please write code of this program in java language, without do the part with have line and color is red

Address file

16916 62493 30198 53683 40185 28781 24462 48399 64815 18295 12218 22760 57982 27966 54894 38929 32865 64243 2315 64454 55041 18633 14557 61006 62615 7591 64747 6727 32315 60645 6308 45688 969 40891 49294 41118 21395 6091 32541 17665 3784 28718 59240 40178 60086 42252 44770 22514 3067 15757 31649 10842 43765 33405 44954 56657 5003 50227 19358 36529 10392 58882 5129 58554 58584 27444 58982 51476 6796 21311 30705 28964 41003 20259 57857 63258 36374 692 43121 48128 34561 49213 36922 59162 50552 17866 18145 3884 54388 42932 46919 58892 8620 38336 64357 23387 42632 15913 15679 22501 37540 5527 63921 62716 32874 64390 63101 61802 19648 29031 44981 28092 9448 44744 61496 31453 60746 12199 62255 21793 26544 14964 41462 56089 52038 47982 59484 50924 6942 34998 27069 51926 60645 43181 10559 4664 28578 59516 38912 63562 64846 62938 27194 28804 61703 10998 6596 37721 43430 22692 62971 47125 52521 34646 32889 13055 65416 62869 57314 12659 14052 32956 49273 50352 49737 15555 47475 15328 34621 51365 32820 48855 12224 2035 60539 14595 13853 24143 15216 8113 22640 32978 39151 19520 58141 63959 53040 55842 585 51229 64181 54879 28210 10268 15395 12884 2149 53483 59606 14981 36672 23197 36518 13361 19810 25955 62678 26021 29409 38111 58573 56840 41306 54426 3617 50652 41452 20241 31723 53747 28550 23402 21205 56181 57470 39933 34964 24781 41747 62564 58461 20858 49301 40572 23840 35278 62905 56650 11149 38920 23430 57592 3080 6677 50704 51883 62799 20188 1245 12220 17602 28609 42694 29826 13827 27336 53343 11533 41713 33890 4894 57599 3870 58622 29780 62553 2303 51915 6251 38107 59325 61295 26699 51188 59519 7345 20325 39633 1562 7580 8170 62256 35823 27790 13191 9772 7477 44455 59546 49347 36539 12453 49640 28290 44817 8565 16399 41934 45457 33856 19498 17661 63829 42034 28928 30711 8800 52335 38775 52704 24380 19602 57998 2919 8362 17884 45737 47894 59667 10385 52782 64416 40946 16778 27159 24324 32450 9108 65305 19575 11117 65170 58013 61676 63510 17458 54675 1713 55105 65321 45278 26256 64198 29441 1928 39425 32000 28549 46295 22772 58228 63525 32602 46195 55849 46454 7487 33879 42004 8599 18641 49015 26830 34754 14668 38362 38791 4171 45975 14623 62393 64658 10963 9058 51031 32425 45483 44611 63664 54920 7663 56480 1489 28438 65449 12441 58530 63570 26251 15972 35826 5491 54253 49655 5868 20163 51079 21398 32756 64196 43218 21583 25086 45515 12893 22914 58969 20094 13730 44059 28931 13533 33134 28483 1220 38174 53502 43328 4970 8090 2661 53903 11025 26627 18117 14505 61528 20423 26962 36392 11365 50882 41668 30497 36216 5619 36983 59557 36663 36436 37057 23585 58791 46666 64475 21615 41090 1771 47513 39338 1390 38772 58149 7196 9123 7491 62616 15436 17491 53656 26449 34935 19864 51388 15155 64775 47969 16315 1342 51185 6043 21398 3273 9370 35463 28205 2351 28999 47699 46870 22311 22124 22427 49344 23224 5514 20504 376 2014 38700 13098 62435 48046 63464 12798 51178 8627 27083 47198 44021 32792 43996 41126 64244 37047 60281 52904 7768 55359 3230 44813 4116 65222 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1. Designing and simulating a contiguous region of memory Design and implement a program that simulates some of managing a contiguous region of memory of size MAX where addresses may range from 0...MAX1. Your program must respond to four different requests: a. Request for a contiguous block of memory. b. Release of a contiguous block of memory. c. Compact unused holes of memory into one single block. d. Report the regions of free and allocated memory. 2. Allocating Memory Your program will allocate memory using one of the three approaches: first fit, best fit, and worst fit depending on the flag that is passed to the RQ command. The flags are: - F-first fit - B-best fit - W-worst fit This will require that your program keep track of the different holes representing available memory. When a request for memory arrives, it will allocate the memory from one of the available holes based on the allocation strategy. If there is insufficient memory to allocate to a request, it will output an error message and reject the request. Your program will also need to keep track of which region of memory has been allocated to which process. This is necessary to support the STAT command and is also needed when memory is released via the RL command, as the process releasing memory is passed to this command. If a partition being released is adjacent to an existing hole, be sure to combine the two holes into a single hole. 3. Compaction If the user enters the C command, your program will compact the set of holes into one larger hole. For example, if you have four separate holes of size 550KB,375KB,1,900KB, and 4,500 KB, your program will combine these four holes into one large hole of size 7,325KB. 4. Implementation hints (Pat I) Your program will be passed the initial amount of memory at startup. For example, the following initializes the program with 1MB(1,048,576 bytes) of memory: ./allocator 1048576 Once your program has started, it will present the user with the following prompt: allocator > It will then respond to the following commands: RQ (request), RL (release), C (compact), STAT (status report), and X (exit). A request for 40,000 bytes will appear as follows: allocator>RQP040000W The first parameter to the RQ command is the new process that requires the memory, followed by the amount of memory being requested, and finally the strategy. (In this situation, "W" refers to worst fit.) Similarly, a release will appear as: allocator > RL P0 This command will release the memory that has been allocated to process P0. The command for compaction is entered as: allocator >C This command will compact unused holes of memory into one region. Finally, the STAT command for reporting the status of memory is entered as: allocator > STAT Given this command, your program will report the regions of memory that are allocated and the regions that are unused. For example, one possible arrangement of memory allocation would be as follows: Addresses [0:315000] Process P1 Addresses [315001: 512500] Process P3 Addresses [512501:625575] Unused Addresses [625575:725100] Process P6 Addresses [725001] ... 5. Designing a Virtual Memory Manager This project consists of writing a program that translates logical to physical addresses for a virtual address space of size 216=65,536 bytes. Your program will read from a file containing logical addresses and, using a TLB and a page table, will translate each logical address to its corresponding physical address and output the value of the byte stored at the translated physical address. Your learning goal is to use simulation to understand the steps involved in translating logical to physical addresses. This will include resolving page faults using demand paging, managing a TLB, and implementing a page-replacement algorithm. 6. Specifications Your program will read a file containing several 32-bit integer numbers that represent logis addresses. However, you need only be concerned with 16-bit addresses, so you must mask t rightmost 16 bits of each logical address. These 16 bits are divided into (1) an 8-bit page number and (2) an 8-bit page offset. Hence, the addresses are structured as shown as: Other specifics include the following: - 28 entries in the page table - Page size of 28 bytes - Frame size of 28 bytes - 256 frames - Physical memory of 65,536 bytes ( 256 frames 256-byte frame size) Additionally, your program need only be concerned with reading logical addresses and translating them to their corresponding physical addresses. You do not need to support writing to the logical address space. 7. Address Translation Your program will translate logical to physical addresses tsing a TLB and page table as outlined in Section 9.3. First, the page number is extracted from the logical address, and the TLB is consulted. In the case of a TLB hit, the frame number is obtained from the TLB. In the case of a TLB miss, the page table must be consulted. In the latter ease, either the frame number is obtained from the page table, or a page fault occurs. A visual representation of the address-translation process is: 9. Test File Create the file addresses.txt, which contains integer values representing logical addresses ranging from 0 to 65535 (the size of the virtual address space). Your program will open this file, read each logical address and translate it to its corresponding physical address, and output the value of the signed byte at the physical address. 10.Implementation Hints (Part II) a. How to Begin Extracting page number and offset from the logical address: First, write a simple program that extracts the page number and offset based on: from the following integer numbers: 1,256,32768,32769,128,65534,33153 Perhaps the easiest way to do this is by using the operators for bit-masking and bit-shifting. Once you can correctly establish the page number and offset from an integer number, you are ready to begin. Initially, we suggest that you bypass the TLB and use only a page table. You ean integrate the TLB ence your page table is werking properly. Remember, address transtation ean work without a TLB; the TLB just makes it faster. When you are ready to implement the TLB, reeall that it has only sixteen entries, so you will need to use a replacement strategy when you update a full TLB. You may use either a FIFO or an LRU poliey for updating your TLB. Populating the physical memory and page table: Read 100 logical addresses from addresses.txt file one-by-one. For each logical address do the following: 1. Make sure that the page number is not in the Page Table (if it is in the page table use it and move on to step 3:computing the physical address) 2. Give a random number as a frame number (make sure it is with the range of frame numbers) 3. Compute the physical address, make sure it is not already allocated. If it is previously allocated find a new random number as a frame number, and so on PS. Since the logical addresses in the file are unique, then no need to make this check for page numbers that are already in the Page Table. 4. Add the new page number \& frame number to the page table, add the value of the corresponding signed byte to the new location in the physical memory created by you. b. How to Run Your Program Your program should run as follows: ./a.out addresses.txt Your program will read in the file addresses.txt, which contains 1,000 logical addresses ranging from 0 to 65535 . Your program is to translate each logical address to a physical address and determine the contents of the signed byte stored at the correct physical address. (Recall that in the C language, the char data type occupies a byte of storage, so we suggest using char values.) Requirement: make FIVE test cases, in each test case run the "address translation" command and provide the following output: a. The logical address being translated (the integer value being read from addresses.txt). b. The corresponding physical address (what your program translates the logical address to). c. The signed byte value stored in physical memory at the translated physical address. d. The signed byte value that corresponds to the logical address with the correct.txt file. e. Display a message stating that the test is successful when the two values generated in c and d matches, or there is an error other wise. We also provide the file correct.txt, which contains the correct output values for the file addresses.txt. You should use this file to determine if your program is correctly translating logical to physical addresses. PS. To avoid sequential accesses within the files, read first 30 values in an array or any other kind of data structure for fast access. Then run the test cases by RANDOMELY choosing an address from this list. 11. Statistics Run the following test case: - Choose a logical address string consisting of 50 logical addresses that are chosen from the list 100 logical addresses that were previously selected and 30 logical addresses that are different from these 100 addresses. The 80 logical addresses should not be divided into two separate groups within and without the original list, both types of addresses should be intermixed. - Run the page number extraction function developed previously on the 80 addresses: - If the page number is found in the page table output the corresponding unsigned byte value. - If not output page not in memory and increment the page fault counter. - At the end, output the length of the addresses string and the Page-fault rate-The percentage of address references that resulted in page faults. 12.Page Replacement (BONUS 3 points) Thus far, this project has assumed that physical memory is the same size as the virtual address space. In practice, physical memory is typically much smaller than a virtual address space. This phase of the project now assumes using a smaller physical address space with 128 page frames rather than 256 . This change will require modifying your program so that it keeps track of free page frames as well as implementing a page-replacement policy using either FIFO or LRU to resolve page faults when there is no free memory