Fill in multiple blanks. Consider the snapshot of this memory starting from the top cell at Address 1E1AH. 10011000 00010101 11011010 10100000 |11000111 01010101 When performing a read of the 4th cell from the top (the first cell is at Address 1E1AH):
Q: the AMAT (in number of clock pulses)?
A: The AMAT (in number of clock pulses)
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- Write an MPI program segment for all-to-all personalized broadcast on a ring of p processors, eachprocessor Pi, 0 ≤ i ≤ p − 1, holding M(i, 0), M(i, 1), ·M(i,(p − 1)). Message M(i, j), an integer, is destinedfrom Pi to Pj . Show only the p − 1 iterative loop. Ensure proper buffer size, MPI message size, howitems are organized and ordered at the destination buffer, etc., through each loop.Now, we consider a 16-byte, four-way, fully-associative cache. Since the capacity of the cache is 16 bytes, the array "a" in our example (does/does not) fit inside the cache. We can deduce that the block size for this cache is ( ? ) bytes per block. So the block index size b=2 bits. For a memory trace record such as: L 1fff000116,2 the 2-bit block offset is (0b01/0b10/0b11/0b00). The tag bits are all the rest of the bits not part of the block offset.Consider the following hexadecimal readout: 000000 8A00 8E00 CFA1 48BF 7900 3202 9015 AD34 000010 0218 6D30 028D 3402 AD35 0288 3102 8D35 000020 0E30 0290 DAEE 3102 4C00 0200 0040 004B Refer to the first byte of memory shown above, address 000000. Assume that this byte is used to store an 8-bit unsigned integer. What is the decimal value stored in this byte? Group of answer choices 138 -27 22,842 66
- 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).Suppose a disk has N cylinders, numbered from 0 to P. At some time, the disk arm is at cylinder X, and there is a queue of disk access requests for cylinders R1, R2, R3, R4. Explain which Disk scheduling algorithm should be the best to be used between the following a) FIFO b) CSCAN c) SSTF d) FSCAN e) CLOOK N: B • • Choose any acceptable numeric numbers for N, P, X, R1, R2, R3 and R4 • • Plagiarism Will be penalized heavilySolve the 8085 Write a program to load twenty memory locations starting from 8005H, where each location's content should increases by 2 over the previous one, however, the first location should contain 04H, assuming that the programs start at memory location 9009H?
- What are the contents of the 25 memory bytes starting at address A, in hex, on a machine that uses Little Endian? (Hint: “a”=061h) Consider the following .data segment: A dw 0AAFFh B db 051h, 0CCh, 0EEh C times 3 dw -23 D dd -177 E db "e", -5, "c", 0 F times 2 dw -17 G dw 0EEhInput file sample.txt contains a hex dump of some data in the following format:<address> <byte1> <byte2> ... <byte16>Example Input File:00000000 54 68 69 73 20 69 73 20 61 6e 20 65 78 61 6d00000010 70 6c 65 20 6f 66 20 68 65 78 20 64 75 6d 70...Construct a pipeline using Ubuntu bash to discard the first column (address) and to reformat bytesinto a single column:Suppose the Vole communicates with a printer using the technique of memory-mapped I/O. Suppose also thataddress 0xFF is used to send characters to the printer, and address 0xFE is used to receive information about theprinter’s status. In particular, suppose the least significant bit at the address 0xFE indicates whether the printeris ready to receive another character (with a 0 indicating “not ready” and a 1 indicating “ready”). Starting ataddress 0x00, write a machine language routine that waits until the printer is ready for another character and thensends the character represented by the bit pattern in register 0x5 to the printer.
- (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 GBA computer uses virtual memory, and a new solid-state drive (SSD) as space for paging. Refer to the last ppt file. In the case presented there, the hard disk drive (HDD) required 25 ms to read in a page, and a rate of 1 page fault per 1000 references introduced a 250 slowdown. If the SSD offers a time of only 80 µs, what is the slowdown in performance caused by 1 pf per 1000 references (you are not concerned with dirty vs. clean pages). What is the maximum rate of page faults you can accept if you want no more than a 5% slowdown in execution using virtual memory? Know your metric prefixes and symbols for time: s for seconds, ms for milliseconds, µs for microseconds, ns for nanoseconds.What the code below does? Consider the following code used to implement a new instruction: foo $t3,$t1,$t2: mask : . word 0xFFFFF83Fs t a r t : l a $t0 , masklw $t0 , 0 ( $ t 0 )l a $t3 , s h f t rlw $t3 , 0 ( $ t 3 )and $t3 , $t3 , $ t 0a ndi $t2 , $t2 , 0 x 0 0 1 fs l l $t2 , $t2 , 6o r $t3 , $t3 , $ t 2l a $t5 , s h f t rsw $t3 , 0 ( $ t 5 )s h f t r : s l l $t3 , $t1 , 0 What does foo does?