] + bt[i];wtavg = wtavg + wt[i];tatavg = tatavg + tat[i];}printf("\nPROCESS\t\tPRIORITY\tBURST TIME\tWAITING TIME\tTURNAROUND TIME");for(i=0;i<n;i++)printf("\n%d \t\t %d \t\t %d \t\t %d \t\t %d ",p[i],pri[i],bt[i],wt[i],tat[i]);printf("\nAverage Waiting Time is --- %f",wtavg/n);printf("\nAverage Turnaround Time is ---…(Practice) a. Using Figure 2.14 and assuming the variable name rate is assigned to the byte at memory address 159, determine the addresses corresponding to each variable declared in the following statements. Also, fill in the correct number of bytes with the initialization data included in the declaration statements. (Use letters for the characters, not the computer codes that would actually be stored.) floatrate; charch1=M,ch2=E,ch3=L,ch4=T; doubletaxes; intnum,count=0; b. Repeat Exercise 9a, but substitute the actual byte patterns that a computer using the ASCII code would use to store characters in the variables ch1, ch2, ch3, and ch4. (Hint: Use Appendix B.)(Practice) Although the total number of bytes varies from computer to computer, memory sizes of millions and billions of bytes are common. In computer language, the letter M representsthe number 1,048,576, which is 2 raised to the 20th power, and G represents 1,073,741,824, which is 2 raised to the 30th power. Therefore, a memory size of 4 MB is really 4 times 1,048,576 (4,194,304 bytes), and a memory size of 2 GB is really 2 times 1,073,741,824 (2,147,483,648 bytes). Using this information, calculate the actual number of bytes in the following: a. A memory containing 512 MB b. A memory consisting of 512 MB words, where each word consists of 2 bytes c. A memory consisting of 512 MB words, where each word consists of 4 bytes d. A thumb drive that specifies 2 GB e. A disk that specifies 4 GB f. A disk that specifies 8 GB
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