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- Consider the following segment table:Segment Base Length Base Limit0 1200 5001 2200 10242 6400 2563 8000 700What are the physical addresses for the following logical addresses?a. 2, 250b. 3, 690c. 0, 515Consider the following segment table:Segment Base Length0 219 6001 2300 142 90 1003 1327 5804 1952 96What are the physical addresses for the following logical addresses? segment # limita. 0 430b. 1 10c. 2 500d. 3, 400e. 4 112Convert the physical addresses to logical using the combined paging segmentation scheme Logical Address (Seg#, Page#, offset) 256 Physical Address 544 850 ? 2 672 1588 736 1630 ? 500 ? 1248 600 ? 1504 4 1632 1792 Hint: phy, add.= base_address + (page# x page_size) + offset **You must also explain how you computed your answer. Shows details of solution. 3.
- The memory location at address of 0X003FB01 contains 1-byte memory variable J (0010_0001), and the memory location at the address of 0X003FB02 contains 1-byte memory variable K (0001 0010), see figure below. There is a 2-byte variable M which hold binary information M (1110 0101 0000 1i11). What is the address in hexadecimal format for 2-byte memory variable M, following little Endian computer? 7 Address in Data in Hex. Format Hex. Format 0X003FBF04 1110 0101 M OX003FBF03 0000 1111 0X003FBF02 0001 0010 0X003FBF01 0010 0001 J Its address in hexadecimal is 0X003FBF02. а. Its address in hexadecimal is 0×003FBF03. O b. Its address in hexadecimal is 0X003FBF04. Its address in hexadecimal is 0×003FBF01. d.adres veri 01h 5x9 02h 5x8 8715683 b190100564 7156 08h 6715988 5x2 bis010054 71 09h 5x1 Write the asm code that will create the address and contents given in the table. (It is mandatory to use loop and indirect addressing.) b10100564 159 190100564-7I5A system uses pure segmentation (no paging). A segment table (see below) contains the following segment sizes and starting physical addresses. Segment Table Address Size 5 2048 2000 6 1024 400 2048 6000 1024 8000 11 Translate each of the following logical addresses (LA) into physical addresses (PA) or enter "illegal" if the LA is not within the segment. (1) LA = (5, 2000) (2) LA = (6, 2000) (3) LA = (7, 1000) (4) LA = (11, 500) Put your answer in the following blanks, one per blank, from left to right, case insensitive.
- Consider the following table that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes (like LC-3): ADDRESS OxFFFF OXOCOE OXOCOD Ox0C0C OXOCOB OXOCOA 0x0C09 0x0000 CONTENTS 1111 1111 1111 1111 1111 1110 1101 1100 0001 1011 1100 0101 0110 0101 1000 0111 1100 0000 0100 0000 0011 0001 0101 0010 0000 1100 0000 1101 0000 0000 0000 0000 The table above shows the addresses in hex (base 16) and the contents at the corresponding address in binary (base 2). A.) What are the contents in hex of the memory location at following address in binary: 0000 1100 0000 1110? (Enter hex like the following example: Ox2A3F)Question 2 Given an array containing 286 telephone area codes assigned to the United States of America as shown below, and with the aid of illustration, show how the binary search can be used to locate the following area codes: 1. 740 2. 364 3. 564 201 202 203 205 206 207 208 209 210 212 213 214 215 216 217 218 219 224 225 228 229 231 11 3. 6 10 12 13 14 15 16 17 18 19 20 21 1 2. 4 6. 234 239 240 248 251 252 253 254 256 260 262 267 269 270 276 281 283 301 302|303 304 305 22 23 24 25 26 27 28 30 31 33 34 35 36 37 38 39 40 41 42 43 307 308 309 310 312 313 314 315 316 317 318 319 320 321 323 325 330 331 334 336 337 339 44 47 52 54 55 56 57 58 59 60 61 62 63 64 65 45 46 48 49 50 51 53 347 351 352 360 361 364 385 386 401 402 404 405 406 407 408 409 410 412 413414 415 416 S7 71 72 73 74 76 78 79 80 81 82 83 $4 $5 $6 66 67 68 69 70 417 419 423 424 425 430 432 434 435 440 443 445 469 470 475 478 479 480 484 501 502 503 99 96 97 98 100 101 102 103 104 105 106 107 108 109 88 89 90 91 92 93 94 95…Part A For each byte sequence listed, determine the Y86 instruction sequence it encodes. If there is some invalid byte in the sequence, show the instruction sequence up to that point and indicate where the invalid value occurs. For each sequence, the starting address, then a colon, and then the byte sequence are shown. 0x100: 30f3fcfffff40630008000000000000 0x100: 30f3fcfffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rrmovq %rsi,0×80A(%rcx) O0x115: 00 halt Ox100: 30f3fcffffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox114: 00 halt 0x100: 30f3fcfffffffff rrmovq $-8,%rbx Ox109: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) 0x200: a06f800c020000000000000030f30a00000000000000 0x113: 00 halt 0x100: 30f3fcffffffffff irmovq $-4,%rbx 0x200: a06f pushq %rsi 0x10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox202: 800c02000000000000 call proc Ox116: 00 halt 0x20b: 00 halt 0x20c: proc: Submit Request Answer 0x20c: 30f30a00000000000000 | irmovq $10,%rbx…
- Variable x has 4-byte representation 0x01234567 Address given by &x is 0x100 If the machine uses Big endian byte ordering which of the following table presents variable in machine memory Select one: a. 0x100 : 100x001: 320x102: 540x103: 76 b. 0x100 : 670x001: 450x102: 230x103: 01 c. 0x100 : 760x001: 540x102: 320x103: 10 d. 0x100 : 010x001: 230x102: 450x103: 67Variable x has 4-byte representation 0x01234567 Address given by &x is 0x100 If the machine uses little endian byte ordering which of the following table presents variable in machine memory Select one: O a. Ox100 : 10 Ox001: 32 Ox102: 54 Ox103: 76 O b. Ox100: 67 Ox001: 45 Ox102: 23 Ox103: 01 O c. Ox100:01 Ox001: 23 0x102: 45 0x103: 67 O d. 0x100:76 Ox001: 54 Ox102: 32 0x103: 10Do you know how many distinct domains you can represent using a 32-bit representation? The number of computers that may be part of a domain is currently unspecified.