a. Can Job 4 be accommodated? Why or why not? b. If relocation is used, what are the contents of the relocation registers for Job 1, Job 2, and Job 3 after compaction? c. What are the contents of the relocation register for Job 4 after it has been loaded into memory?
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a. Can Job 4 be accommodated? Why or why not?
b. If relocation is used, what are the contents of the relocation registers for Job 1, Job 2, and Job 3 after compaction?
c. What are the contents of the relocation register for Job 4 after it has been loaded into memory?
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- Carefully read each sentence in this question. You may agree/disagree with each sentence (consider each sentence separately) if you have a good justification for doing so (not in more than 2 lines for each sentence). “A daisy chaining technique could result in starvation.” “The received data is saved in contiguous memory regions in DMA transfer.” “The most difficult method of device identification is multiple interrupt lines.” “An interrupt request takes longer for the CPU to handle than a DMA request.” “The architecture of InfiniBand operation is layered.”Consider the free memory layout shown below (the blocks are ordered from top to bottom). To the right, draw to the what the free memory layouts would be after satisfying requests for (in order) memory blocks of size 55K, 210K, 381K, and 395K. Draw one memory layout for the best-fit algorithm, and one for first-fit. 500K 300K 200K 300KIn a system where a single-level page table is maintained and no other support mechanism exists, exactly two memory accesses will be required to complete the process of adding a constant number to a value in a register of the CPU. (a) True (b)False
- Consider a CPU which operates with 20Mbyte/s operating speed. The CPU is operating on program control mode of I/O and it has to transfer data of 20 bytes from it. The data is transferred byte wise. Size of the Status register is 2 bytes. What is the total time needed to perform the data transfer?Consider a word-addressable computer with 32 bits per word. The instruction set consists of 30 different All instructions have an operation code field, a mode field to specify one of 7 possible addressing modes, a register address field to specify one of 60 available registers, and a memory address field. Each instruction is stored in one word of memory. What is the maximum allowable size for memory? Use KB, MB, or GB as a unit.You are given the following data about a virtual memory system:(a) The TLB can hold 512 entries and can be accessed in 1 clock cycle (1nsec).(b) A page table entry can be found in 100 clock cycles or 100 nsec.(c) The average page replacement time is 9 msec.If page references are handled by the TLB 99% of the time, and only 0.01%lead to a page fault, what is the effective address-translation time?
- A 64KB memory system is organized with 8-way interleaving. Match each of the following memory addresses (in Hex) with the module number and the address within the module. 2468 A000 1234 140A 1. 2. 3. 4. module=4, address within module=246 module=0, address within module=48D module=0, address within module=1400 module=2, address within module=281A given computer has a single cache memory (off-chip) with a 2 ns hit time and a 98% hitrate. Its main memory has 40 ns access time. What is the computer’s effective access time? If an on-chip cache with a 0.5 ns hit time and a 94% hit rate is added to it, what isthe computer’s new effective access time? How much of a speedup does the on-chip cache give the computer?Consider a computer with cache, DRAM, HDD memory hierarchy. The hit rate of cache is 90% and DRAM is 95%. Read latencies of cache, DRAM, and HDD are 5ns, 100ns, and 1ms respectively. What is the average latency of executing an instruction involving a memory read? Express your answer in micro-seconds. Round it to the nearest integer. Enter your answer here
- According to the memory view given below, if RO = 0x20008000, then LDRSB r1, r1 = ?(data overlay big endian)? Memory address Data Øx20008002 ØXA1 Øx20008001 ØXB2 Øx20008000 ØxC3 ØX20007FFE ØXD4 ØX20007FFE OXE5 (Ctrl) A-R1 = 0XC3 B-R1 = 0x000000C3 C-R1 = OXC3000000 D-R1 = 0xffffffC3 E-R1 = OxC3ffffffThe microinstructions stored in the control memory of a processor have a width of 26 bits. Each microinstruction is divided into three fields: a micro-operation field of 13 bits, a next address field (X), and a MUX select field (Y). There are 8 status bits in the inputs of the MUX.Calculate the number of bits in the X and Y fields. Also find the size of the control memory in number of words?Suppose a specific MCU has the following size of memories: 2 M byte of flash, starting from 0x0800_0000, 256 k byte of SRAM starting from 0x2000_0000, and 8 k byte registers for GPIOs, start at 0x4001_0000. (Note that 0x is the prefix for hexadecimal numbers.) Draw the memory map based on your calculations for the addresses.