I am planning to buy an SSD. In these two pictures, which do you think is better or faster?
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I am planning to buy an SSD. In these two pictures, which do you think is better or faster?
![CrystalDiskMark 8.0.4 x64 [Admin]
File Settings Profile Theme Help Language
All
SEQ1M
Q8T1
SEQ1M
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RND4K
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1 GiB
Read (MB/s)
2129.79
1170.74
486.38
41.60
C: 32% (76/238GiB)
X
MB/s
Write (MB/s)
1229.81
1204.54
452.77
131.89](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8cab4459-86d0-49ba-bee1-199ba265b33c%2F6307034e-37f5-455a-9d79-0a04c93f70d7%2F6etsbwk_processed.jpeg&w=3840&q=75)
![CrystalDiskMark 8.0.4 x64 [Admin]
File Settings Profile Theme Help Language
3
All
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SEQ128K
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1 GiB
Read (MB/s)
1978.21
1971.24
373.61
48.35
C: 43% (101/238GiB)
MB/s
Write (MB/s)
X
1024.09
1030.03
494.82
125.14](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8cab4459-86d0-49ba-bee1-199ba265b33c%2F6307034e-37f5-455a-9d79-0a04c93f70d7%2Frtq3axw_processed.jpeg&w=3840&q=75)
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- Address Instruction Ox242 CALL sort Ох262 sort: MOVLW Ох90 Fill in the following blanks based on the above code segment: First word of machine code stored in locations 0x243 and Ox242: Ox * Second word of machine code stored in locations Ox245 and Ox244: Oxphyscal 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?.data var1 SBYTE -4,-2,3,1 var2 WORD 1000h,2000h,3000h,4000h var3 SWORD -16,-42 var4 DWORD 1,2,3,4,5 mov al,var1 ; [a] a. mov ax,[var2+2] ; [b] b. mov ah,[var1+3] ; [c] c. mov ax,[var3 - 4] ; [d] d. mov edx,[var4+8] ; [e] e. movzx edx,var2 ; [f] f. mov edx,[var4+12] ; [g] g. movsx edx,var1 ; [h] h. What are the values in a-h
- Paging: Select all of the following statements that are true. Systems that use paging but do not support inverted page tables maintain at least one separate page table for each process. The frame table is a system-wide data structure. When paging is applied, the selected page size determines which part of a virtual address belongs to the page number and which to the offset. The page size may differ from the frame size. The Translation Look-aside Buffer (TLB) represents a page directory for all pages in the system. Paging is prone to internal fragmentation.Code a descriptor that describes a memory segment that begins at location 0005CF00h and ends at location 00060EFFh. The memory segment is a data segment that grows upward in the memory system and can be written. The segment has a user level privilege (lowest) and has not been accessed. The descriptor is for an 80386 microprocessor.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 M
- Direct Mapping Example: CPU is searching an Instruction stored at RAM address 1110011010 in cache and doesn't find it. What happens then? Tag 0000000000 ?? 0000000001 ?? Block:0, (j=0) 0000000010 Instruction-1 Line1 = 1 Tag Tag 0000000011 Instruction-2 0000000100 Instruction-3 Instruction-4 Block:1, G=1) Solution: 0000000101 Line 2 (= 2) Tag 0000000110 Instruction-5 0000000111 Instruction-6 0000001100 Instruction-7 0000001101 Instruction-8 Block:2 0000001110 Data-1 (j=2) 0000001111 Data-2 Line.3 (3) 0000011100 Data-3 0000011101 Data-4 Tag 0000011110 Data-5 0000011111 Data-6 Cache Size = 64B Size of a line = 4B Total number of Lines in Cache: m 16 RAM Size = 1KB %3D 1111111100 %3D Size of a Block= 4B Total number of Blocks, M = 256 j = 0, 1, 2, ..255 (M-1) Block 255. G =255) 1111111101 1111111110 i = 0, 1, 2, ... 15 (m-1) 1111111111True or false: Temporal locality is the tendency for a program to access a memory address shortly after accessing a 'nearby' memory addressRead-only memory (usually known by its acronym, ROM) is a class of storage media used in computers and other electronic devices that is non-volatile (non-changeable). Because data stored in ROM cannot be modified (at least not very quickly or easily), it is mainly used to distribute firmware (software that is very closely tied to specific hardware, and unlikely to require frequent updates). Figure 3.1 depicts a ROM block diagram which have k-bit address and an n-bit data. The contents of a ROM chip are defined once. Hence, we can use a constant array to model a ROM in VHDL. Basically, to name the size of a ROM is by stating ROM k x n, which represents ROM with k-bit address andn-bit data size. For example, ROM 4x8 (16 address location with each location containing 8-bit data). Question. Develop a code to allow the ROM operation works in synchronous when a read enable is asserted.• Generate and simulate the VHDL codes in Altera Quartus II.• Demonstrate and obtain the instructor…
- 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)Memory address Data According to the memory view given below, if RO = Ox20008002 then LDRSB r1, [r0, #-4] is executed as a result of r1 = ?(data overlay big endian)? Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 Øx73 ØX20007FFE ØXD4 ØX20007FFE Lütfen birini seçin: O A. R1 = 0X7F O B. R1 = Oxffffffd4 O C. R1 = Oxffffff7F O D. R1=0XD4000000 O E. R1 = 0XD4Computer 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?