subroutine stack segment 3 1400 segment o 2400 segment 0 symbol table limit base Sqrt segment 4 1 main program 0 1000 1400 400 6300 2 400 4300 3200 3 1100 3200 4 1000 4700 segment 3 segment table segment 1 segment 2 4300 segment 2 4700 Virtual address space segment 4 5700 6300 segment 1 6700 physical memory What physical address does <4,152> resolve to? Error • 4852 • 4851 4853
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- 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?For the instruction (0x6479), select all data paths that are used from the beginning of the Decode Instruction phase through the end of the Store Result phase. FYI: Be certain; Canvas deducts points for incorrect choices. OL tol OH to J OK to N OL to E OH to F OM to B OC to N OH to L A to F ON to O OM to N O to MThe 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 be
- The address bus consists parallel signal lines while data bus consists separate lines. True O False Hardwired control units are generally faster than micro - programmed designs. * True False The main property of a bus is that it is a shared transmission medium. * True O FalseConsider 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)Below is a list of 64-bit memory address references given as word address. Ox03, Oxb4, Ox2b, Ox01, Oxb7, Ox58, Oxbe, Ox02, Oxb5, Ox2e, Oxb6, Ox5b 0000 4 0100 8 1000 1100 1 0001 5 0101 9 1001 d 1101 2 0010 6 0110 a 1010 e 1110 3 0011 7 0111 b 1011 f 1111 Given a direct-mapped cache with 16 word blocks, what is the hit ratio? O 0.5 O 1 O 0.75 O 0.25
- What are the benefits of segmented memory address translation over a straight translation?Q:Find the value of SP address if SP= (2000) hex after execute the =following instructions SP PUSH A PUSH B PUSH C PUSH D إجابتكShow how the decimal value 2019111310 would be stored by byte-addressable machines with 32-bit words, using little endian format and then big endian format. Assume the value starts at address 201916. Draw a diagram of memory and place the appropriate values in the correct (and labeled) memory locations.
- 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]TRUE OR FALSEMost architectures today are accumulator based.TRUE OR FALSEThe term “endian” refers to an architecture’s byte ordering.