Three levels
Every degree programme that has courses at Levels 1, 2 and 3 has three weights associated
with it. These are called w
1, w
2 and w
3.
They denote the relative importance given to each course at Levels 1, 2 and
3 respectively. The weights are recommended by the relevant sub-board but approved
by the College Board of Examiners. The weights must always satisfy 0<w
1
<
w
2
<
w
3;
normally they must also satisfy 2w
1
<
w
2 and w
3
<6w
1.
Each such degree programme also has a number z associated with it. This
is the number of courses that will be weighted most highly. Normally z is
equal to the minimum number of course-units at Level 3 that a student on the
given degree programme may take, but z may be lower than this. Thus z is
equal to one of ½, 1, 1 ½, 2, 2 ½,
3, 3 ½, 4. The value of z is recommended
by the relevant examination board but approved by the College Board of Examiners.
Three levels; full course-units only
If the degree programme has three levels and all courses are full course-units,
then M is calculated by the formula
Where:
| D = 3w1 + yw2 + zw3 |
| y = 7 - z |
| X = total of the marks on the best three course units at level 1 |
| Z = total of the marks on the best z course units at level 3 |
| Y = total of the marks on the best y remaining course units at levels 2 and 3 |
Example of Rule
Suppose that
| w1 = 1, w2= 3 w3 = 5, and z = 3. | Then y = 4, D = 3 + 12 + 15 = 30 and |
| M = X + 3Y + 5Z
30 |
Example of Candidate
Suppose that a candidate's marks are:
| Level 1: | 58, 53, 46, 39 |
| Level 2: | 55, 67, 47, 50, |
| Level 3: | 68, 56, 54, 49 |
Then:
| X = 58 + 53 + 46 = 157 |
| Z = 68 + 56 + 54 = 178 |
| Y = 67 + 55 + 50 + 49 = 221 |
| M = 1710 = 57.00 30 |
Three levels; half course-units only
If the degree programme has three levels and all courses are half course-units,
then M is calculated by the formula
where:
| D = 6w 1 + yw 2 + 2zw 3 |
| y = 14 - 2z |
| X = total of the marks on the best 6 half course-units at Level 1 |
| Z = total of the marks on the best 2z half course-units at Level 3 |
| Y = total of the marks on the best y remaining half course-units at Levels 2 and 3 |
Note that w
1, w
2, w
3, y, z and D are
the same for the whole degree programme; only X, Y and Z change
with each candidate.
Example of Rule
Suppose that w
1 = 1, w
2 =
3, w
3 = 5 and z = 2½. Then y =
9, D = 6 + 27 + 25 = 58 and
Example of Candidate
Suppose that a candidate's marks are:
| Level 1: | 58, 41, 68, 55, 58, 53, 39, 28 |
| Level 2: | 59, 66, 76, 55, 56, 68, 55, 64, 66 |
| Level 3: | 67, 50, 76, 88, 58, 78, 73 |
Then:
| X = 68 + 58 + 58 + 55 + 53 + 41 = 333 |
| Z = 88 + 78 + 76 + 73 + 67 = 382 |
| Y = 76 + 68 + 66 + 66 + 64 + 59 + 58 + 56 + 55 = 568 |
| X + 3Y + 5Z = 3947 |
| M = 3947 = 68.05 58 |
Three levels; full and half course-units
If the degree programme has three levels, and some courses are full course-units
and some are half course-units, then either of the following procedures may
be used. They should give the same answer.
(a) Count each full course-unit as two half course-units at the same Level
and with the same mark.
Then use formula for three levels; half course units only.
(b) Decide which are the best courses equivalent to 3 full course-units at
Level 1, and the best courses equivalent to z full course-units at Level
3, and the best courses equivalent to 7 - z full course units among
the remaining courses at Levels 2 and 3. Make allowance for the fact that it
is possible for only half of a full course-unit to contribute to a total. Then
divide all marks on half course-units by 2. Then use formula for three levels;
full course units only.
Example of Rule using (b)
Suppose that w
1 = 1 and w
2 =
3, w
3 = 5 and z = 3. Then y = 4, D =
3 + 12 + 15 = 30 and
Example of Candidate
Suppose that a candidate's marks are:
| Level 1: | 50, 55, 60, 65 | full course-units |
| Level 2: | 40, 47, 58, 63, | full course-units |
| 56 | half course-units | |
| Level 3: | 50, 60, 80 | full course-units |
| 70 | half course-units |
Then:
| X = 65 + 60 + 55 = 180 |
| Z = 80, 70/2 + 60 + 50/2 = 200 |
| Y = 63 + 58 + 56/2 + 50/2 + 47 = 221 |
| X + 3Y + 5Z = 1843 |
| M = 1843 = 61.43 30 |
If you have any queries about the way in which your final result is calculated, please contact your Departmental Examinations Officer.