A-Calculate the offset address for the following. Assume all numbers in hexadecimal and the first part represent a physical address. 1- 12340:1000 3- ABCD0:0100 2- A2000:12CF 4- FA120:B2C0.
Q: he contents of the following registers are: CS = 2212 H DS = 4040 H SS = 3025 H IP = 2230 H SP =…
A: We are going to find out physical addresses of CS,DS, and SS. I have uploaded image for the solution…
Q: To address 1KB, 2KB, 4KB, 1MB, 1GB, and 4GB of RAM, how many bit addresses are required? How many…
A: INtroduction To address 1KB, 2KB, 4KB, 1MB, 1GB, and 4GB of RAM, how many bit addresses are…
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A: The Answer is
Q: 5. Assume 32 bit memory addresses, which are byte addresses. You have a 2-way cache with a block…
A: Introduction :
Q: Given the following memory requirements, determine the minimum number of address bits required.…
A: Introduction: Here we are required to determine the minimum number of address bits required, also we…
Q: A system using segmentation provides the segment below. Compute of the logical addresses. If the…
A: a. 0, 323 Segment = 0 Offset = 323 Offset < Length (323<400) Physical Address = Base + Offset…
Q: blocks of 16 words are there in a 256 Gig memory space? Draw the logical organization of the full…
A: How many blocks of 16 words are there in a 256 Gig memory space? Draw the logical organization of…
Q: Data bytes are stored in memory locations from 8050h to 805FH. Write 8085 ALP to transfer the data…
A: Write ALP to transfer the data to the location 8150h to 815Fh in the reverse order ..
Q: Suppose, A6BA1H is a particular physical location, and 1234 is the base address of that segment. So…
A: Answer is given below .
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:
Q: Complete the following table given the value %rdx = Oxe800, %rcx = 0x0400 (5 Points) Address Address…
A: %rdx = 0xf800%rcx = 0x0200Expression: 0x80(%rdx)Address Computation: 0xf800 + 0x80Address: 0xf880…
Q: For each of the following decimal virtual address : 32768, 60000. Compute the virtual page number…
A: The solution for the above given question is given below:
Q: (b) Calculate the offset address for the following. Assume all numbers in hexadecimal and the first…
A: Answer is given below .
Q: If logical address is (0001100001011) binary find page number p: displacement
A: Logical address: 0001100001011 Page number: 00011000 Displacement: 01011 Suppose the page addressed…
Q: Suppose we have a byte-addressable computer using direct mapping with 16-bit main memory addresses…
A: Step 1:- Memory address size=16 bit Number of cache block=32 a) The number of bits of the offset…
Q: 2. Use the defined GPR of TMP (it is a DATA) and RST (It is an ADDRESS), present your ASM codes that…
A: Answer : a). TMP - 0XDF -> Result in W Operation: SP-2→SP, PC+2→@SP dst→PC. Temp→ 0X (.W)…
Q: 19-The MSB in the 20 bits of physical address specifies the segment. Select one: C True False
A: Given question are true or false based question.
Q: Given a 4-way set associative cache, which has 256 blocks and 64 bytes per block. Assume a 32-bit…
A:
Q: Q) MOV CX, [481d] ; assuming DS= 2162H, logical address will be?
A: The instruction has Direct addressing mode.
Q: If logical addresses are represented using m bits as shown below, where m=4 and n=2. What is the…
A: Actually, memory is used to stores the data.
Q: 2. Answer all the questions given below. (a) -For the given code segment, find the value of ax [CO2]…
A: a.(i) From the question we know that ax register is in hexadecimal form and hence a register is…
Q: 3. A12C bus transmits signals as shown in figure below. a. Write the correct names of A and B…
A: a) The correct names of A and B signals :- Signal A is NRZ(Non-return-to-zero) signal ans Signal B…
Q: Problem 2. Convert the following virtual addresses to physical addresses, and indicate whether the…
A: Answer: Our instruction is answer the first three part from the first part and . I have given…
Q: a) In the SRAM region, what is the corresponding bit-band alias address for the bit [4] of the…
A: #define BITBAND_SRAM_REF 0x20000008 #define BITBAND_SRAM_BASE 0x22000008 #define BITBAND_SRAM(a,b)…
Q: What are the physical addresses for the following logical addresses (Segment ID, Segment Offset):
A: The physical address of following logical address are: i) 0 200 This falls under segment 0, first…
Q: How convert 16 bit address to 20 bit address explains with the help of example. Which Flag is…
A: I have provided a solution in step2.
Q: Solve the following segmentation address translation Assume that GDT has the following content GDT:…
A: The solution for the above given question is given below:
Q: 3- Suppose that ECx=12345678h , EBx=87654321h ,and DS=1100h. Determine the contents of each address…
A: Suppose that ECx =12345678h, EBx=87654321h,and DS =1100h .
Q: Given the page table shown below. What is the physical address (in the binary-number format)…
A: The logical address is <5, 17> The logical address is given in format <page#, offset>
Q: Assume 32 bit memory addresses , which are byte addresses. You have direct mapped cache with…
A: Block size = 8 bytes So block offset bits = 3 bits Total # of cache blocks = 128 So index bits = 7…
Q: Difference between logical and physical addresses is in:
A: Given: We have to discuss Difference Between logical and physical Address is .
Q: Suppose, 3BD15 H is a particular physical location, and 1234 is the value of the offset.What should…
A: Answer is given below .
Q: In a 1 MB memory divided into 64 KB segments, if a segment starts at the address 1234A find the last…
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Q: Suppose the data segment (DS) holds the base address as 1100h and the data you need is present in…
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Q: In a system, with 10 bit addresses of which 4 bits is for the page number, give following page…
A: Here number of bits for page number=4 Thus number of bits for offset = 10-4 = 6. These are least…
Q: (i) The contents of the following segment registers are as given. CS = 1111H, DS = 3333H, SS =…
A: Segmentation is the process in which the main memory of the computer is logically divided into…
Q: Calculate the physical address. Assume DS = 1234h, AX = 4523h with the help of memory diagram. MOV…
A: S. No. 8086 microprocessor 8088 microprocessor 1 The data bus is of 16 bits. The data bus is of…
Q: How many bits are required for addressing (i.e. what is the size, in bits, of an address)? 30 bits O…
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Q: a) A paging system with 512 pages of logical address space, a page size of 2* and number of frames…
A:
Q: Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory…
A: The logical address is represented as combination of page number and page offset. physical address…
Q: 4. Assume 32 bit memory addresses, which are byte addresses. You have a direct-mapped cache with a…
A: Block size = 8 bytes So block offset bits = 3 pieces All out # of cache blocks = 128 So index…
Q: 11. Translate the following code into MIPS. Assume that a=$a0 and b=$al are integer arrays whose…
A: # a = $a0, b = $a1, n = $a2 li $t0, 0 # i = 0 for: bge $t0, $a2, end # for (i < n) ble…
Q: Determine the number of page table entries (PTES) that are needed for the following combinations of…
A: A page table is the data structure used by a virtual memory system in a computer operating system to…
Q: Determine the number of page table entries (PTES) that are needed for the following combinations of…
A: The connection between page table entries, virtual location size and page size is given as Page…
Q: Data bytes are stored in memory locations from 8050h to 805Fh. Write 8085 ALP to transfer the data…
A: We need to write a 8085 alp for the given scenario.
Q: Assume that the base address of 8255-PPI chip is OC00H and the address of port C of the chip is…
A: Assembly level language is a low-level programming language, that's used to communicate directly…
Q: Write the binary translation of the logical address 0001010010111010 to physical address, under the…
A:
Q: a) In the SRAM region, what is the corresponding bit-band alias address for the bit [4] of the…
A: a) In the SRAM region, what is the corresponding bit-band alias address for the bit [4] of the…
Q: Determine the number of page table entries (PTES) that are needed for the following combinations of…
A: The relation between page table entries, virtual address size and page size is given as page table…
Step by step
Solved in 2 steps
- a. Find the address accessed by each of the following instructions. If DS = 0100H, BX= 0120H, DATA = 0140H, and SI = 0050H and real mode operation:1. MOV DATA[SI], ECX2. MOV BL, [ BX+SI]b. Descriptor contains a base address of 00260000H, a limit of 00110H, and G = 1,determine starting and ending locations are addressed by the descriptor for aCore2.Solve the following segmentation address translation Assume that GDT has the following content GDT: Ox1 -> Өx100 Ox2 -> Ox300 Ox3 -> Өx400 Ox4 -> Ox800 What would be the address of ds:eax if DS register contains (Ox3 << 3, remember the lowest 3 bits of the segment selector are flags) and EAX contains value Ox3245. Please write answer in HEX format (example: Ox1234): Enter your answer hereliomework: Identify the addressing modes used for the source and destination operands, and then compute the physical address for the specified orx•rand in each of the following instructions: a) Mov +XYZ,AH (Destination operand) b) Mov tBX) (Destination operand) Suppose that BX=030016, 010016.
- Evaluate the expression: F = (c -a)*d +e-b focusing 1- address through 3-address format. Also explain which addressing mechanisms bears maximum number of instructions and why?The 32-bit number 52AB43FC (in hexadecimal) is stored in abyte-addressable memory starting at physical address FE08 (in hex) using Little-Endian notation. The byte(value ni hexadecimal) stored at theaddress FE0B will beIdentify if possible the type of addressing modes for the followinginstructions. 1. MOV [EBX + EDI + ABCDH], EDX
- 5 - Provide the format and assembly language instruction for the following hex values: Address 1000: 13 Address 1001: 03 Address 1002: C5 Address 1003:00 Hint: first consider big-endian vs. little-endian as you convert to binary. Then, divide the bits up into the appropriate fields, decipher the opcode, and so on.Q5 Please convert the following C/C++ function to RISC-V assembly (RV321). Function: swap Registers: ra = x1 (return address) &v[0] -> a0 = x10 (arguments/return values) k-> a1 = x11 (arguments/return values) to = x5 (temporary) t1 = x6 (temporary) t2 = x7 (temporary) hint: void swap(int v[], int k) { int temp; temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; %3D }The 8-bit register AR, BR, CR, and DR initially have the following values: [5]AR = 11010010; BR = 11111111; CR = 10101001; DR = 10101010Determine the 8-bit values in each register after the execution of the following sequence ofmicrooperations.AR AR + BR Add BR + ARCR CR AND DR, BR BR + 1 AND DR to CR, Increment BRAR AR - CR Subtract CR from AR
- B- Calculate the offset address for the following. Assume all numbers in hexadecimal and the first part represent a physical address. 1-23451:2222 2- 6B000:6000 3- 22550:2132 4-31234:3000 5- 42234:4100Q2- Write a program in assembly language for the 8085 microprocessor to receive one byte of data via the SID and store it at the memory address (3000H to 3009H) using a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz . When receive the required bytes, you must adhere to the following: The bits of two high bits will be received at the beginning of the reception(start bits 1 1 ), after that the data bits will be received, after that the low bit of the stop bit will be received (stop bit 0 ). The following flowchart will help you. The solution must be integrated and include the calculation of the baudrate delay timeQ1- Write a program in assembly language for the 8085 microprocessor to receive 10 bytes of data via the SID and store it at the memory address (3000H to 3009H) using a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz. When you receive each byte of the required bytes, you must adhere to the following: The bits of two high bits will be received at the beginning of the reception (start bits), after that the data bits will be received, after that the low bit of the stop bit will be received (stop bit). The following flowchart will help you, but you should notice that this flowchart deals with one byte, and you are required to deal with 10 bytes The solution must be integrated and include the calculation of the baudrate delay time Of+CD!HID+[00 Yes SIDATA Read SID Start Bit? Wait for Half-Bit Time Set up Bit Counter Wait Bit Time Read SID Save Bit Decrement Bit Counter All Bits Received? Add Bit to Previous Bits Go Back to Get Next Bit Return IMUNI