Computer systems

This page describes and explains how we mapped raw points to grades out of 6.0.

First, we picked the number of raw points that corresponds to a 4, a 5, and a 6:

• 46,00 --> 4
• 67,00 --> 5
• 90,00 --> 6

In between these points we used linear scales:

Suppose a student received X raw points. We mapped X to a grade out of 6.0, Y, as follows:

• If X<46, Y = (X/46)*3 + 1.
• Else if X<67, Y = (X-46)/(67-46) + 4.
• Else if X<90, Y = (X-67)/(90-67) + 5.
• Else, Y = 6.

To pick the number of raw points mapping to 4, 5, and 6, we considered the content of the exam, and what we think a student should know in order to achieve the corresponding grade; and we also took into account that the exam turned out to be longer than intended.

For instance, consider Question 1:

• To achieve a 4, a student should be able to answer the first three sub-questions, plus at least half of sub-question 4. So, a total of 10/20 points. This reflects basic understanding of a process, a thread, the IP register, the text segment, CPU user and kernel mode, and syscalls, plus at least partial understanding of the relationship between files, blocks, and inodes.
• To achieve a 5, a student should be able to completely answer sub-question 4, which reflects good understanding of the relationship between files, blocks, and inodes. So, a total of 15/20 points.
• To achieve a 6, a student should be able to also answer sub-question 5, which reflects good understanding (of at least one aspect) of file-system operation and where the limit on file-system size comes from. So, a total of 20/20 points.

We did this for every Question, then adjusted for the duration of the exam. This is how we picked the transition points -- 46, 67, and 90.