1. 1000000 – 1100011 – 100100 2. 111+ 101101 + 110010 + 111 3. 101111 + 101011 - 101
Q: Address Content 50 10 51 57 52 53 21 OA 54 55 52 01 56 32 57 СО 58 59 00 Suppose the memory cells at…
A: Answer: I have given answer in the brief examples
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A: We solve the question in figure: Figure 1:
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A: Here word size=64bit = 64/8=8 Bytes.
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A: Physical Address The complete physical address is 20-bits long. It is generated using segment and…
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A: Actually, binary numbers are nothing but a 0's and 1's.
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A: Please Check step 2 for the answers. I have provided the correct answer.
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A: Answer: b. Real Memory
Q: Address Content 50 10 51 57 52 21 OA 53 54 52 55 01 56 32 57 58 CO 59 00 Suppose the memory cells at…
A: Here, I have to answer the above question.
Q: 2. The 8-bit registers AR, BR, CR, and DR initially have the following values: AR= 11010010, BR=…
A: Actually, AR,BR,CR and DR are registers. The 8-bit registers AR, BR, CR, and DR initially have the…
Q: [5] The following table represents a small memory with 4 locations: Address Data 0001 1110 0100 0011…
A: Answer:)
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A: given an operating system that uses Segmentation. Given an address [S,D], where S is the segment and…
Q: Address Content 50 10 51 57 52 21 53 0A 54 52 55 01 56 32 57 CO 58 CO 59 00 Suppose the memory cells…
A: Find the required answer given as below:
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A: Lets see the solution.
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A: As per our guidelines, only 3 sub parts will be answered. So, please repost the remaining questions…
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A: Answer:-
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A: The answer given as below:
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A: This question comes from Computer Organization/Architecture which is a paper of Computer Science.…
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A: It is defined as programs can be asked from control statements, array, string, oops etc. Java basic…
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A: Actually, register is a used to stores the data\information.
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A: Answer :- ( d ) 16,777,216
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A: According to Bartleby Guidelines we need to answer only 3 sub questions so I have answered first…
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A: Note: As per the guidelines can answer the first three sections. Please resubmit the question for…
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Q: Address Content 50 10 51 57 52 21 53 OA 54 52 55 01 56 32 57 58 59 Со 00 Suppose the memory cells at…
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A: The program to find 2's complement of 8 bit number is given below:-
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A: Given a string of length 9, and each bit can be either 0 or 1.
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A: THe solution for the above given question is given below:
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A: Solution:-
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A: Answer : - option A) 65536
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Q: 10110; BR = 11100111; CR = 10110001; DR = 10111010 Determine the 8-bit values in each register after…
A: AR = AR + BR 11010110 + 11100111 = 0110111101 CR = CR AND DR 10110001 and 10111010…
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A: According to the Question below the Solution:
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Microprocessor(Binary number)
1. 1000000 – 1100011 – 100100
2. 111+ 101101 + 110010 + 111
3. 101111 + 101011 - 101
4. 11100 + (1011001 – 101011)
5. 101000 + 110110 - 11111
6. 1100110 – 1111 – 110110
7. 101010 + 100001 - 10011
8. 100100 + 111000 - 11110
9. 1100011 - 11011 – 111111
10. 1010011 – 101001 + 111111
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- The main memory of a computer is addressed by bytes and is managed by two-level paging storage. The address structure is as follows: Outer Page (10 bits) Inner Page (10 bits) Page offset (12 digits) The outer page number and inner page number corresponding to the logical address 0x20501225 are ____ , ____ respectively. 0x081, 0x101 0x081, 0x401 0x201, 0x101 0x201, 0x401Calculate Physical Address of Main memory • Given the contents of CS:IP = [3000H]:[100H] • Given the contents of SS:SP = [50A0H]:[300H] • Given the contents of ES:DI = [60C0H]:[200H] • Given the contents of DS:SI = [A1C0H]:[2F00H] • Given the contents [CS]:[IP]=[1F0AH]:[15A6H] • Given the contents [SS]:[SP]=[82FFH]:[0FA1H]Questions are from Microprocessor (8086)REAL MODE MEMORY ADDRESSING 1. In the real mode, show the starting and ending address of the segment located by the following segment register values (in hex): a) SR= DC28b) SR=FA91 2. Find the memory location addressed by the microprocessor, when operated in the real mode, for the following segment register and 80286 register combinations: a) DS=8EBC & DX=A3D7b) CS=DCAF & IP=FAC8
- Show how the following values would be stored by machines with 32-bit words, using little endian and then big endian format. Assume each value starts at address 1016. Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations. 1234567816 ABCDEF1216 87654321161.What is the difference between bit address 7CH and byte address7CH? What is the specific location of bit address 7CH in memory?1 A computer has main memory access time of 500 ns or 500 x 10-9 seconds and cache memory access of 100 ns or 100 x 10-9 seconds and the hit ratio of the cache memory is 0.8, find the average access time2Calculate the largest decimal number that can be represented by TU bitsUsing 2’ complement only subtract 52136 from 54126Add 100100102 to 10001012 and convert your answer to base 4Explain why 8790229 is not a good number system
- REAL MODE MEMORY ADDRESSING 1. In the real mode, show the starting and ending address of the segment located by the following segment register values (in hex): a) SR= DC28b) SR=FA91 2. Find the memory location addressed by the microprocessor, when operated in the real mode, for the following segment register and 80286 register combinations: a) DS=8EBC & DX=A3D7b) CS=DCAF & IP=FAC8 Appreciate your help. Thanks!A CPU is equipped with a cache. Accessing a word takes 40 clock cycles if the data is not in the cache and 5 clock cycles if the data is in the cache. What is the effective memory access time in clock cycles if the hit ratio is 80%?Carefully read each sentence in this question. You may agree/disagree with each sentence (consider each sentence separately) if you have a good justification for doing so (not in more than 2 lines for each sentence). “A daisy chaining technique could result in starvation.” “The received data is saved in contiguous memory regions in DMA transfer.” “The most difficult method of device identification is multiple interrupt lines.” “An interrupt request takes longer for the CPU to handle than a DMA request.” “The architecture of InfiniBand operation is layered.”
- Computer Science This question is about paging-based virtual memory A computer has a virtual-momory space of 250MB (megabytes) The computer has 325) of primary memory. The pige som s-4000 by the address is 1011 0001 0101 1110 0010 1010 0010 a. How many frames can it have? b. Which of the bits in the virtual address correspond to the Page number? c. Which of the bits correspond to the page offset?True or False. A computer has a 16 bit address space A[15:0]. If all addresses having bits A[15:13]=111 are reserved for I/O device registers, then the maximum number of actual word addressable memory locations is 2^16 through 2^14.TOPIC: MARIE Assembly/MARIE Simulator Pls answer only if u know, correct answers only pls! will rate you if correct 1.) What is the address of the last (bottom-rightmost) memory location? My answer is 4095 but it is wrong. I thought MARIE starts at 0000 and ends with 4095 so I need help on this. 2.) What is the content of memory address 001? (No need to write 0x from hereon; we already assume hexadecimal unless otherwise stated.) My answer is 3000 but it is wrong.