Tutorial 10 - Solutions (Operation scheduling)

Universitätsprofessor
Dr.-Ing. Bernd Hellingrath
Lehrstuhl für Wirtschaftsinformatik
und Logistik

Leonardo-Campus 3
48149 Münster

Tel. +49 251 83-38000
Fax +49 251 83-38009





Tutorial Operations Management
Please note: Round off to two digits, if not indicated otherwise.

Operation scheduling and production control

Exercise 1: Route Planning
The initial route costs in total: 207 + 308 + 667 + 637 + 194 + 553 = 2566 [€].
Δij,nm = -cij - cnm + cin + cjm
Formula cost changes:

1. Iteration: Calculation of possible cost savings through pairwise permutation.
From \ To 01 12 23 34 45 50
01 -226 8 422
12 -289 354 -229
23 -62 -270
34 -208
45
50

Please note that we have only calculated all necessary values in this and the following tables.
Since we assume symmetric costs, a calculation of pairwise permutations below the table-
diagonal would lead to identical values. Therefore, a double calculation is unnecessary. Fur-
thermore, a calculation of pairwise permutations of connected route sections (for example a
change of 01 ↔ 12) does not make sense, as this step wouldn’t cause a change of route (and
therefore costs wouldn’t change).

We use pairwise permutation, which leads to a maximal cost reduction. Through pairwise
permutation of route 12 (Kassel-Hamburg) and route 34 (Stuttgart-Berlin), the reduced costs
are of 289 €. Consequently, our new route costs 2566 − 289 = 2277 [€] and goes from
Münster via Kassel, Stuttgart, Hamburg, Berlin and Dresden back to Münster.

, 2


2. Iteration: Calculation of possible cost reductions through pairwise permutation.
From \
To 01 13 32 24 45 50
01 -93 164 422
13 289 328 -96
32 253 -429
24 107
45
50

After the second iteration it becomes clear that a pairwise permutation of route 32 (Stuttgart-
Hamburg) and route 50 (Dresden-Münster) reduces costs of 429 €. Hence, the new route goes
from Münster, via Kassel, Stuttgart, Dresden, Berlin and Hamburg back to Münster and costs
2277 − 429 = 1848 [€] in total.

3. Iteration: Calculation of possible cost reductions through pairwise permutation.
From \ To 01 13 35 54 42 20
01 107 531 284
13 427 390 133
35 315 429
54 474
42
20

There are no further possible cost reductions. Consequently, the algorithm stops at this point.
Hence, the optimal route is found (Münster - Kassel - Stuttgart - Dresden - Berlin - Hamburg -
Münster) with costs of 1848 €.

Costs of the initial route: 2566 [€]

Costs of the optimal route: 1848 [€]

Therefore, a total of 718 € can be saved.