We approach this project in three steps: **1. Transforming Data; 2. Building the Model; 3. Running the solution.** The first step is transforming data we got from OAA. There are two initial resources from OAA:

**one excel file containing:**A matrix with students (rows) and majors + interests(columns); there are total__1615 “true ” students and 36 majors and 4 concerns__including whether they defined themselves as transfer students, Neurodiverse, student-athletes and first-generation or not__.__“True students” means students who will be matched with advisors using the matching model, and there are also students who will be*pre-matched*to certain advisor, so they will not go through the algorithm; and there are 1666 students in total, with 51 pre-matched students. In addition, each student will choose at least 2 majors and at most 5 majors among 36 majors, and majors they choose will ranked from “1st major” to “5th major”.**So it is a 1615 x 40 matrix.****one excel file containing:**Detail information about advisors. Each advisor (343 in total) lists*(a) interested majors*(maximum of 5 and minimum of 1), (*b) demand/maximum students to be received*and (*c) their attitude to four concerns including willingness*to receive transfer students, Neurodiverse, student-athletes and first-generation or not. Thus, the excel file is also a matrix with advisors (rows) and academic interests (columns); there are total 343 advisors; 36 majors and 4 concerns.**So it is a 343 x 40 matrix**

We define student matrix be S matrix and Advisor matrix be A matrix. Since our ultimate goal is to maximize the satisfaction level of matching between students, __we must first define “the satisfaction level”.__ Because student *i* is matched to advisor *j*, we use* “pij “ *to represent satisfaction level. Specifically,

*pij*takes the number of common majors and concerns between students and advisor into consideration, so make sure the algorithm takes a global perspective. The idea is to first get

*pij*value individually for S matrix (

__students’ satisfaction level toward each major__) and A matrix (

__advisors’ satisfaction toward each major + attitude toward each concern__), and then

**we use matrix multiplication to get the final**

**pij****between student and advisors**. (Thinking of matrix multiplication, S matrix and transpose of A matrix could be multiplied because of their common “40” majors + concerns.)

First, in S matrix, students could choose at least 2 and at most 5 majors and the __corresponding pij values are listed__:

__1st major__

__pij__

__=2.5, 2nd major__

__pij__

__=2.0, 3rd major__

__pij__

__=1.5, 4th major__

__pij__

__=1.0, 5th major__

__pij__

__=0.5____.__Then, for questions that whether they defined themselves as transfer students, Neurodiverse, student-athletes and first-generation or not,

__pij__

__=1 if they answer “Yes” or “Unsure”;__

__pij__

__= 0 if they answer “No”.__Second, in A matrix, advisor will have *pij=2* for their home department; for instance, a professor in math department will have *pij=2* for math major. And __for each other major advisor choose, ____pij__* =1*. For 4 concerns, advisors could response “1”, “2” or “3”. The response “1” means “don’t want to match with students having following concerns”, response “2” means neutral attitude and response “3” means advisors are willing to match with students having following concerns.

__Therefore, response “1”__

__pij__

__= – 100 (large negative value to prevent matching), response “2”__

__pij__

__= 0.25 and response “3”__

__pij__

__= 1____.__

**For example**, for student who choose Math, Art, Biology and History, the excel file and corresponding S matrix are:

For advisor who choose Math, Computer Science and Biology, the excel file and corresponding A matrix are:

Therefore, the *pij* value, i.e., the satisfaction level, between student 1 and advisor 1 is (2×0) +(1.5×1) +(0x1)+(1×0)+(2.5×2)+(1×0.25)+(0x0.25)+(0x0.25)+(0x1)=6.75.

Therefore, by converting the information into *pij* value, we could obtain S matrix and A matrix. Then we get the __transpose of A matrix (T matrix),__**a 40 x 343 matrix**, we multiply the S matrix and T matrix. **The final P matrix, using students as rows and advisors as columns, is a 1615 x 343 matrix**. Each cell of P matrix contains the *pij* value between each student and advisor. We could thus easily evaluate satisfaction level between each advisor and student, and then select the best one. __The highest possible pij value is 14 and the lowest possible pij value is -400.__

After we transforming the data, we could go to next step — building model.

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