A significant problem with threshold marking are students who are close to, but below, the threshold. For example, students who achieve a grade of less than 40 may not have laboratory marks counted into their final grade.

Projects/Labs Not Counted into Grade with Poor Examination Result

The following is an alternative which has similar properties to threshold marking but where a change of grade on the final of 1% will affect their overall average by at most 2-3%.

Let F, M, and PL be the student's mark (out of 100) on the final, midterm, and projects/labs, respectively,
Let f, m, and pl be the normal proportions of the grade for final, midterm, and projects/labs, respectively, where f + m + pl = 1,
Select a lower bound L such that if F ≤ L then the projects/labs do not count,
Select an upper bound U such that if F ≥ U then the projects/labs count for the full weight of pl, and
For L < F < U, the weight of the projects/labs increases in a linear manner.

Then the formula for the final grade when L < F < U is:

(f·F + m·M)·(U − F)/(U − L)/(f + m) + (f·F + m·M + pl·PL)·(F − L)/(U − L)

Example 1

For example, in ECE 250, f = 0.5, m = pl = 0.25, L = 40, and U = 60. In this case, the formula evaluates to:

If F ≥ 60, use (0.5·F + 0.25·M + 0.25·PL),
If F ≤ 40, use (0.5·F + 0.25·M)/0.75, and
Otherwise, use (0.5·F + 0.25·M)·(60 − F)/15 + (0.5·F + 0.25·M + 0.25·PL)·(F − 40)/20.

Example 2

Another example could be f = 0.5, m = 0.2, pl = 0.3, L = 40, and U = 50, in which case, the formula evaluates to:

If F ≥ 50, use (0.5·F + 0.2·M + 0.3·PL),
If F ≤ 40, use (0.5·F + 0.2·M)/0.7, and
Otherwise, use (0.5·F + 0.2·M)·(50 − F)/7 + (0.5·F + 0.2·M + 0.3·PL)·(F − 40)/10.

Example 3

Finally, another example could be where the projects/labs do not count unless the student achieves F = 50 on the final. Let us assume that f = 0.5, m = 0.25, pl = 0.25, L = 50, and U = 60, in which case, the formula evaluates to:

If F ≥ 60, use (0.5·F + 0.25·M + 0.25·PL),
If F ≤ 50, use (0.5·F + 0.25·M)/0.75, and
Otherwise, use (0.5·F + 0.25·M)·(60 − F)/7.5 + (0.5·F + 0.25·M + 0.25·PL)·(F − 50)/10.

Full Weight on Final with Poor Final Result

An alternate scheme would be to require that the final examination is 100% of the grade if F ≤ L and the mid-term and projects/laboratories count for their full weight only if F ≥ U. In this case, the formula is:

F·(U − F)/(U − L) + (f·F + m·M + pl·PL)·(F − L)/(U − L)

In this case, the formulae for the intermediate grading in the above examples would change to

F·(60 − F)/20 + (0.5·F + 0.25·M + 0.25·PL)·(F − 40)/20,
F·(50 − F)/10 + (0.5·F + 0.2·M + 0.3·PL)·(F − 40)/10, and
F·(60 − F)/10 + (0.5·F + 0.25·M + 0.25·PL)·(F − 50)/10,

respectively.

The ECE 250 web site documents this nicely with the image in Figure 1.

Figure 1. Pictorial representation of the ECE 250 grade.

Maple

The following is Maple code which calculates the formulae for Example 1.

L := 40:
U := 60:
f := 0.5;
m := 0.25;
pl := 0.25;
# Midterm and Final for F < L
(f*F + m*M)*(U - F)/(U - L)/(f + m) + (f*F + m*M + pl*PL)*(F - L)/(U - L) ;
lprint(%);
convert( %, 'rational' );
lprint(%);
# Final only for F < L
(f*F + m*M)*(U - F)/(U - L)/(f + m) + (f*F + m*M + pl*PL)*(F - L)/(U - L) ;
lprint(%);
convert( %, 'rational' );
lprint(%);

Maximum Derivative

The partial derivative with respect to F is of the general formula is

(2fF + mM + plPL - fL + (fU - 2fF - mM)/(f + m))/(U - L).

The coefficient of F in this equation is 2f(1 - 1/(f + m))/(U - L) which is clearly negative, and consequently, the greatest increase will be when F = L.