Example problem data
Data for the airlift scheduling model
airlift1.dat:
A: [(1 1) 24 (1 2) 14 (2 1) 49 (2 2) 29]
Aswitch: [(1,1,2) 19 (1,2,1) 29 (2,1,2) 18 (2,2,1) 26]
F: [7200 7200]
b: [(1,1) 50 (1,2) 75 (2,1) 60 (2,2) 40]
Cost: [(1,1) 7200 (1,2) 6000 (2,1) 7200 (2,2) 4000]
CostSwitch: [(1,1,2) 7000 (1,2,1) 8200 (2,1,2) 5500 (2,2,1) 8700]
cplus: [500 250]
cminus: [0 0]
DValues: [
(1,1) 927.758357
(1,2) 982.516248
(1,3) 961.404897
(1,4) 922.915716
(1,5) 986.342969
(1,6) 999.134104
(1,7) 970.324386
(1,8) 949.613106
(1,9) 991.773703
(1,10) 979.491162
(1,11) 979.679661
(1,12) 964.052640
(1,13) 957.691777
(1,14) 930.372603
(1,15) 933.799027
(1,16) 995.204085
(1,17) 957.344884
(1,18) 923.484318
(1,19) 959.026809
(1,20) 946.706588
(1,21) 991.897924
(1,22) 956.965721
(1,23) 981.616042
(1,24) 957.688286
(1,25) 1000.035618
(2,1) 1433.626750
(2,2) 1149.727635
(2,3) 1492.817415
(2,4) 1250.557154
(2,5) 1353.935057
(2,6) 1226.355338
(2,7) 1378.148830
(2,8) 1200.624255
(2,9) 1045.317699
(2,10) 899.933450
(2,11) 1439.677972
(2,12) 1170.804834
(2,13) 1474.838349
(2,14) 1572.684651
(2,15) 1207.662826
(2,16) 1368.931017
(2,17) 1327.981462
(2,18) 943.075132
(2,19) 1226.555028
(2,20) 1543.605354
(2,21) 1243.379144
(2,22) 1302.917735
(2,23) 1122.900897
(2,24) 1355.585501
(2,25) 1255.185201]
Probabilities: [0.04 0.04 0.04 0.04 0.04 0.04 0.04 0.04 0.04 0.04
0.04 0.04 0.04 0.04 0.04 0.04 0.04 0.04 0.04 0.04
0.04 0.04 0.04 0.04 0.04]
Data for the forest planning problem
forest.dat:
yield:
[0 0 16 107 217 275 298 306]
v: [320.3417 356.1874 398.4370 448.2349 506.9294 564.9294 587.9294 595.9294]
alpha: 0.9
beta: 1.1
delta: 0.905
gamma: 50
ProbDiscret:
[1.0000 0.0000 0.0000 0.4616 0.5384 0.0000 0.1847 0.2769 0.5384]
ValuesDiscret:
[0.06258 0.00000 0.00000 0.08612 0.04240 0.00000 0.10499 0.07354 0.04240]
s1: [241 125 1404 2004 9768 16385 2815 61995]
Asset liability management model
 |
Z* | ZEV | ZPI | Z*- ZEV | Z*- ZPI |
| 1.2 | -0.202383 | -0.622399 | -1.08452 | 0.420016 | 0.882137 |
|---|---|---|---|---|---|
| 1.25 | -0.0470575 | -0.484575 | -0.965948 | 0.4375175 | 0.9188905 |
| 1.3 | 0.108268 | -0.34675 | -0.847378 | 0.455018 | 0.955646 |
| 1.35 | 0.263594 | -0.208925 | -0.728808 | 0.472519 | 0.992402 |
| 1.4 | 0.418919 | -0.0711 | -0.610238 | 0.490019 | 1.029157 |
| stocks\ |
1.2 | 1.25 | 1.3 | 1.35 | 1.4 |
|---|---|---|---|---|---|
| 1 | 0 | 0 | 0 | 0 | 0 |
| 2 | 0.127183 | 0.132482 | 0.137781 | 0.143081 | 0.14838 |
| 3 | 0.128142 | 0.133481 | 0.13882 | 0.144159 | 0.149499 |
| 4 | 0.622716 | 0.648663 | 0.674609 | 0.700556 | 0.726502 |
| 5 | 0.007406 | 0.007715 | 0.008023 | 0.008332 | 0.008641 |
| 6 | 0 | 0 | 0 | 0 | 0 |
| 7 | 0 | 0 | 0 | 0 | 0 |
| 8 | 0 | 0 | 0 | 0 | 0 |
| 9 | 0.002992 | 0.003116 | 0.003241 | 0.003366 | 0.00349 |
| 10 | 0.003805 | 0.003964 | 0.004123 | 0.004281 | 0.00444 |
| 11 | 0.031253 | 0.032555 | 0.033857 | 0.03516 | 0.036462 |
| 12 | 0.000323 | 0.000336 | 0.000349 | 0.000363 | 0.000376 |
| m | No. of scenarios after deletion | Optimal solution |
|---|---|---|
| 1 | 209 | 0.0560891 |
| 1.5 | 1614 | 0.0849195 |
| 2 | 3357 | 0.0947222 |
| 2.5 | 4524 | 0.0995853 |
| 3 | 4899 | 0.107498 |
| 3.5 | 4975 | 0.107498 |
| 4 | 4995 | 0.108268 |
Code for deleting the `extreme' scenarios
The scenarios falling outside the intervals [
n
-m
n, n
+ mn]
can be deleted from the scenario tree by adding the following code
into the initial Mosel program after the line Spgentree and before
the model definition.
Declarations ! Statistical characteristics of the scenarios set
means, stdev: array(Stocks) of real
Eset: set of integer ! Set of "extreme" scenarios
m: real ! Length of the interval
end-declarations
initializations from 'means.dat'
means ! Mean returns of each of 12 stocks
end-initializations
initializations from 'stdev.dat'
stdev ! Standard deviations of the returns
end-initializations
Eset:={} ! Initialize the set of extreme scenarios as an empty set
m:=2
forall(s in Scen,p in Stocks)
if abs(Values(s,p)-means(p)) m*stdev(p) then
Eset:=Eset+{s} ! Update the set of extreme scenarios
break
end-if
spdelscen(Eset) ! Delete the set of extreme scenarios
Changing the value of m allows one to delete different numbers of scenarios. Instead of deleting the extreme scenarios, one can aggregate them using spaggregate(Eset) instead of spdelscen(Eset). In our case, the solutions of the problem with aggregated extreme scenarios are identical to the solutions with deleted extreme scenarios presented above.
ALM.dat:
Initial_Asset: 851826105 Initial_payment: 22170020 Total_wages: 211097880
Description of other data files
- data.dat: matrix with 500 rows and 12 columns, where each column represents 500 realizations of the return of a given stock
- data_L.dat: array of 500 elements representing the realizations of the amount of liabilities
- means.dat: array of 12 elements representing the mean return of each stock
- stdev.dat: array of 12 elements representing standard deviations of the returns.
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