Break the class C address 216.21.5.0/24 into five (5) subnets.
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- A machine has a memory of 64 frames, with each frame being 1K bytes. Current free-frame list is: O×2E, Ox27, Ox37, Ox25, OX0C, Ox04, OXOE, Ox09, O×1D, Ox14, Ox16, Ox07, 0x22, 0×3E, and Ox30. You just scheduled a process that requires 10 frames. Show the resulting page table. Show the translation of logical address Ox240B and OX5A32 into physical addresses using your page table. Express your result in hex.Consider the following code: #pragma omp parallel for for(int i = 1: i <= 18; i++) a[i] = i Rewrite the code so the values of i assigned to each thread are the following: P1: 1, 2, 3, 10, 11. 12 P2: 4, 5, 6, 13, 14, 15 P3: 7. 8, 9, 16. 17. 18 Use the editor to format your answerAssume that a large number of consecutive IP addresses are available starting at 198.16.32.0 and suppose that two organizations, A, B, C, and D, request 1000, 3000, 2000, and 4500 addresses, respectively, in that order. For each of these, give the first possible IP address within the assigned subnet, the last possible IP address within the assigned subnet, and the subnet representation in slash notation (w.x.y.z/s). In the imagine I have the answers but I am very confused on how to get them. Specially after A and B I do not undertand how the initial address of C is 198.16.64.0 how did that 64 happen??? Please explain all in detail. Thank you!
- 4. The address 192.192.0.255 /24 is considered: a) Private Address b) Last Host Address c) First Host Address d) Broadcast Address10.7 - What are the values in cells C5, C7, C10, and C13 in the output layer? Con(Layer1 < 2, 0, 1) Layer1 2 |N 1| 2 1 N1 3 4 1 2 |0 Output 2 N C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 | C11 |C12 С13 | С14 С15 | С16Write a C program Producer – Consumer as a classical problem of synchronizationStep 2. Write a program* that solves the producer - consumer problem. You may use the following pseudo code for implementation.*program to write: produce -consumer problem to produce and consume the alphabet.//Shared data: semaphore full, empty, mutex;//pool of n buffers, each can hold one item//mutex provides mutual exclusion to the buffer pool//empty and full count the number of empty and full buffers//Initially: full = 0, empty = n, mutex = 1//Producer thread do {…produce next item…wait(empty); wait(mutex);…add the item to buffer…signal(mutex); signal(full);} while (1);//Consumer thread do {wait(full) wait(mutex);…remove next item from buffer…signal(mutex); signal(empty);…consume the item } while (1);
- Given the information above: Give the address of each of the following variables assuming the base address of V is 20. Eg: V[0].x -> 20 + 0 + 0 = 20 V[5].y V[0].anotherShort V[5].aUnion.secondPart.next V[5].y = [ Choose ] V[0].anotherShort = [ Choose ] [Choose ] 97 V[5].aUnion.secondPart.next = 101 14 34 81 102 12 110 80 90 Use this Prolog program for the 92 >Breakdown each part of the following rules and tell what each segment does iptables -t filter -A INPUT -s 221.98.146.40 -j DROP i. What would happen here if we omitted the ‘-t filter’option? iptables -A INPUT -p tcp --dport ssh -j DROP c. iptables -A INPUT -p udp --dport ftp -j REJECT d. iptables -A INPUT -p tcp --dport 443 -j ACCEPT e. iptables -A INPUT -m iprange --src-range 192.168.40.25-192.168.40.35 -j DROPIKT204-G H21, Routing tables 13 In this assignment, the objective is to determine the correct forwarding link given the routing table below. A router has the following entries in its forwarding table: Link1: 00001010.10101000.00000100.00000000/22 Link2: 00001010.10101000.00000110.00000000/23 Link3: 00001010.10101000.00000111.00000000/24 Link4: 00001010.10101000.00000000.00000000/16 Link5: All other addresses Assume the router receives IPV4 datagrams destined to the following addresses and decide which link they are forwarded to: A: 00001010.11111000.00000101.00001111 B: 00001010.10101000.00000111.00011000 C: 00001010.10101000.00000011.01010000 D: 00001010.10101000.00000111.11111110 E: 00001010.10101000.00000110.10000010 On which link will they be forwarded? A: link B: link C: link D: link E: link
- Consider the following runtime stack: BEFORE 0000100 0 00000006 ESP 00000FFC 00000FF 8 00000FF 4 00000FF 0 What would be the value of ESP after pushing the 32-bit value shown below onto the stack? 000000A5 000000A5 00000FF8 00001020 00000FFC7.12. Consider a simple paging system with the following parameters: 232 bytes of physical memory; page size of 210 bytes; 216 pages of logical address space. a. How many bits are in a logical address? b. How many bytes in a frame? c. How many bits in the physical address specify the frame? d. How many entries in the page table? e. How many bits in each page table entry? Assume each page table entry contains a valid/invalid bit.An ISP is granted a block of addresses starting with 130.50.4.0/22. The ISP wants to distribute these blocks to 100 organizations with each organization receiving just eight addresses. Design the subblocks and give the slash notation for each subblock. Find out how many addresses are still available after these allocations.
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