Bitmap data format lets the OS track free and full partitions. How do bitmaps show allocation unit processes or holes? Answers: *process bit *hole bit *free bit *flag bit
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Bitmap data format lets the OS track free and full partitions. How do bitmaps show allocation unit processes or holes?
Answers: *process bit *hole bit *free bit *flag bit
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- True or false: Temporal locality is the tendency for a program to access a memory address shortly after accessing a 'nearby' memory addressThe memory location at address 00002001 contains the memory variable in binary form. What is the data memory variable in hexadecimal form? MEMORY 1110 1011 00002001 1110 1010 00002000 1110 1001 00001999 1110 1000 00001998 1110 0111 00001997 1110 0110 00001996 DATA ADDRESS The data memory variable in hexadecimal form is E7. a. b. The data memory variable in hexadecimal form is EA. The data memory variable in hexadecimal form is EB. C. The data memory variable in hexadecimal form is E9. Od.Assembly Language x86: (Micro Macro): templet: .386.model flat, stdcall.stack 4096ExitProcess PROTO, dwExitCode: DWORD.data day BYTE 0month BYTE 0year WORD 0 .codemain PROC main ENDPINVOKE ExitProcess, 0END main Question: What bit string repersents April 1, 2024? Examples: Date: 00111 (year) 1100 (month) 10010 (day)00111 = 18, 1100 = 12, 10010 = 30, 1980 + 30 = 2010 = 2010 Dec 18mov ax, 00111110010010b; 2010 Dec 10; 30 12 18 DAY:mov dx, ax and dx, 0000000000011111bmov day, dl (18) ; 00111 Month: mov dx, ax and dx, 00000001111000000b shr dx, 5 ; 000000000000 1100 (12) mov month, dl Year:mov dx, axshr dx, 9 ; 011110and dx, 0000000001111111bmov year, dl (30) 10010
- The OS uses a bitmap data structure to keep track of whether or not a partition is empty or full. What bitmap conventions are used to depict the procedure or the void in allocation units? Answer alternatives that fall under this cluster include: *process bit *hole bit *free bit *flag bitIn a main memory-disk virtual storage system, the page size is 1KByte and the FIFO algorithm is used for page replacements. A given program has been allocated three page frames in the main memory and it makes the following 16 memory references when it starts executing (the addresses are given in decimal):500, 2000, 2500, 800, 4000, 1000, 5500, 1500, 2800, 400, 5000, 700, 2100, 3500, 900, 2400 Fill in the contents of the three page frames after each memory reference in a table and calculate the hit ratio. Hint: denote by 'a' the page consisting of locations 0 through 1023 in memory. Similarly, b: 1024-2047, c: 2048-3071, d: 3072-4095, e: 4096-5119 and f: 5120-6143. Round to three decimal places.lices Font Peragrsoh Diraving Example: • Move a block of N consecutive bytes of data starting at offset address BIK1ADDR in memory to another block of memory locations starting at address Blk2ADDR. Assume that both blocks are in the same data segment whose starting point is defined by the data segment value DATASEGADDR. 11 Steter
- An application loads 100 libraries at start-up. Loading each library requires exactly one disk access. The seek time of the disk to a random location is given as 10 ms. Rotational speed of disk is 6000 rpm. If all 100 libraries are loaded from random locations on the disk, how long does it take to load all libraries? (The time to transfer data from the disk block once the head has been positioned at the start of the block may be neglected.)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?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 = 0XD4
- Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Page Table Physical Memory Physical Address (starting) Oxppppdddd Page Frame Frame Size (hex) Size (dec) Ox10000 Ox10000 2 Охс000 65536 PPpp: page number dddd: page offset 1 1 Оxd000 65536 3 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x00011119 What is the physical address for 0x00000001 What is the logical address for Oxd0000001 ? What is the logical address for Oxc0000002 ?It is easy to explain how dynamic memory allocation works.LCPU assignment Multiply the number by 1.25: A = X * 1.25X = 0x3C (direct). Then save the result to main memory using direct addressing. Example: Multiply the value by 0.75: X * 0.75 = 0.5 * X + 0.25 * X X = 100 (0x64)100 * 0.75 = 750x64 = 01100100 00110010 [Shift 1x to the right - add 0 to the left and cut off one bit from the right]00011001 [Shift 2x to the right - add 00 to the left and truncate 2 bits from the right]01001011 [Sum of previous two transactions] 01001011 = 0x4b = 75