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For the forecast rank computation in the 0-value singularity case, a special solution was developed. All the 0 ensemble member values (all below 0.1 m3/s) get an evenly-representing rank assigned from any of the percentiles that have 0 values (i.e. below 0.1 m3/s) in the model climatology. In practice, this will mean, the 'rank-undefined' section of the ensemble forecast is going to be spread evenly across the 'rank-undefined' section of the climatology during the rank computation. Figure 3 demonstrates the process on an idealised example, where the lowest 77 percentiles are 0 in the climatology and 23 out of 51 ensemble members are also 0 (see Figure 3a). The 23 ensemble members with 0 value then are spread across the 0-value range of the climatology from 1 to 77 (see Figure 3b). This way the ranks of the 23 members will be assigned from 1 to 77 with equal as possible spacing in between (see Figure 3c). Finally, the remaining non-zero ensemble members also get their ranks in the usual way, as described above in Figure 2. Finally, the schematic of ranks of all 51 members are provided in Figure 3d.
Figure 3. Schematic of the forecast extremity ranking calculation for areas with 0 river discharge values. In the extreme case of all climate percentiles being 0, which happen over river pixels of the driest places of the world, such as the Sahara, the ensemble forecast member ranks can either be 100 for any non-zero value, regardless of the magnitude of the river discharge, or the evenly spread ranks from 1 to 100, as a representation of the totally 0 climatology. In the absolute most extreme case of all 99 climate percentiles being 0 and all 51 members being 0 in the forecast, the ranks of the forecast will be from 1 to 100 in equal representation. This means, this forecast will be a perfect representation of the climatological distribution, or with another word a forecast showing climatologically expected probabilities for all anomaly categories from Extreme low to Extreme high. |
Expected forecast anomaly category computation
The ensemble forecasts have 51 members, which will be assigned an extremity rank each. Using these 51 ranks, the forecasts will be put in one of the 7 anomaly categories (as described in Table 1). This is done based on the arithmetic mean of the 51 ensemble member rank values (rank-mean) (see Figure 4). This rank-mean will also be a number between 1 and 100, but this time a real (not integer) number. If the anomaly is 50.5, that is exactly the normal (median) condition, i.e. no anomaly whatsoever. If the anomaly is below 50.5, then drier than normal conditions are forecast, if above 50.5, then wetter than normal. The lower/higher the anomaly value is below/above 50.5, the drier/wetter the conditions are predicted to be. The lowest/highest possible value is 1/100, if all ensemble members are 1/100 (the most extremely dry/wet). Then, based on this rank-mean, we define the expected forecast anomaly category (one of the 7 categories in Table 1) for the whole ensemble forecast, by placing the rank-mean into the right categories, as defined in Table 1 above. For example, all rank-mean values from 40.0 to 60.0, interpreted as 40.0<= <60.0, will be assigned to 'Normal', or category-4.
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Figure 4. Schematic of the forecast extremity ranking of the 51 ensemble members and the calculation of the expected forecast anomaly category for the whole ensemble.
Forecast uncertainty category computation
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In addition to the expected forecast anomaly computation for the whole ensemble, as one of 7 predefined categories, the forecast uncertainty is also represented in some of the sub-seasonal and seasonal products, namely on the new river 'Seasonal outlook - River network' and basin 'Seasonal outlook - Basin summary' products. The forecast uncertainty is defined by the standard deviation (std) of the ensemble member ranks (rank-std). If the ensemble member ranks cluster well, and the spread of the ranks is low, then the forecast uncertainty will be low and conversely the confidence will be high.
The standard deviation of the even distribution of ranks, with values ranging from 1 to 100, which is the perfect match to the climatologically expected spread, is going to be (100-1)/sqrt(12) = 28.86. On the other hand, the most extreme rank-std value is when half of the members are with rank 1 and the other half with rank 99 (perfectly split ensemble), in which case the rank-std is 49.5, while the lowest rank-std value is 0, when all ranks are the exact same, so there is no variability in the ranks at all.
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