Transportation issue is one of the significant zones of utilization of Linear Programming Model. In this paper, transportation model is utilized to decide an ideal answer for the transportation issue in a run of the mill world class university utilizing Covenant University as a contextual analysis. Covenant University is a potential world class University. The quick development of Covenant University Campus over the most recent fourteen years affects its transportation framework. This paper particularly takes a gander at streamlining the time spent by the students moving from their lodgings to lecture rooms. Google guide was utilized to figure the separation and time between every cause and every goal. North-west corner technique, Least Cost strategy and Vogel’s estimation technique were utilized to decide the underlying fundamental plausible arrangement (initial feasible solution) and MODI strategy was utilized to locate the ideal arrangement (optimal solution). The last outcome demonstrates that the development of understudies from hostel to lecture rooms can be streamlined if the total time spent is decreased.
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 [
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.
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)).
In this paper ten origins representing the hostels and eight destinations representing the lecture rooms were considered.
Origin | Destination | ||
---|---|---|---|
Esther Hall | O_{1} | College of Science and Technology | D_{1} |
Mary Hall | O_{2} | College of Developmental Studies | D_{2} |
Deborah Hall | O_{3} | Lecture Theatre One | D_{3} |
Lydia Hall | O_{4} | Lecture theatre two | D_{4} |
Dorcas Hall | O_{5} | Chemical and Petroleum Engineering Block | D_{5} |
Peter Hall | O_{6} | Mechanical Engineering Block | D_{6} |
Paul Hall | O_{7} | Civil Engineering Block | D_{7} |
John Hall | O_{8} | Electrical and Electronics Engineering Block | D_{8} |
Joseph Hall | O_{9} | ||
Daniel Hall | O_{10} |
D_{1} | D_{2} | D_{3} | D_{4} | D_{5} | D_{6} | D_{7} | D_{8} | Actual Distance (m × 100) | |
---|---|---|---|---|---|---|---|---|---|
O_{1} | 10 | 8 | 11 | 11 | 11 | 16 | 17 | 20 | 9 |
O_{2} | 14 | 11 | 15 | 14 | 15 | 19 | 20 | 23 | 11 |
O_{3} | 12 | 10 | 13 | 13 | 13 | 18 | 19 | 22 | 10 |
O_{4} | 13 | 11 | 14 | 14 | 14 | 19 | 20 | 23 | 11 |
O_{5} | 13 | 11 | 14 | 14 | 14 | 19 | 20 | 23 | 10 |
O_{6} | 12 | 11 | 14 | 14 | 14 | 22 | 24 | 23 | 11 |
O_{7} | 15 | 14 | 16 | 16 | 17 | 21 | 23 | 25 | 12 |
O_{8} | 16 | 14 | 17 | 17 | 17 | 22 | 23 | 26 | 13 |
O_{9} | 15 | 13 | 16 | 16 | 16 | 21 | 22 | 25 | 11 |
O_{10} | 14 | 13 | 16 | 16 | 16 | 20 | 22 | 24 | 12 |
Expected Distance (m × 100) | 11 | 10 | 12 | 12 | 13 | 16 | 17 | 19 |
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
Solving transportation problems involves a lot of methods, some of them heuristic in nature [
Using the North West corner method to obtain an initial basic solution the values derived are shown below (
Using the least cost method we obtain the following values as shown below (
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 (
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 (
This method involves two formulas in arriving at an optimal solution.
U i + V j = C i j , 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 i j = K i j , which is used in determining the best empty route that can
D_{1} V_{1}= | D_{2} V_{2}= | D_{3} V_{3}= | D_{4} V_{4}= | D_{5} V_{5}= | D_{6} V_{6}= | D_{7} V_{7}= | D_{8} V_{8}= | Actual Distance (Km) | ||
---|---|---|---|---|---|---|---|---|---|---|
U_{1}= | O_{1} | ^{10 } | 9^{8 } | ^{11 } | ^{11 } | ^{11 } | ^{16 } | ^{17 } | ^{20 } | 9 |
U_{2}= | O_{2} | ^{14 } | ^{11 } | ^{15 } | 11^{14 } | ^{15 } | ^{19 } | ^{20 } | ^{23 } | 11 |
U_{3}= | O_{3} | 9^{12 } | 1^{10 } | ^{13 } | ^{13 } | ^{13 } | ^{18 } | ^{19 } | ^{22 } | 10 |
U_{4}= | O_{4} | ^{13 } | ^{11 } | 5(-)^{14 } | ^{14 } | ^{14 } | ^{19 } | 6(+)^{20 } | ^{23 } | 11 |
U_{5}= | O_{5} | ^{13 } | ^{11 } | 1^{14 } | 1^{14 } | 8^{14 } | ^{19 } | ^{20 } | ^{23 } | 10 |
U_{6}= | O_{6} | 2^{12 } | ^{11 } | ^{14 } | ^{14 } | 5^{14 } | ^{22 } | ^{24 } | 4^{23 } | 11 |
U_{7}= | O_{7} | ^{15 } | ^{14 } | 6(+)^{16 } | ^{16 } | ^{17 } | 4^{21 } | (-)^{23 } | 2^{25 } | 12 |
U_{8}= | O_{8} | ^{16 } | ^{14 } | ^{17 } | ^{17 } | ^{17 } | ^{22 } | ^{23 } | 13^{26 } | 13 |
U_{9}= | O_{9} | ^{15 } | ^{13 } | ^{16 } | ^{16 } | ^{16 } | ^{21 } | 11^{22 } | ^{25 } | 11 |
U_{10}= | O_{10} | ^{14 } | ^{13 } | ^{16 } | ^{16 } | ^{16 } | 12^{20 } | ^{22 } | ^{24 } | 12 |
Expected Distance (Km) | 11 | 10 | 12 | 12 | 13 | 16 | 17 | 19 |
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 i j
O 1 D 2 : U 1 + V 2 = 8 (1)
O 2 D 4 : U 2 + V 4 = 14 (2)
O 3 D 1 : U 3 + V 1 = 12 (3)
O 3 D 2 : U 3 + V 2 = 10 (4)
O 4 D 3 : U 4 + V 3 = 14 (5)
O 5 D 3 : U 5 + V 3 = 14 (6)
O 5 D 4 : U 5 + V 4 = 14 (7)
O 5 D 5 : U 5 + V 5 = 14 (8)
O 6 D 1 : U 6 + V 1 = 12 (9)
O 6 D 5 : U 6 + V 5 = 14 (10)
O 6 D 8 : U 6 + V 8 = 23 (11)
O 7 D 6 : U 7 + V 6 = 21 (12)
O 7 D 7 : U 7 + V 7 = 23 (13)
O 7 D 8 : U 7 + V 8 = 25 (14)
O 8 D 8 : U 8 + V 8 = 26 (15)
O 9 D 7 : U 9 + V 7 = 22 (16)
O 1 0 D 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),
O 1 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 i j − U i − V j = K i j
O 1 D 1 : 10 − U 1 − V 1 = 10 − 0 − 10 = 0
O 1 D 3 : 11 − U 1 − V 3 = 11 − 0 − 12 = − 1
O 1 D 4 : 11 − U 1 − V 4 = 11 − 0 − 12 = − 1
O 1 D 5 : 11 − U 1 − V 5 = 11 − 0 − 12 = − 1
O 1 D 6 : 16 − U 1 − V 6 = 16 − 0 − 17 = − 1
O 1 D 7 : 17 − U 1 − V 7 = 17 − 0 − 19 = − 2
O 1 D 8 : 20 − U 1 − V 8 = 20 − 0 − 21 = − 1
O 2 D 1 : 14 − U 2 − V 1 = 14 − 2 − 10 = 4
O 2 D 2 : 11 − U 2 − V 2 = 11 − 2 − 8 = 1
O 2 D 3 : 15 − U 2 − V 3 = 15 − 2 − 12 = − 1
O 2 D 5 : 15 − U 2 − V 5 = 15 − 2 − 12 = − 1
O 2 D 6 : 19 − U 2 − V 6 = 19 − 2 − 17 = 0
O 2 D 7 : 20 − U 2 − V 7 = 20 − 2 − 19 = − 1
O 2 D 8 : 23 − U 2 − V 8 = 23 − 2 − 21 = 0
O 3 D 3 : 13 − U 3 − V 3 = 13 − 2 − 12 = − 1
O 3 D 4 : 13 − U 3 − V 4 = 13 − 2 − 12 = − 1
O 3 D 5 : 13 − U 3 − V 5 = 13 − 2 − 12 = − 1
O 3 D 6 : 18 − U 3 − V 6 = 18 − 2 − 17 = − 1
O 3 D 7 : 19 − U 3 − V 7 = 19 − 2 − 19 = − 2
O 3 D 8 : 22 − U 3 − V 8 = 22 − 2 − 21 = − 1
O 4 D 1 : 13 − U 4 − V 1 = 13 − 12 − 10 = − 9
O 4 D 2 : 11 − U 4 − V 2 = 11 − 12 − 8 = − 9
O 4 D 4 : 14 − U 4 − V 4 = 14 − 12 − 12 = − 10
O 4 D 5 : 14 − U 4 − V 5 = 14 − 12 − 12 = − 10
O 4 D 6 : 19 − U 4 − V 6 = 19 − 12 − 17 = − 10
O 4 D 7 : 20 − U 4 − V 7 = 20 − 12 − 19 = − 11
O 4 D 8 : 23 − U 4 − V 8 = 23 − 21 − 10 = − 10
O 5 D 1 : 13 − U 5 − V 1 = 13 − 2 − 10 = 1
O 5 D 2 : 11 − U 5 − V 2 = 11 − 2 − 8 = 1
O 5 D 6 : 19 − U 5 − V 6 = 19 − 2 − 17 = 0
O 5 D 7 : 20 − U 5 − V 7 = 20 − 2 − 19 = − 1
O 5 D 8 : 23 − U 5 − V 8 = 23 − 2 − 21 = 0
O 6 D 2 : 11 − U 6 − V 2 = 11 − 2 − 8 = 1
O 6 D 3 : 14 − U 6 − V 3 = 14 − 2 − 12 = 0
O 6 D 4 : 14 − U 6 − V 4 = 14 − 2 − 12 = 0
O 6 D 6 : 22 − U 6 − V 6 = 22 − 2 − 17 = 3
O 6 D 7 : 24 − U 6 − V 7 = 24 − 2 − 19 = 3
O 7 D 1 : 15 − U 7 − V 1 = 15 − 4 − 10 = 1
O 7 D 2 : 14 − U 7 − V 2 = 14 − 4 − 8 = 2
O 7 D 3 : 16 − U 7 − V 3 = 16 − 4 − 12 = 0
O 7 D 4 : 16 − U 7 − V 4 = 16 − 4 − 12 = 0
O 7 D 5 : 17 − U 7 − V 5 = 17 − 4 − 12 = 1
O 8 D 1 : 16 − U 8 − V 1 = 16 − 5 − 10 = 1
O 8 D 2 : 14 − U 8 − V 2 = 14 − 5 − 8 = 1
O 8 D 3 : 17 − U 8 − V 3 = 17 − 5 − 12 = 0
O 8 D 4 : 17 − U 8 − V 4 = 17 − 5 − 12 = 0
O 8 D 5 : 17 − U 8 − V 5 = 17 − 5 − 12 = 0
O 8 D 6 : 22 − U 8 − V 6 = 22 − 5 − 17 = 0
O 8 D 7 : 23 − U 8 − V 7 = 23 − 5 − 19 = − 1
O 9 D 1 : 15 − U 9 − V 1 = 15 − 3 − 10 = 2
O 9 D 2 : 13 − U 9 − V 2 = 13 − 3 − 8 = 2
O 9 D 3 : 16 − U 9 − V 3 = 16 − 3 − 12 = 1
O 9 D 4 : 16 − U 9 − V 4 = 16 − 3 − 12 = 1
O 9 D 5 : 16 − U 9 − V 5 = 16 − 3 − 12 = 1
O 9 D 6 : 21 − U 9 − V 6 = 21 − 3 − 17 = 1
O 9 D 8 : 25 − U 9 − V 8 = 25 − 3 − 21 = 2
O 1 0 D 1 : 14 − U 10 − V 1 = 14 − 3 − 10 = 1
O 1 0 D 2 : 13 − U 10 − V 2 = 13 − 3 − 8 = 2
O 1 0 D 3 : 16 − U 10 − V 3 = 16 − 3 − 12 = 1
O 1 0 D 4 : 16 − U 10 − V 4 = 16 − 3 − 12 = 1
O 1 0 D 5 : 16 − U 10 − V 5 = 16 − 3 − 12 = 1
O 1 0 D 7 : 16 − U 10 − V 7 = 16 − 3 − 19 = − 6
O 1 0 D 8 : 24 − U 10 − V 8 = 24 − 3 − 21 = 0
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
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
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.
Agarana, M.C., Udoh, A.I. and Ehigbochie, A.I. (2018) Optimizing Movement of Students from Hostels to Lecture Rooms in a Potential World Class University Using Transportation Model. Applied Mathematics, 9, 633-646. https://doi.org/10.4236/am.2018.96044