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$(0.5)_{10}$ in IEEE 754 Single precision floating point representation $(0.5)_{10}$ = $(0.1)_{2}$ here $(0.1)_{2}$ should be represented in SUBNORMAL form Or Normalized for ..??? here $(0.1)_{2}$ * 2^{0}$...E = 0 and no leading 1 ..will it be represented in SUBNORMAL form ?? 0 votes 2 How can we convert a signed number to it's equivalent unsigned number? Let's say I have signed number as -2 now what would be it's equivalent unsigned representation. Thanks! 0 votes 3 What is the reason for Belady’s Anomaly,I am aware that it is not a stack based algorithm and for a certain set of pages it shows this anomaly where the increase in page frame increases the page fault rate. 0 votes 4 Q. Suppose you have a computer system with a 48-bit logical address, page size of 16KB and 4 bytes per page table entry. If we have a 48MB program such that the entire program and all necessary page tables are in memory. Assume that each page table at diff level fits in a single page.How much memory is used by program, including its page tables? 0 votes 5 Please Explain. 0 votes 6 Whenever Questions like:- (Example is shown below) is given and asked that say what is the turnaround time for$6k$Process then we will always assume memory to have fixed length partition right??? cuz they have given different partition sizes in$KB$so say if a process allocates$2k$... 0 votes 7 Explain Last In First Out (LIFO) and MRU with this example and find a number of a page fault. Question: Let the number of the frame in main memory is 5. Reference String: 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 0 votes 8 Given 64 bit logical space, Page Size =$4KB$. Do the three level paging? My Solution:- 1. Since address space is =$ 2^{64} Bytes$Page size =$2^{12}$Bytes. Number of entries in third level Page table is :-$2^{64} / 2^{12}$=$2^{52}$. Size of third level ... three levels as:-$1st$level$2nd$level$3rd$level Offset$ 32101012$Please explain how did they find this out ? Please :) 1 vote 9 What is the difference between Concurrency and Parallelism? 0 votes 10 Consider a uniprocessor system with four processes having the following arrival and burst times: Calculate the average waiting time and also the average turnaround time if shortest (remaining) job first (SJF) scheduling policy is used with pre-emption. Assume that the context switching ... burst times of$2$units each. In this case, what will be the turnaround time of$P1\$? Justify your answer
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Round robin will always result in better output compared to FCFS. Is this statement TRUE? I know statement is FALSE when time quantum of RR is greater than longest CPU burst time of all the process, so isn't its answer will be FALSE because it talking of "always".