1. Party-by-Night Problem
Shortest Path Problem
Party-by-Night Problem

2. To illustrate the Shortest Path Problem, we will find the shortest path from node 1 (Engineering Building) to node 14 (Springbok’s).
Our network will have fourteen nodes, node i (i = 1;2;3;…;14). For i<j, an arc length, (i, j), corresponds to the distance (call it lij) between the nodes.
lij = The direct length (distance) of a single arc between Node i and Node j

3. Applying this formula to the information in the problem, yields:
l89 = l OR l
l13 = 1400
l911 = l OR l OR l
l14 = 1100
l912 = l OR l OR l
l25 = l
l913 = l OR l OR l
l26 = l
l1011 = l OR l
l27 = l
l1012 = l OR l
l36 = l
l1114 = l
l38 = l
l1214 = l OR l
l45 = l
l1314 = l
l47 = l
l48 = l
l59 = l OR l
l610 = l OR l
l79 = l OR l
l710 = l OR l

4. Analysing the formulas, the following 26 paths are available:
= 2510
= 3280
= 1910
= 2530
= 170
= 2080
= 3270
= 2530
= 2520
= 1930
= 2070
= 3190
= 2730
= 3290
= 2680
= 2540
= 3730
= 2090
= 3830
= 2190
= 2520
= 2490
= 1920
= 3540
= 3180
= 3640

5. From the paths available it is clear that the three shortest paths are as follow:
= 1930
= 1920
= 1910
From the paths available it is clear that the three shortest paths are as follow:
Node1 - Node4 - Node5 - Node9 - Node11 - Node14
Node1 - Node4 - Node5 - Node9 - Node12 - Node14
Node1 - Node4 - Node5 - Node9 - Node13 - Node14
They all have the following first three nodes in common:
[Node1 Node4 Node5 Node9]

6. [Node1 Node4 Node5 Node9 Node13 Node14]
The single shortest path can be deducted by inspection.
= 1910
[l14, l45, l59, l913, l1314]
[Node1 Node4 Node5 Node9 Node13 Node14]
The Shortest Path length (distance) from node 1 to node 14 is thus 1910m.