Q2.For the SIC/XE dumpA=000000X=000005L=000000B=000005S=FFFFFFT=FFFFFFP=00000DCC=LT01A069100001 54FFFFFF FE434F50 3336303101во32FFFF4B 9F123456 000001A0 FFFFF1FF(а)What SIC integer is stored beginning at 01A6?(b) What are the 4 ASCII characters beginning at 01AB?What SIC/XE instruction is stored beginning at 01B3?(c)(d)What does the instruction at 01A0 do?

The solution is given below for the above given question:

For the SIC/XE dump A=000000 X=000005 L=000000 B=000005 S=FFFFFF T=FFFFFF P=00000D CC=LT 01A0 69100001 54FFFFFF FE434F50 33363031 01B0 32FFFF4B 9F123456 000001A0 FFFFF1FF What SIC integer is stored beginning at 01A6? (b) What are the 4 ASCII characters beginning at 01AB? What SIC/XE instruction is stored beginning at 01B3? (a) (c) (d) What does the instruction at 01A0 do?

For the SIC/XE dump A=000000 X=000005 L=000000 B=000005 S=FFFFFF T=FFFFFF P=00000D CC=LT 01A0 69100001 54FFFFFF FE434F50 33363031 01B0 32FFFF4B 9F123456 000001A0 FFFFF1FF What SIC integer is stored beginning at 01A6? (b) What are the 4 ASCII characters beginning at 01AB? What SIC/XE instruction is stored beginning at 01B3? (a) (c) (d) What does the instruction at 01A0 do?

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

See similar textbooks

Similar questions

Money Transfer Max. score: 100 SeaMoney has a feature to transfer money between users. Let's imagine a simple scenario. N usernames--consisting of lowercase Latin characters-each starts with a balance b. Amongst them, T transactions happened, where user ua transfers x amount of money to ub. If x is larger than the ua's balance when the transaction happens, the transaction is automatically rejected. Output the final balance of each user. Input The first line contains two numbers N and T (1 < N,T < 1000). Each of the next N following lines contain username u; (1< ]u;| < 10) and integer b; (1 < b; < 10°), denoting the balance of user u;. Following that (1
Answer Question 3
Assuming 4K clustering for a 500-byte file, what is the size (in bytes) of the File Slack? O 12 bytes O 1536 bytes O 3584 bytes O 4096 bytes
Load infile using one byte per channel and save it as a PNG, using outfile for the file name. Use one byte per channel. graphics.cpp 1 #define STB IMAGE IMPLEMENTATION 2 #define STB IMAGE WRITE IMPLEMENTATION #include "stb_image.h" #include "stb_image_write.h" 3 4 #include #include 7 using namespace std; 8 9. int main() 10 { string infile = "pete.png"; string outfile = "out.png"; 11 12 13 14 15 16 17 } CodeCheck Reset
Given the following data, write a MIPS program to sort the list in ascending order and output thetwo lists. Use the -1.0000 as the termination of the input data list.0.0238-0.01870.3451-0.0045-0.01110.0599-1.0000
flowdata: 01/04/2021 10:15:46 50.622101/04/2021 10:15:47 50.609801/04/2021 10:15:48 50.535301/04/2021 10:15:49 50.585601/04/2021 10:15:50 50.591301/04/2021 10:15:51 50.618101/04/2021 10:15:52 50.492701/04/2021 10:15:53 50.331401/04/2021 10:15:54 50.344201/04/2021 10:15:55 50.301601/04/2021 10:15:56 50.225301/04/2021 10:15:57 50.243101/04/2021 10:15:58 50.28501/04/2021 10:15:59 50.317901/04/2021 10:16:00 50.321801/04/2021 10:16:01 50.397201/04/2021 10:16:02 50.388301/04/2021 10:16:03 50.394201/04/2021 10:16:04 50.525401/04/2021 10:16:05 50.601301/04/2021 10:16:06 50.638901/04/2021 10:16:07 50.461101/04/2021 10:16:08 50.484201/04/2021 10:16:09 50.616801/04/2021 10:16:10 50.829801/04/2021 10:16:11 50.836201/04/2021 10:16:12 50.918501/04/2021 10:16:13 50.817401/04/2021 10:16:14 50.849801/04/2021 10:16:15 50.792801/04/2021 10:16:16 50.790301/04/2021 10:16:17 50.856301/04/2021 10:16:18 50.810601/04/2021 10:16:19 50.758101/04/2021 10:16:20 50.673501/04/2021 10:16:21 50.708401/04/2021 10:16:22…
2. Translate the lines from ORG 12 to ORG 64 in Table 7.2 to binary. Microoperation None AC - AC + DR AC+0 F3 Microoperation None AC + AC O DR AC - AC Symbel NOP ADD CLRAC INCAC DRTAC DRTAR PCTAR WRITE Condition T Symbal Always 1 DR(15) AC(15) AC0 000 001 F2 000 Microoperation None AC+ AC-DR Symbol NOP Symbol NOP XOR 000 001 SUB OR AND READ ACTDR INCDR PCTDR 001 010 010 AC- AC V DR AC- AC A DR DR + MAR DR + AC DR+ DR +1 DR(0-10) + PC 010 COM 011 100 101 110 AC + AC +1 AC + DR 011 AC + shl AC AC+ nhr AC PC + PC +1 PC + AR Reserved SHL 100 100 101 110 111 SHR AR + DR(O-10) AR + PC MJAR + DR 101 110 111 INCPC ARTPC 111 Comments Unconditional branch Indirect address bit (conditional branch) Sign bit of AC (conditional branch) Zero value in AC (conditional branch) CD 00 01 10 11 BR 00 Symbol Punction JMP CAR AD if condition T jump) CAR CAR+1 condition 0 CAR + AD, SBR- CAR +1 if condition1 (subroutine call) branch, save address CAR+ CAR+1 if condition 0 CAR + SBR (return from subroutine)…
Student information is stored in a data area, and each student entry is formatted as follows:The first nine bytes are the student ID in ASCII format.The following byte is the course score.The following phrase is the section identifier. There are over 300 such student records in memory, which have been loaded consecutively beginning at address $10000. To indicate the conclusion of the records, the last record loaded is a dummy record with a section identifier of $FFFF. Create an assembler subroutine GetMax that searches the complete list for the highest mark. If there is a tie, the student with the highest score on the list should be found. The beginning address of the subroutine should be returned.
Please Help ASAP!!!
What is the range of Unsigned quadword type?
me remaining: 00:09:34 Computer Science Create a MPI version of the program below that uses a striped decomposition. Each process is responsible for some number of columns of the square and also maintains two columns of ghost cells to mirror the columns on the neighboring processes. The ghost cells are needed in order to update the cells along the process boundary. At each time step, a ghost cell exchange takes place, then the update takes place. #include <stdlib.h> #include <assert.h> #include "anim.h" const int n = 200; // number of discrete points including endpoints int nstep = 200000; // number of time steps int wstep = 400; // time between writes to file const double m = 100.0; // initial temperature of rod interior const int h0 = n/2 - 2, h1 = n/2 + 2; // endpoints of heat source const double k = 0.2; // diffusivity constant char * filename = "diff2d.anim"; // name of file to create double ** u, ** u_new; // two copies of the temperature function ANIM_File…
Having the following rdd that contains:('37.1242586781', '-119.424203758')('37.7374074392', '-121.790061298')('45.0929780547', '-117.473248072')('34.9786739919', '-111.700983823')('35.8736386531', '-120.421476019')('34.0912863966', '-118.144076285')('35.6720438532', '-120.416438975')('36.284116521', '-119.030325609')('44.4092246446', '-122.891970809')('38.4770167448', '-122.296973442')('38.4296857989', '-121.418135072')('37.3822266494', '-122.016158235') Implement an iterative algorithm (k-means) in Spark to calculate k-means fora set of points that are in a file(rdd), a k-means algorithm in python. Do not use use K-means in MLib of Spark to solve the problem. Follow this pattern: 1. Choose k = 5 random points as starting centers2. Find all points closest to each center (use groupByKey or reduceByKey)3. Find the new center (mean) of each cluster4. If the centers changed by more than convergeDist (e.g.convergeDist = 0.1), iterate again starting from step 2;otherwise, terminate For…