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Optimizing Movement of Students from Hostels to Lecture Rooms in a Potential World Class University Using Transportation Model ()

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*Applied Mathematics*,

**9**, 633-646. doi: 10.4236/am.2018.96044.

1. Introduction

Transportation issue is an extraordinary class of linear programming issue and it is viewed as imperatively as a vital angle that has been considered in an extensive variety of operations including research areas. All things considered, it has been utilized as a part of recreation of a few genuine issues. In this way, streamlining transportation issue of factors has amazingly been critical to different controls [1] . Transportation issues are being confronted in urban regions because of increment in human exercises and most world class universities resemble little urban areas in light of their size and structures [2] . Additionally, today’s grounds have moved toward becoming as mind boggling as urban zones, in view of their interesting attributes and between related impacts, most principle university’s campuses are arranged as little urban communities/groups [2] [3] . Covenant University is a potential world class university and accordingly is considered as a little city. The university has been chosen as one of the best universities in West Africa. The university is located in Ota, Ogun State, Nigeria. As a potential world class university with its refined frameworks, all around prepared labs, all around prepared and qualified teachers, skilled students and an examination driven establishment the university flourishes for more brilliance [4] [5] . Aside the exceptionally normal or understood criteria of turning into a world class university like the one specified over, the transportation organized inside the campus is likewise vital to make campus life favorable and transportation less demanding for their students. In this paper time limiting transportation issue will be considered as against the standard cost limiting transportation issue [5] [6] . The essential contrast between the two is that the cost of transportation changes with varieties in the amount yet the time included stays unaltered and independent of the amounts [6] . Time limiting transportation model will be utilized to decide how the aggregate time spent by students making progress toward move from their hostels to their distinctive lecture rooms can be limited or enhanced to lessen exposure to un-conducive climate conditions and different variables [7] [8] [9] . The outcome that will be gotten from this work can be utilized by the University to deal with the transportation organized inside the campus. Arranging real university campus has been a genuine test as of late [10] [11] . Feasible transportation approaches have turned into a fundamental device to manage road transportation issues in most world class universities [11] . The call for manageable versatility is picking up fame worldwide for little and extensive groups including real university campus [11] [12] . Transportation issue is an extraordinary sort of linear programming issue where the goal is to limit the cost of dispersing an item from various sources or birthplaces to various destinations [13] [14] [15] . Transportation Problem (TP) has broad pragmatic application esteems [16] [17] [18] . In this manner, TP is a subject deserving of further research, in spite of the fact that it has been considered from various points of view by various researchers [19] [20] [21] [22] . The first formulation and discussion of a planar transportation model was introduced by Hitchcock (1941). Likewise, various scientists have done reviews on time transportation issue, for example, Sharma and Swarup (1977). As a general rule, because of changes in free market activity, climate conditions, street conditions and other instability variables, vulnerability transportation issue is especially essential [23] [24] [25] .

2. Problem Formulation

Covenant University is used in this paper as a case study. The aim is to reduce the transportation time of students’ movement from hostels to their respective lecture rooms. The Hostels represent the origins while the lecture rooms represent the destinations.

2.1. Assumptions

The following assumptions were made in this paper:

1) The time spent to move from one origin to a destination is always the same.

2) All the routes mentioned in this paper are always accessible.

3) Movement of students is always from the named origins to the named destinations.

4) Each student at the origin (hostel) visits each destination (lecture rooms).

5) Students attend lectures in each lecture room at least once a week (that is, from origin (hostel) to destination (lecture rooms)).

2.2. Transportation Tableau

In this paper ten origins representing the hostels and eight destinations representing the lecture rooms were considered.

Table 1 shows the different origins and destinations while Table 2 shows the transportation problem tableau. The distance between the origins and destinations are shown in the cells. The total time and distance covered in each route was determined using goggle map. The mathematical modeling of this transportation problem is a special linear programming problem in which the objective function is to minimize the time of transportation subject to the demand and supply constraints. The values in each cell represents the time per unit distance between one origin and one destination. They are measured in (meter ×100). Along the outer rim in the tableau is the actual distance and expected distance

Table 1. Origins and destinations considered.

Table 2. Transportation tableau.

constraint quantity values which are referred to as rim requirements. In a time transportation problem, the time of transportation from m origins to n destinations is minimized, satisfying certain condition in respect of availability at sources and requirements at the destinations. From Table 2, we have a balanced transportation problem in which the total demand (expected distance) equals the total supply (actual distance). The distance is measured in meter ×100 while the time or duration is measured in minutes.

3. Problem Solution

Solving transportation problems involves a lot of methods, some of them heuristic in nature [26] - [31] . In this paper, the transportation problem is solved based on the data collected, which is relatively reliable. The first stage of the solution process was to find the initial basic feasible solution [32] . The following three methods of obtaining the initial solution were used. These methods are North-west corner method, least cost method and Vogel’s approximation method. Based on the Least cost method that gave the minimum time spent an optimality test was carried out, after checking for degeneracy, to determine an optimal solution using the MODI (Modified Distribution) method.

3.1. North-West Corner Method

Using the North West corner method to obtain an initial basic solution the values derived are shown below (Table 3).

3.2. Least Cost Method

Using the least cost method we obtain the following values as shown below (Table 4).

Table 3. North-west corner method matrix.

Table 4. Least cost method matrix.

3.3. Test for Degeneracy

The only optimality test condition is transportation degeneracy. Out of the three methods used to find the initial basic feasible solution to the transportation problem under consideration, the least cost method gave the minimum value of 1916. Therefore the least cost matrix is used to check for degeneracy:

$m+n-1=10+8-1=17$

The basic cells are also 17. This implies there is no problem of degeneracy thus optimality test is carried out (Table 5).

3.4. Test for Optimality

From the results obtained using the three methods of finding the initial basic feasible solution, least cost method gave the least amount of total time spent. Thus, its solution will be used to test for optimality in order to obtain the optimal solution (Table 6).

3.5. Modified Distribution Method (Modi Method)

This method involves two formulas in arriving at an optimal solution.

${U}_{i}+{V}_{j}={C}_{ij}$ , which is used in determining the values of U_{i} and V_{j} using the values in the occupied cells.

${U}_{i}-{V}_{j}-{C}_{ij}={K}_{ij}$ , which is used in determining the best empty route that can

Table 5. Vogel method.

Table 6. Modified least cost method matrix.

be followed to reduce the total time spent and it is only the route with the most negative K_{ij} that will be considered.

U_{i} and V_{j} are the row and column values which will be determined while C_{ij} and K_{ij} are the unit time and time increase or decrease respectively.

Calculating for occupied cells gives:

${U}_{i}+{V}_{j}={C}_{ij}$

${\text{O}}_{\text{1}}{\text{D}}_{\text{2}}:{U}_{1}+{V}_{2}=8$ (1)

${\text{O}}_{\text{2}}{\text{D}}_{\text{4}}:{U}_{2}+{V}_{4}=14$ (2)

${\text{O}}_{\text{3}}{\text{D}}_{\text{1}}:{U}_{3}+{V}_{1}=12$ (3)

${\text{O}}_{\text{3}}{\text{D}}_{\text{2}}:{U}_{3}+{V}_{2}=10$ (4)

${\text{O}}_{\text{4}}{\text{D}}_{\text{3}}:{U}_{4}+{V}_{3}=14$ (5)

${\text{O}}_{\text{5}}{\text{D}}_{\text{3}}:{U}_{5}+{V}_{3}=14$ (6)

${\text{O}}_{\text{5}}{\text{D}}_{\text{4}}:{U}_{5}+{V}_{4}=14$ (7)

${\text{O}}_{\text{5}}{\text{D}}_{\text{5}}:{U}_{5}+{V}_{5}=14$ (8)

${\text{O}}_{\text{6}}{\text{D}}_{\text{1}}:{U}_{6}+{V}_{1}=12$ (9)

${\text{O}}_{\text{6}}{\text{D}}_{\text{5}}:{U}_{6}+{V}_{5}=14$ (10)

${\text{O}}_{\text{6}}{\text{D}}_{\text{8}}:{U}_{6}+{V}_{8}=23$ (11)

${\text{O}}_{\text{7}}{\text{D}}_{\text{6}}:{U}_{7}+{V}_{6}=21$ (12)

${\text{O}}_{\text{7}}{\text{D}}_{\text{7}}:{U}_{7}+{V}_{7}=23$ (13)

${\text{O}}_{\text{7}}{\text{D}}_{\text{8}}:{U}_{7}+{V}_{8}=25$ (14)

${\text{O}}_{\text{8}}{\text{D}}_{\text{8}}:{U}_{8}+{V}_{8}=26$ (15)

${\text{O}}_{\text{9}}{\text{D}}_{\text{7}}:{U}_{9}+{V}_{7}=22$ (16)

${\text{O}}_{\text{1}0}{\text{D}}_{\text{6}}:{U}_{10}+{V}_{6}=20$ (17)

From the equations above there are eighteen (18) unknowns but seventeen (17) equations. To solve this, 0 is assigned to one of the unknowns.

Let ${U}_{1}=0$

From Equation (1),

${\text{O}}_{\text{1}}{\text{D}}_{2}:{U}_{1}+{V}_{2}=8$ (18)

$0+{V}_{2}=8$

Therefore, ${V}_{2}=8$

Substituting Equation (18) into Equation (4) gives

${U}_{3}=2$ (19)

Substituting Equation (19) into Equation (3) gives

${V}_{1}=10$ (20)

Substituting Equation (19) into Equation (9) gives

${U}_{6}=2$ (21)

Substituting Equation (21) into Equation (10) gives

${V}_{5}=12$ (22)

Substituting Equation (21) into Equation (11) gives

${V}_{8}=21$ (23)

Substituting Equation (23) into Equation (14) gives

${U}_{7}=4$ (24)

Substituting Equation (23) into Equation (15) gives

${U}_{8}=5$ (25)

Substituting Equation (22) into Equation (8) gives

${U}_{5}=2$ (26)

Substituting Equation (26) into Equation (7) gives

${V}_{4}=12$ (27)

Substituting Equation (21) into Equation (2) gives

${U}_{2}=2$ (28)

Substituting Equation (26) into Equation (6) gives

${V}_{3}=12$ (29)

Substituting Equation (30) into Equation (5) gives

${U}_{4}=2$ (30)

Substituting Equation (24) into Equation (12) gives

${V}_{6}=17$ (31)

Substituting Equation (24) into Equation (13) gives

${V}_{7}=19$ (32)

Substituting Equation (33) into Equation (16) gives

${U}_{9}=3$ (33)

Substituting Equation (32) into Equation (17) gives

${U}_{10}=3$ (34)

Calculating for the unoccupied cells gives:

${C}_{ij}-{U}_{i}-{V}_{j}={K}_{ij}$

${\text{O}}_{\text{1}}{\text{D}}_{\text{1}}:10-{U}_{1}-{V}_{1}=10-0-10=0$

${\text{O}}_{\text{1}}{\text{D}}_{\text{3}}:11-{U}_{1}-{V}_{3}=11-0-12=-1$

${\text{O}}_{\text{1}}{\text{D}}_{\text{4}}:11-{U}_{1}-{V}_{4}=11-0-12=-1$

${\text{O}}_{\text{1}}{\text{D}}_{\text{5}}:11-{U}_{1}-{V}_{5}=11-0-12=-1$

${\text{O}}_{\text{1}}{\text{D}}_{\text{6}}:16-{U}_{1}-{V}_{6}=16-0-17=-1$

${\text{O}}_{\text{1}}{\text{D}}_{\text{7}}:17-{U}_{1}-{V}_{7}=17-0-19=-2$

${\text{O}}_{\text{1}}{\text{D}}_{\text{8}}:20-{U}_{1}-{V}_{8}=20-0-21=-1$

${\text{O}}_{\text{2}}{\text{D}}_{\text{1}}:14-{U}_{2}-{V}_{1}=14-2-10=4$

${\text{O}}_{\text{2}}{\text{D}}_{\text{2}}:11-{U}_{2}-{V}_{2}=11-2-8=1$

${\text{O}}_{\text{2}}{\text{D}}_{\text{3}}:15-{U}_{2}-{V}_{3}=15-2-12=-1$

${\text{O}}_{\text{2}}{\text{D}}_{\text{5}}:15-{U}_{2}-{V}_{5}=15-2-12=-1$

${\text{O}}_{\text{2}}{\text{D}}_{\text{6}}:19-{U}_{2}-{V}_{6}=19-2-17=0$

${\text{O}}_{\text{2}}{\text{D}}_{\text{7}}:20-{U}_{2}-{V}_{7}=20-2-19=-1$

${\text{O}}_{\text{2}}{\text{D}}_{\text{8}}:23-{U}_{2}-{V}_{8}=23-2-21=0$

${\text{O}}_{\text{3}}{\text{D}}_{\text{3}}:13-{U}_{3}-{V}_{3}=13-2-12=-1$

${\text{O}}_{\text{3}}{\text{D}}_{\text{4}}:13-{U}_{3}-{V}_{4}=13-2-12=-1$

${\text{O}}_{\text{3}}{\text{D}}_{\text{5}}:13-{U}_{3}-{V}_{5}=13-2-12=-1$

${\text{O}}_{\text{3}}{\text{D}}_{\text{6}}:18-{U}_{3}-{V}_{6}=18-2-17=-1$

${\text{O}}_{\text{3}}{\text{D}}_{\text{7}}:19-{U}_{3}-{V}_{7}=19-2-19=-2$

${\text{O}}_{\text{3}}{\text{D}}_{\text{8}}:22-{U}_{3}-{V}_{8}=22-2-21=-1$

${\text{O}}_{\text{4}}{\text{D}}_{\text{1}}:\text{13}-{U}_{4}-{V}_{1}=13-12-10=-9$

${\text{O}}_{\text{4}}{\text{D}}_{\text{2}}:11-{U}_{4}-{V}_{2}=11-12-8=-9$

${\text{O}}_{\text{4}}{\text{D}}_{\text{4}}:14-{U}_{4}-{V}_{4}=14-12-12=-10$

${\text{O}}_{\text{4}}{\text{D}}_{\text{5}}:14-{U}_{4}-{V}_{5}=14-12-12=-10$

${\text{O}}_{\text{4}}{\text{D}}_{\text{6}}:19-{U}_{4}-{V}_{6}=19-12-17=-10$

${\text{O}}_{\text{4}}{\text{D}}_{\text{7}}:20-{U}_{4}-{V}_{7}=20-12-19=-11$

${\text{O}}_{\text{4}}{\text{D}}_{\text{8}}:23-{U}_{4}-{V}_{8}=23-21-10=-10$

${\text{O}}_{\text{5}}{\text{D}}_{\text{1}}:13-{U}_{5}-{V}_{1}=13-2-10=1$

${\text{O}}_{\text{5}}{\text{D}}_{\text{2}}:11-{U}_{5}-{V}_{2}=11-2-8=1$

${\text{O}}_{\text{5}}{\text{D}}_{\text{6}}:19-{U}_{5}-{V}_{6}=19-2-17=0$

${\text{O}}_{\text{5}}{\text{D}}_{\text{7}}:20-{U}_{5}-{V}_{7}=20-2-19=-1$

${\text{O}}_{\text{5}}{\text{D}}_{\text{8}}:23-{U}_{5}-{V}_{8}=23-2-21=0$

${\text{O}}_{\text{6}}{\text{D}}_{\text{2}}:11-{U}_{6}-{V}_{2}=11-2-8=1$

${\text{O}}_{\text{6}}{\text{D}}_{\text{3}}:14-{U}_{6}-{V}_{3}=14-2-12=0$

${\text{O}}_{\text{6}}{\text{D}}_{\text{4}}:14-{U}_{6}-{V}_{4}=14-2-12=0$

${\text{O}}_{\text{6}}{\text{D}}_{\text{6}}:22-{U}_{6}-{V}_{6}=22-2-17=3$

${\text{O}}_{\text{6}}{\text{D}}_{\text{7}}:24-{U}_{6}-{V}_{7}=24-2-19=3$

${\text{O}}_{\text{7}}{\text{D}}_{\text{1}}:15-{U}_{7}-{V}_{1}=15-4-10=1$

${\text{O}}_{\text{7}}{\text{D}}_{\text{2}}:14-{U}_{7}-{V}_{2}=14-4-8=2$

${\text{O}}_{\text{7}}{\text{D}}_{\text{3}}:16-{U}_{7}-{V}_{3}=16-4-12=0$

${\text{O}}_{\text{7}}{\text{D}}_{\text{4}}:16-{U}_{7}-{V}_{4}=16-4-12=0$

${\text{O}}_{\text{7}}{\text{D}}_{\text{5}}:17-{U}_{7}-{V}_{5}=17-4-12=1$

${\text{O}}_{\text{8}}{\text{D}}_{\text{1}}:16-{U}_{8}-{V}_{1}=16-5-10=1$

${\text{O}}_{\text{8}}{\text{D}}_{\text{2}}:14-{U}_{8}-{V}_{2}=14-5-8=1$

${\text{O}}_{\text{8}}{\text{D}}_{\text{3}}:17-{U}_{8}-{V}_{3}=17-5-12=0$

${\text{O}}_{\text{8}}{\text{D}}_{\text{4}}:17-{U}_{8}-{V}_{4}=17-5-12=0$

${\text{O}}_{\text{8}}{\text{D}}_{\text{5}}:17-{U}_{8}-{V}_{5}=17-5-12=0$

${\text{O}}_{\text{8}}{\text{D}}_{\text{6}}:22-{U}_{8}-{V}_{6}=22-5-17=0$

${\text{O}}_{\text{8}}{\text{D}}_{\text{7}}:23-{U}_{8}-{V}_{7}=23-5-19=-1$

${\text{O}}_{\text{9}}{\text{D}}_{\text{1}}:15-{U}_{9}-{V}_{1}=15-3-10=2$

${\text{O}}_{\text{9}}{\text{D}}_{\text{2}}:13-{U}_{9}-{V}_{2}=13-3-8=2$

${\text{O}}_{\text{9}}{\text{D}}_{\text{3}}:16-{U}_{9}-{V}_{3}=16-3-12=1$

${\text{O}}_{\text{9}}{\text{D}}_{\text{4}}:16-{U}_{9}-{V}_{4}=16-3-12=1$

${\text{O}}_{\text{9}}{\text{D}}_{\text{5}}:16-{U}_{9}-{V}_{5}=16-3-12=1$

${\text{O}}_{\text{9}}{\text{D}}_{\text{6}}:21-{U}_{9}-{V}_{6}=21-3-17=1$

${\text{O}}_{\text{9}}{\text{D}}_{\text{8}}:25-{U}_{9}-{V}_{8}=25-3-21=2$

${\text{O}}_{\text{1}0}{\text{D}}_{\text{1}}:14-{U}_{10}-{V}_{1}=14-3-10=1$

${\text{O}}_{\text{1}0}{\text{D}}_{\text{2}}:13-{U}_{10}-{V}_{2}=13-3-8=2$

${\text{O}}_{\text{1}0}{\text{D}}_{\text{3}}:16-{U}_{10}-{V}_{3}=16-3-12=1$

${\text{O}}_{\text{1}0}{\text{D}}_{\text{4}}:16-{U}_{10}-{V}_{4}=16-3-12=1$

${\text{O}}_{\text{1}0}{\text{D}}_{\text{5}}:16-{U}_{10}-{V}_{5}=16-3-12=1$

${\text{O}}_{\text{1}0}{\text{D}}_{\text{7}}:16-{U}_{10}-{V}_{7}=16-3-19=-6$

${\text{O}}_{\text{1}0}{\text{D}}_{\text{8}}:24-{U}_{10}-{V}_{8}=24-3-21=0$

Table 7. MODI method iteration showing the new allocation.

From the calculations above, O_{4}D_{7} is the only cell with the most negative K_{ij} with −11. With this result it means that the total time will be reduced by −11 if we allocate to that cell. The Table 7 illustrates this.

4. Result Discussion

From the analysis carried out with the data collected, it can be seen that least cost method gave the best initial basic feasible solution. Optimality test carried out, using MODI method, gave an optimal value of 1896. This was an improvement on the result, 1916, given by the least cost method. Also, it can be seen that by allocating maximally to route O_{4}D_{7}, the total time will be reduced by −11. From Table 7 above, it can be seen that the total time can be minimized if 900 m is covered using route O_{1}D_{2}, 11 km covered using O_{2}D_{4}, 9 km covered using O_{3}D_{1}, 1 km covered using O_{3}D_{2}, 5 km covered using O_{4}D_{3}, 6 km covered using O_{4}D7, 1 km covered using O_{5}D_{3}, 1 km covered using O_{5}D_{4,} 8 km covered using O_{5}D_{5}, 2 km covered using O_{6}D_{1}, 5 km covered using O_{6}D_{5}, 4 km covered using O_{6}D_{8}, 6 km covered using O_{7}D_{3}, 400 m covered using O_{7}D_{6}, 13 km covered using O_{8}D_{8}, 11 km covered using O_{9}D_{7}, 1200 m covered using O_{10}D_{6}. Also, from the table it can be seen that some requires less time to cover than the others. For instance, route O_{1}D_{2} requires just 8 minutes to cover 900 m while some other routes, like route O_{5}D_{3} and route O_{5}D_{4}, require a lot of time. This could be as a result of different factors relating to the transportation system generally which the university authority should look into. Also, this analysis and the result would help the university to plan in order to optimize its transportation system within the campus, especially as regards the movement of students to and fro their lecture rooms.

5. Conclusion

In this paper time minimizing transportation model was used to determine how the movement of students by road from their hostels to their different lecture rooms can be optimized. Three methods were used to arrive at an initial basic feasible solution and Least Cost method was found to give the least total time as compared to the other two methods. MODI method was used to obtain an optimal solution. From the result derived, it shows that the total time spent on the movement of students to their lecture rooms, from their different hostels, can be minimized thereby improving campus life and reducing lateness to lectures. Covenant University can use this to plan and manage the transportation system of the university, as it thrives on becoming one of the best universities in the world.

Conflicts of Interest

The authors declare no conflicts of interest.

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