Has anyone's project been remarked? I emailed the professor on the 10th, and it seems that there are still no updates. I understand that there is a large amount of exams to mark, just worried if this will be looked at before the marks release on the 2...

Same here, we got 2 10 and checked that our code passed all the test cases on help forum - and still waiting for reply from Amitava

I m in a similar boat and would love some clarification on it )

Hi everyone, I m really frustrated with the grading of my recent project submission, where I received a 0 10. I just verified if my submission was proper with the appropriate names and student IDs mentioned in the code and despite the code passing m...

Even I have the same issue. We got a zero score.

ANONYMOUS wrote Hi Anon, I would recommend sending an email to Amitava as soon as possible, it might have been a mistake, especially if your project met the requirements. It seems very odd to get 0, usually that's a grade reserved for submitting nothi...

Hello everyone, I'm reaching out to seek advice and support regarding the recent grading of our Project 2. Our team received a zero score, which we find difficult to accept given the time, effort, and energy we invested. We carefully rechecked our cod...

Hi, I was just wondering when the Project 2 marks would be released.

Pretty sure this is not correct. The primary page table size is 2 8 256, where each of these 256 entries points to a secondary page table. Therefore there are 2 8 256 secondary page tables, one for each index of the primary page table. Each of these ...

meant to say the amount of frames is determined by the maximum frame number which is everything left of the offset. so in a 32 bit computer if 10 were for offset then it means that there are up to 2 22 pages and if each of those pages are 2 10 bytes ...

an off set of 10 bits means the page size is 2 10 Bytes which is 1 KB. the amount of pages is determined by the maximum page number. Also, each digit in a memory address represents a byte so a 32 bit computer would have 2 32 unique addresses which is...

Thank you for your reply, I'd like to ask why there'd be 2 14 secondary page tables instead of the size of secondary page table being 2 14. As I understand, primary page table will hold the address to a single secondary page table. Perhaps I'm mistak...

Sorry for the confusion, when the offset is 10 bits, does it mean there are 2 10 bytes, or 2 10 32-bit addresses?

Dear All, I will sort out all problems related to project 1 marks on Monday, as I had an eye operation and was on leave yesterday and today. The exam paper is fairly easy, and I am sure you will do well. all the best, Amitava

ANONYMOUS wrote good answer, thanks

No, we haven't started marking the second project.

ANONYMOUS wrote It needs to be consistent so that it is not misused by malicious processes.Complicated interfaces may be easier to manipulate to pass malicious parameters. Also parameter checking is more efficient if they are consistent.

ANONYMOUS wrote Each process starts with three open file descriptors, stdin, stdout, stderr - these will require system calls to open.

ANONYMOUS wrote page size is 2 10 primary page table size is 2 8 There are 2 14 secondary page tables, each entry of the primary page table holds the starting address of a secondary page table. One can bring the secondary page tables on demand (the sec...

ANONYMOUS wrote Sort of, but you can ignore that question. I don't know what Chris meant, may be segmented paging (I haven't explained that).