3.3 Consider a simple segmentation system that has the following segment table: Segment |Starting Address 500 Length (bytes) 350 1700 550 2 200 118 3 900 654 For each of the following logical addresses, determine the physical address or indicate if a segment fault occurs: a. 0, 180 b. 1, 152 C. 2, 114 d. 3, 702 0, 212 e.
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Q: 3.3 Consider a simple segmentation system that has the following segment table: Segment Starting…
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Q: Answer only 3 and 4 Suppose memory has 256KB, OS use low address 20KB, there is one program…
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Q: we have not provided for any type of error handling, such as if the address on the address lines…
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Q: Suppose that DS=300H, SS=200H, CS=100H, CX=AF39H, ARRAY=D3A4H, IP=5A76H, DI=245AH, SP=5489H,…
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Q: Let's tabulate that, and then extend the concept to fill in the blank cells: Memory |сарacity…
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- Answer only 3 and 4 Suppose memory has 256KB, OS use low address 20KB, there is one program sequence: (20) • Prog1 request 80KB, prog2 request 16KB, • Prog3 request 140KB • Prog1 finish, Prog3 finish; • Prog4 request 80KB, Prog5 request 120kb • Use first match and best match to deal with this sequence • (from high address when allocated) • (1)Draw allocation state when prog1,2,3 are loaded into memory? • (2)Draw allocation state when prog1, 3 finish? • (3)use these two algorithms to draw the structure of free queue after prog1 , 3 finish(draw the allocation descriptor information,) • (4) Which algorithm is suitable for this sequence ? Describe the allocation process?Q3 Consider a swapping system in which main memory contains the following hole sizes in memory order: 10K, 4K, 20K, 18K, 7K, 9K, 12K, and 15K. Which hole is taken for successive segment requests of (a) 12K, (b) 10K and (c) 9K for Next-Fit? Assume the last allocated hole is 20K.Consider a virtual memory system that can address a total of 32 bytes. You have unlimited hard disk space, but are limited to only 16MB of semiconductor (physical) memory. Assume that virtual and physical pages are each 4 KB in size. What is the total size of the page table in bytes? (Assume that, in addition to the physical page number, each page table entry also contains some status information in the form of a valid bit (V) and a dirty bit (D)).
- Let's pretend for a moment that we have a byte-addressable computer with fully associative mapping, 16-bit main memory addresses, and 32 blocks of cache memory. The following holds true if each block is 16 bits in size:a) Determine how many bytes the offset field is.Measure the tag field's width and height in pixels (b).1. Suppose 8 bit registers have following contentsX=00001111Y=10101010Z= 11011011W=00110011What will be the 8 bit values of each register after execution of following sequences ofmicrooperations ?X ← ? + ?Z←Z⋀ ?, ? ← ? + 1X←X-ZSuppose we have a byte-addressable computer using direct mapping with 16-bit main memory addresses and 32 blocks of cache. If each block contains 16 bytes. a. Determine the number of bits of the offset field. b. Determine the number of bits of the block (or slot) field. c. Determine the number of bits of the tag field. d. To which cache block would the hexadecimal address 0x2468 map? e. What is the tag of the hexadecimal address 0x2468 f.To which cache block would the hexadecimal address 0x864A map? g. What is the tag of the hexadecimal address 0x864A?
- Consider a swapping system in which memory consists of the following hole sizes inmemory order: 12 MB, 4 MB, 20 MB, 12 MB,18 MB, 7 MB, 9 MB and 12 MB.Which hole is taken for successive segment requests of(a) 10MB(b) 15MB(c) 9 MBfor first fit, worst fit, and next fit?Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. 8 bits are used for offset, 8 bits are used for page # and the max number of pages a process can have is 256. e. Translate the following virtual addresses to physical addresses, and show how you obtain the answers. (Hint: You do not need to convert hexadecimal numbers to decimal ones.) 0x0389 0xDF78 0x0245 0x8012 f) Now, suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) g)What is the size of the two-level page table…Suppose we have a system with the following properties:The memory is byte addressable.Memory accesses are to 1-byte words (not to 4-byte words).Addresses are 13 bits wide.The cache is 4-way set associative (E = 4), with a 4-byte block size(B = 4) and eight sets (S = 8).Consider the following cache state. All addresses, tags, and valuesare given in hexadecimal format. The Index column contains the set index for each set of four lines. The Tag columns contain the tag value for each line. The V columns contain the valid bit for each line. The Bytes 0−3 columns contain the data for each line, numbered left to right starting with byte 0 on the left. A. What is the size (C) of this cache in bytes?B. The box that follows shows the format of an address (1 bit perbox). Indicate (by labeling the diagram) the fields that would beused to determine the following:CO. The cache block offsetCI. The cache set indexCT. The cache tag
- (Practice) Although the total number of bytes varies from computer to computer, memory sizes of millions and billions of bytes are common. In computer language, the letter M representsthe number 1,048,576, which is 2 raised to the 20th power, and G represents 1,073,741,824, which is 2 raised to the 30th power. Therefore, a memory size of 4 MB is really 4 times 1,048,576 (4,194,304 bytes), and a memory size of 2 GB is really 2 times 1,073,741,824 (2,147,483,648 bytes). Using this information, calculate the actual number of bytes in the following: a. A memory containing 512 MB b. A memory consisting of 512 MB words, where each word consists of 2 bytes c. A memory consisting of 512 MB words, where each word consists of 4 bytes d. A thumb drive that specifies 2 GB e. A disk that specifies 4 GB f. A disk that specifies 8 GBCA_10 Let the virtual address be V bits and the virtual addtess space be byte-addressable, the page size be P KB (and P is a power of 2), and the the main memory size be MM MB(where [MM MB]) is divide into [P KB]). (d)How many of the virtual memory bits need to be translated? (e) How many bits will be produced if the virtual-to-pyysical address translation is "successful" (f) How many bits does a physical address have, and how are each of these bits obtained?Fill in blank Suppose that linear page table is used where the memory addresses are 12-bit binary numbers and the page size is 256 bytes. If a virtual address in binary format is 101000011100, then the VPN (virtual page number) in binary format will be ---------