## Non-Threshold Marking

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 FL then the projects/labs do not count,
• Select an upper bound U such that if FU 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 FL and the mid-term and projects/laboratories count for their full weight only if FU. 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.