Assigning Different Assignment Question to Student Automatically.

 

Problem.

In large classes with over 100 students, assessment plagiarism is a problem. One way to resolve it is to have examination in an exam hall. Anything that is not done in an exam hall, there is a tendency for some student to copy.

The situation can be so bad that in some private universities, student will only do work during contact hours (lab, lecture and tutorial hour). The only way for them to learn more is to give after contact hour assessment such as take home test or assignment. It may not be graded but during lab hour they will be tested/quizzed on what they did for their assignment. However, this is less effective, as most student will not put in their effort in anything that is not graded and in this scenario, they may just simply do their quiz.

One effective way for student to learn is to give them an assignment where the student is forced to work out the solution for a pre-long period of time so that they can indirectly learn from it. This assignment should be zebra type assignment, where the student can self-test whether their solution is correct or wrong. For example in Computer Science,

a)      built a network using Cisco Packet Tracer with 5 routers.

b)     Build a 4 bit counter with a count sequence of 5->10->15->2 (repeat).

However, the student will copy their assignment from each other. Thus, the examiner must give a lot of different questions. If an examiner has 500 students with 100 questions, each question is done by a group of 2 students, keeping a record of which student doing which question can be tedious. There is a better solution to solve this problem.

Solutions.

Each student indirectly has some random variable such as birthday, student ID, National ID, Passport No, Gender, Race, Home State, HomeTown, Country etc. Hash function can also be used to ensure uniform distribution. If it is a single student, the question assignment can be in the following form

a)      Sha256(STudentID, Birthday) % (No of question) = Question No

b)     (studentID)%No of question =Question No. Variable on the question will depend on the student Birthday.

A complete example will be as followed.

If your student ID ends with "0", you suppose to do 5R10.pka If your student ID ends with "1", you suppose to do 5R1.pka If your student ID ends with "2", you suppose to do 5R2.pka If your student ID ends with "3", you suppose to do 5R3.pka If your student ID ends with "4", you suppose to do 5R4.pka If your student ID ends with "5", you suppose to do 5R5.pka If your student ID ends with "6", you suppose to do 5R6.pka If your student ID ends with "7", you suppose to do 5R7.pka If your student ID ends with "8", you suppose to do 5R8.pka If your student ID ends with "9", you suppose to do 5R9.pka   All PCs are connected to f0/0 port. All IP addresses must start with "2" The first 2 digits of your IP address is based on your birthday Let said your birth day is 4th Dec 2000 Then the IP address for the whole network will be 204.12.X.X   Let said your birth day is 12th July 2000 Then the IP address for the whole network will be 212.7.X.X All the last digit of the Router IP for all interfaces will follow the Router No!. R1 f0/0 = 212.7.1.1 R2 f0/0 = 212.7.2.2 R3 f0/0 = 212.7.3.3 R4 f0/0 = 212.7.4.4 R5 f0/0 = 212.7.5.5   All PC IP will end with 100. E.g. PC1 IP = 212.7.1.100   If R1 is Connected to R2, then the IP will be 212.7.12.1 (R1) and 212.7.12.2 (R2)   If R3 is Connected to R2, then the IP will be 212.7.23.1 (R1) and 212.7.23.2 (R2) (start with the smaller digit). The network mask is 255.255.255.0   Special Cases, If your birthday falls on 24 - 31st, Then you must deduct 20 days from your birthday. For example., your birthday is 26th Feb 2000, (minus 20 days will be 6th feb 2000) then the IP address you use should be 206.2.X.X

 

If it is a group assignment consisting of two person, a lot of variation can be done

a)      Question No = (StudentID 1 [+,x,^] StudentID2 )%NoOfQuestion

b)     Question No = sha256(studentID1,studentID2)%NoOfQuestion

Etc.

In this way, student cannot copy directly from their friend. However, they can look at their friend’s solution and try to implement the same approach into their assignment. They are also encouraged to discuss their assignment with their friends. Given, the enormous amount of effort required to do the assignment, it is quite unlikely to have someone else to do your assignment.

However, this kind of scenario is only possible for certain subjects in Computer Science such as

Digital Logic Circuits (varies between different Boolean Function, Count Sequence, Flip-Flop Type, IC type)

 Computer Network (Varies the network topology, IP address, interface type and protocol type)

Programming Subject (Varies between different type of programming question for example in MicroP programming classes)

Q22. The 32 bits value stored in 0x2000 till 0x2000 + 36 is actually the date in BCD format. For example, MMDDYYYY,  for example 0x11221972 is represented as 22nd November 1972. If the date format is valid, store “1111” in 0x2100 till (0x2100 + 36). Else store “2222” in 0x2100 till (0x2100 + 36). Q23. The 32 bits value stored in 0x2000 till 0x2000 + 36 is actually the date in BCD format. For example, YYYYMMDD,  for example 0x19230523 is represented as 23rd May 1923. If the date format is valid, store “1111” in 0x2100 till (0x2100 + 36). Else store “2222” in 0x2100 till (0x2100 + 36). Q24. The 32 bits value stored in 0x2000 till 0x2000 + 36 is actually the date in BCD format. For example, YYYYDDMM,  for example 0x19231503 is represented as 15th March 1923. If the date format is valid, store “1111” in 0x2100 till (0x2100 + 36). Else store “2222” in 0x2100 till (0x2100 + 36). Q25. The 32 bits value stored in 0x2000 till 0x2000 + 36 is actually the timestamp in BCD format. For example, 00:HH:MM:SEC. For example 0x00233435 is represented as 23 hour 34 min 35 second. If the time format is valid, store “1111” in 0x2100 till (0x2100 + 36). Else store “2222” in 0x2100 till (0x2100 + 36).

 

For other subjects may require some form of creativity is required to varies different type of question from a similar problem.

Assessment Problem.

1.      The difficulty of each question is not uniform

2.      If the solution is correct it is easy to verify, however if the solution is incorrect then it may be very time consuming to check. Thus, it may be best to give a slightly easier problem so that it is achievable for majority of the students. For example, instead of 4 bit counter, reduced it to 3 bit counter.

3.      It may not be possible to give answer to all solutions. It is only possible after one round of teaching.

4.      The examiner does not want to do extra work or don’t have the creativity to create many questions.

 

Key summary

1.      Student are forced to do their assignment

2.      Most of the assignment cannot be copied but the approach is similar where student can learn from each other by looking at each other solutions.

3.      It also encourages discussion/self-teaching among all student.