Ex 1. Show how the following values would be stored by byte-addressable machines with 32-bit words, using little endian and then big endian format. Assume that each value starts at address 1016. Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations. a) O×456789A1 b) O×0000058A
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A: Answer : Given R0 = 1500, R1 = 4500 and R2=1000 (a) ADD (R0), R2 It means that M(R0) <-…
Q: Part(c) : Assume a hypothetical system with eight 32-bit words cache and small Main memory of 1 KB…
A: the solution of part c is given below :
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Q: Q1: Consider a machine with a byte addressable main memory of bytes and block size of 8 bytes.…
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Q: The hypothetical machine has two I/O instructions: 0011 = Load AC from I/O 0111 = Store AC to I/O In…
A:
Q: QUESTION ONE (1) 1. The hypothetical machine has two I/O instructions: 0011 = Load AC from I/O 0111…
A: 34 Opcode Address (a) Instruction format Magnitude (b) Integer format Program counter (PC) = Address…
Q: 1. The hypothetical machine of figure 3.4 also has two I/O instructions: 0011= Load AC fro I/O 0111=…
A: Given:
Q: Question Show how the following values would be stored bybyte-addressable machines with 32- bit…
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Q: Suppose a computer using fully associative cache has 224 bytes of byte-addressable main memory and a…
A: Please find the answer to the above question below:
Q: Suppose a computer using direct-mapped cache has 2² bytes_of_byte-addressable main memory and a…
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Q: 5. suppose a computer using fully associative cache has 224 bytes of byte-addressable main memory…
A: Given: 5. suppose a computer using fully associative cache has 224 bytes of byte-addressable main…
Q: 1. Show how the following values would be stored bybyte-addressable machines with 32- bit words,…
A: Answer 1: A: 56789ABC16 Address Big Endian Little Endian 301816 56 16…
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Q: . suppose a computer using fully associative cache has 224 bytes of byte-addressable main memory and…
A: Actually, cache is a fast access memory. Which located in between cpu and secondary memory.
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Q: 1. Windows The hypothetical machine of has two I/O instructions: 0011 Load AC from I/O 0011 Store AC…
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Q: Please do a,b,c,d, and e Consider a machine with a byte addressable main memory of Bytes and block…
A: Answer in step2
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- Solve the question in Assembly Language using emu8086. Note: PLEASE do not copy this answer from the ones that are already available on chegg because I already tried them but those are showing errors. So do it by yourself and give me a proper answer please. It is a humble request. TIA Question: Imagine a series starts from 1 with an interval of 5. What will be the summation of 1st 5 values of the series? [**must build the series and save it into an array]Show the content of the individual bytes allocated in memory in hexadecimal for the following declarations. Assuming that the address of I is 404000h, what are the addresses of J, K, and L? What is the total number of allocated bytes?The table below shows a segment of primary memory from a Von Neumann model computer Address Data 10101000 10001000 11001000 10011001 10100000 10101010 10110100 10111011 10001100 11001100 The program counter (PC) contains a value of 11001000. Find the value (in binary) that will be placed in MAR (memory address register)? MAR (in binray) %3D Find the value (in binary) that will be placed in MBR (memory buffer register)? MBR = (binray) %3D
- 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:Let's say that p is a pointer to memory and the next six bytes in memory (in hex) beginning at p's address are: aa bb cc dd ee ff. What value would be in x if the following code is run on a little- endian computer? uint16_t *q uint16_t x = (uint16_t *)p; q[0]; aa aabb bbaa aabbccdd ddccbbaa) Consider the following sequence of virtual memory references (in decimal) generatedby a single program in a pure paging system:100, 110, 1400, 1700, 703, 3090, 1850, 2405, 2460, 4304, 4580, 3640a. Derive the corresponding reference string of pages (i.e. the pages the virtual addressesare located on), assuming a page size of 1024 bytes. (Assume that page numberingstarts at 0)b. For the page sequence derived above, determine the number of page faults for each ofthe following page replacement strategies, assuming that two (2) page frames areavailable to the program.i. LRUii. FIFOiii. OPT (Optimal)
- Find the PA of the memory location and its contents after the execution of the following assuming that DS=1512h. MOVAL, 99H MOV [3518], ALAssume you are given an array of four 16-bit numbers stored in memory with a starting address x3110. Using PC- relative addressing, write an LC-3 machine language program that will copy the same four 16-bit numbers in reverse order, starting at memory address x3120. For example, if the following are the four 16-bit values stored in x3110, x3110 12 x3111 23 х3112 34 х3113 45 then the result of your program execution should write the following starting at memory address x3120: x3120 45 х3121 34 x3122 23 x3123 12 Load your program starting at address x3100. You will need to demonstrate the correct execution of your program by using the LC-3 Simulator.1B. Write three address code and quadruple for the expression a[i] = -b * (a[k--]- y[ k--] /2)
- Consider the following image that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes (like LC-3): A memory location can store an address. We call that memory location's contents a "pointer" since it's an address that "points" to another memory location. G.) Interpret the contents at address 0x0C0B as a pointer. (Enter hex like the following example: 0x2A3F) H.) What are the contents of the memory location that the pointer above is pointing to? (Enter hex like the following example: 0x2A3F) Another reference : LC-3 Opcodes in Hex ADD 0x1 JMP 0xC LDR 0x6Write 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, 75Please solve this problem in emu8086 microprocessor in Assembly Language. Please solve properly, will upvote after getting proper solve Read 10 inputs (range 0-9) from the user and store them in an array. Find the position (or index) of the maximum and minimum values in the array, and swap them (move the biggest element to the position of the smallest, and move the smallest element to the position of the biggest), and print that again. If there are multiple maximum or minimum numbers, then choose the first occurrence, values. Your program should be able to handle negative values. Explanation: Here from the given input we see that 9 is the maximum number and it has occurred twice, first in the 2nd index and again in the 5th index. Similarly here 3 is the least and it also has appeared twice, once in the 3rd index and again in the 6th index. So we swap 9 and 3 from the 2nd and 3rd indices and then after swapping, print the new array. Input: 7935935874 Output: 7395645874 Comment:Range 0…