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
- Below shows the following hexidecimal values:Address 1000: 13Address 1001: 03Address 1002: C5Address 1003: 00 Provide the format and assembly language instruction for the following hexadecimal values.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 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]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]
- 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
- . 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 R15Physical address is formed from segment address & offset address. Select one: O True O FalseWhat are the benefits of segmented memory address translation over a straight translation?