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- Most Intel CPUs use the __________, in which each memory address is represented by two integers.If a microprocessor has a cycle time of 0.5 nanoseconds, what’s the processor clock rate? If the fetch cycle is 40% of the processor cycle time, what memory access speed is required to implement load operations with zero wait states and load operations with two wait states?Consider a memory implemented for 8086 microprocessor Draw the memory block diagram. Determine the values for A0 , /BHE ,address lines(A1..A19) and data lines(D0.. D15) in order to access: A byte at odd address [01FF3H] A byte at even address [01FFCH] A word at even address [01FFEH] A word at odd address [01ABFH]
- A microprocessor has a 32-bit address line. The size of the memory contents of each address is 8 bits. The memory space is defined as the collection of memory position the processor can address. What is the address range (lowest to highest, in hexadecimal) of the memory space for this microprocessor? What is the size (in bytes, KB, or MB) of the memory space? 1 KB = 21010 bytes, 1MB = 22020 bytes, 1GB = 23030 bytes A memory device is connected to the microprocessor. Base o the size of the memory, the microprocessor has assigned the addresses 0xA4000000 to 0xA07FFFFF to this memory device. What is the size (in bytes, KB, or MB) of this memory device? What is the minimum number of bits required to represent the addresses only for this memory device?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?Suppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01 -What is the Value of r0 and r1 after executing LDR r1, [r0, #2] -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]
- . Assume SP=0XE99D, R16=0XE2, R17=0x25, R01=0XFC, R15=0X1F and the following memory information. Address contents (hex) post Address contents (hex) post pre 22 pre 44 OXE996 OXE99C OXE997 46 OXE99D C5 OXE998 17 OXE99E Аб OXE999 21 OXE99F 77 ОхЕ99A F2 OXE9A0 78 OXE99B C3 OXE9A1 A5 Find the values of the registers SP, R01, R16 and R17 after the following operations. РОP R01 РО R16 РОP R17 РOP R20 PUSH R15 SP R16 R17 R01 R20 R15Suppose r0 = ox300010A0, r2 = 0x00000011, and the memory layout is as follows Address: Data: 0x300010A7 0x72 0x300010A6 0XA5 0x300010A5 0x9F 0x300010A4 0x00 0x300010A3 0x50 0x300010A2 0x2B 0x300010A1 0XA5 0x300010A0 0x01 -What are the values of r0 and r1 after executing the following code? Illustrate your process in a memory map. LDR r1, [r0], #3 ADD r1, r1, r2 STR r1, [r0, r#4]The IBM System/370 architecture uses a two-level memory structure and refers to the two levels as segments and pages, although the segmentation approach lacks many of the features described earlier in this chapter. For the basic 370 architecture, the page size may be either 2 Kbytes or 4 Kbytes, and the segment size is fixed at either 64 Kbytes or 1 Mbyte. For the 370/XA and 370/ESA architectures, the page size is 4 Kbytes and the segment size is 1 Mbyte. Which advantages of segmentation does this scheme lack? What is the benefit of segmentation for the 370?
- Q1- Write a program in assembly language for the 8085 microprocessor to send 10 bytes of data located at the memory address (3000H to 3009H) using SOD at a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz. When sending each of the required bytes, you must adhere to the following: The two high bits of the start bits must be sent, after that the data bits are sent, after that the low bit of the stop bit is sent. 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 Transmit No Set up Character Bit Counter Send Start Bit Wait Bit Time Get Character in Accumulator Output Bit Using Do Wait Bit Time Rotate Next Bit in Do Decrement Bit Counter Is It Last Bit? Yes Add Parity if Necessary • Send Two Stop Bits Return (a)physcal addresses are 4s ng 4 Ame dat in a cetain compe, te addresses can be translaled without y TLB entries At most how many ditina vid the address translation peh has 12 vld The Translation Look aside Bulfer (TLB)i sine is kB and the word size iby The memory is word addresible. The pe virtual addresses are 64 bea long d th sine is miss?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)