A computer system with 32 bit word length is implementing paging with 14 bit logical addresses and 4KB pages. If physical memory size is 64K Bytes, mark all correct statements below: O In physical addresses , d is 12 bits long. O Physical memory has 16 frames. O Frame size is 4K words. O In logical addresses , p is 4 bits long. O In physical addresses , f is 6 bits long. O Maximum program memory size is 16 K words. O In logical addresses , d is 10 bits long. O Logical memory size is at most 32 pages per program.
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- Case Study Assume a computer system has a main memory of 256 Bytes. The following is a memory byte-access trace history of a program run on this system. For example, as it is shown, the program first accesses memory address 0000 0000, and then it accesses memory address 0000 0001 and so on. Note the memory addresses are represented in binary: 00000000, 00000001, 00000010, 00000011, 00001000, 00010000, 00010001, 00000100, 00000101, 00000110, 00000111, 00001001, 00001010, 00001011, 00001100, 00001000, 00001001, 00001010, 00011100, 00011101. Q3. Assume the system has a 16-Byte direct mapped unified L1 cache with a block size of 2 Bytes. The following table shows how the cache looks like after the first access to the memory is finished. Please show; in the provided table, how it looks like after the 20th access is finished. You could ignore the "Data" Column. (Add or remove Rows/Columns in the provided table for your answer, as you see fit). Cache contents after the 1st access: Cache Index…A computer system with 32 bit word length is implementing paging with 14 bit logical addresses and 4KB pages. If physical memory size is 64K Bytes, mark all correct statements below: O In logical addresses , p is 4 bits long. O In physical addresses , f is 6 bits long. O Logical memory size is at most 32 pages per program. O In logical addresses , d is 10 bits long. O Physical memory has 16 frames. O Frame size is 4K words. O In physical addresses , d is 12 bits long. O Maximum program memory size is 16 K words.Consider the string of data #. The ASCII value of data # is 23 H which has to be moved from source memory to destination memory. Calculate the 20 bit PA of source memory and destination memory. Assume CS=2000 H, DS=5000 H, SS=6000 H, ES=8000 H. BX=1234 H, BP=2434 H, SP=3500 H, SI=4000 H, DI=5276 H, IP=D1245 H.
- 4. The Single Instruction Computer (SIC) has only one instruction that can do all operations our RISC-V does (you did a homework problem). The instruction has the following format sbn a, b, c # Mem[a]=Mem[a]- Mem[b]; if (Mem[a]- Mem[b]<0) go to PC+c For example, here is the program to copy a number from location a to location b: sbn temp, temp, 1 sbn temp, a, 1 sbn b,b, 1 sbn b, temp 1 Start: Design a single cycle datapath and control for this instruction set architecture.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.Consider a program that uses absolute physical memory references meaning that each reference refers to a specific physical memory location. One part of such a program is below: ( $0x100), ( $0x104), %eax, movl %eax movl %ebx ($0x104) ($0x100) movl movl %ebx, This program works fine when it is loaded at address 0x0, but not when it is loaded at address 0x1000. Why not? Re-write the above code so the program works when it is loaded at memory address 0x1000.
- Suppose that DS=1000H, SS= 2000H, CS=3000H, ES=4000H, BP=FFH, BX=FFFFH and DI=5H.i. Which memory locations are addressed by: MOV DL, [BP]?ii. Which memory locations are addressed by: MOV EAX, [BX+DI]?HD Video C x 3ll al N X Ue G X ogle agi X Microsoft moodle.nct.edu.om/pluginfile.php/67858/mod_resource 100% 212 2. A singer voice is sampled for 4 second by an 8-bit ADC at a sampling frequency of 10 kHz. The samples are stored on a memory chip. What is the minimum memory size both in [bits and bytes] required to store all the samples? i. How long the signal must be sampled to store the samples in 100Mbytes CD- ROM iii. What should be the highest frequency component of the signal? iv.If sampling frequency is doubled then how long the signal must be sampled to store samples in 100Mbytes CD-ROM 14 MarksWrite a code in sim8085 for the following problem: The pressure of two boilers is monitored and controlled by a microcomputer works based on microprocessor programming. A set of 6 readings of first boiler, recorded by six pressure sensors, which are stored in the memory location starting from 2050H. A corresponding set of 6 reading from the second boiler is stored at the memory location starting from 2060H. Each reading from the first set is expected to be higher than the corresponding position in the second set of readings. Write an 8085 sequence to check whether the first set of reading is higher than the second one or not. If all the readings of first set is higher than the second set, store 00 in the ‘D’ register. If any one of the readings is lower than the corresponding reading of second set, stop the process and store FF in the register ‘D’. Data (H): First set: 78, 89, 6A, 80, 90, 85 Second Set:71, 78, 65, 89, 56, 75
- Code Tracing: What is the final value of c if d is equal to 9? mov word[c], 7 mov byte[four], 4 mov ax, word[d] div byte[four] cmp ah, 1 jne next mov ax, word[c] mul word[d] mov word[c], ax next: mov bx, word[d] mov word[c], bxConsider a simple paging system with the following parameters: Size of the program = 256 bytes Page size of 16 bytes 2-Byte addressable memory Main memory is divided into 32 frames Page table for the program is as follows: Page No. 0 1 2 3 4 5 6 7 8 9 10 11 a) 49 b) 22 12 13 14 15 Frame No. 14 22 10 17 12 9 31 Convert the following logical addresses into physical addresses using the mathematical procedure. You are expected to show each and every step of your solution:Assume a computer system has a main memory of 256 Bytes. The following is a memory byte-access trace history of a program run on this system. For example, as it is shown, the program first accesses memory address 0000 0000, and then it accesses memory address 0000 0001 and so on. Note the memory addresses are represented in binary: 00000000, 00000001, 00000010, 00000011, 00001000, 00010000, 00010001, 00000100, 00000101, 00000110, 00000111, 00001001, 00001010, 00001011, 00001100, 00001000, 00001001, 00001010, 00011100, 00011101.