1. Calculate the smallest and fourth largest segment for the following physical addresses also write down the logical address for each of them: a. FB2DEH b. 18C3Dh
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- 6. Write down the three-address codes for the following code segment. S := 0, I := 0 WHILE I < 10 DO S := s + X[I] I := I + 1 ENDWHILEphyscal 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 segment table:Segment Base Length0 219 6001 2300 142 90 1003 1327 5804 1952 96What are the physical addresses for the following logical addresses? segment # limita. 0 430b. 1 10c. 2 500d. 3, 400e. 4 112
- Consider the following segment table:Segment Base Length Base Limit0 1200 5001 2200 10242 6400 2563 8000 700What are the physical addresses for the following logical addresses?a. 2, 250b. 3, 690c. 0, 515IF PC= 682 , AR=123 , DR=A2BF, IR= 5672, then execute the following M[AR] DR IR M[AR] Determine the value of IR after the execution and comment on the type of addressing.QI- a- Write the codes used for segments (of 8086 microprocessor) designation in machine language coding. b- Write a program to subtract two words stored in addresses that start at [1200 H] & store the result in successive addresses.
- The following data segment starts at memory address 1000h (hexadecimal) .data printString BYTE "ASSEMBLY IS FUN",0 moreBytes BYTE 25(DUP)0 dateIssued DWORD ? dueDate DWORD ? elapsedTime Word ? What is the hexadecimal address of dueDate ? a. 1045h b. 1029h c.1010h d. 102DhFor a computer system, which utilizes pure segmentation, segmentation table and list of virtual addresses (a: segment number, b: offset) are given below. Please, write corresponding physical addresses for each virtual address given: Segment No Base Limit 219 600 a. 0,430 1 2300 14 b. 1,10 90 100 c. 2,500 3 1327 580 d. 3,400 4 1952 96 e. 4,112 -----5. Given the descriptor: Determine the following: a. Limit: b. Starting Address: c. Ending Address: d. Segment: e. DPL: f. G: g. AV: h. D: i. Descriptor used: 7 00H 5CH 6 5 9AH 01H 4 5678H 1 1234H