So far we have dealt, simultaneously, with both methods of data processing
corresponding to the different criteria for classifying the plots according to
their ability to host a new school.
Both methods have resulted in some differences, but it is true that with
a little effort, the classical method could be improved and, therefore, we
could get equivalent results.
Now, we will attack the final phase, the aggregations of the four criteria,
and there we will face much more significant differences.
Firstly, let’s see what is proposed by the classical method, following
with the Spatial Analyst ESRI tutorial.
How to merge aptitude maps
The final step in the aptitude model is to consolidate the reclassified Distance
outputs (aptitude maps) vs. recreational facilities, Distance vs. existing schools,
Slope and Land Use.
To take into account that some goals weight more than others in the
aptitude model, you can weight the datasets. You provide to the larger datasets
a higher percentage of weighting. If all datasets are of equal importance, you can
provide the same weighting to all of them.
In our example, you know in advance given the problem examination, that
the first objective is to locate the school near recreation facilities, and the
second is to set it apart from existing schools. The following weighting
percentages will be attributed to the aptitude maps. Values in parentheses correspond
to the percentage divided by 100 to normalize values. These are attributed to
each aptitude map:
- Distance to recreational
- Distance to schools: 25%
- Slope: 12.5%
- Land use types: 12.5%
The Distance vs. Recreational Facilities aptitude map has
an weight of 50% (.5) over the final result, and the Distance from Schools
map has a weight of 25% (.25). Both Slope and land use types have
a weight of 12.5% (0.125). As when assigning aptitude scales, assigning
weights is a subjective process that depends on the most important goals of
What you have just read seems logical, and even clever. Do not be
offended if you’re a fan of weights, but let’s try to see a little deeper.
What does it mean to say this mathematical transformation of a criterion
in weight value?
And more important, remember that the recipient of the result will necessarily
be a decision maker to whom we will provide a map to help him make his
We understand that the distance criteria from existing schools and recreation centres
are more important for the decision-maker than the type of terrain (slope and
By applying these weights we will (we believe) respect the decision maker
So, as an example is better than a long speech, let’s discuss the
following two examples:
1- A site that is excellent according to the three criteria: distance to
schools (value = 10), distance to recreation centres (value = 10) and slope
(value = 10), will have a cumulative result for the three criteria (criterion
value multiplied by the weight):
0.5 x 10 + 0.25 x 10 + 0.125 x 10
if land use is the best (agriculture = 10) we will have as final result
8.75 + 1.25 = 10
if land use is the worst (built = 3) we will have as final result
8.75 + (0.125 x 3) = 9.125
Do you reckon the decision-maker can decide between 9,125 and 10? To the
point of realizing that the second choice involves major drawbacks? Do you
think it appears at first glance?
So let’s consider an even more obvious example:
2- A site that is excellent according to the three criteria: distance to
schools (value = 10), distance to recreation centres (value = 10) and land
occupation (value = 10), it will have the following cumulative result for the
0.5 x 10 + 0.25 x 10 + 0.125 x 10
if the slope is 30 ° it will have a value of 7. The final result will be
8.75 + (0.125 x 7) = 9.625
Here, we border on perfection. We are, almost, at the ideal site!
Yet a slope of 30 ° corresponds to a slope percentage of 57% which is
more than double the suitable limit for a construction project!
How do you think such a proposal will be perceived by the decision
Then, while crossing the criteria we will exclude values from the
Slope criterion, with the Restricted option for the weighted overlay. This will
prevent proposing sites with slopes of more than 30%. Except that now, the only
acceptable values of the criterion will be 10, 9 and 8, others being
excluded. And therefore, all the other criteria being constant, between the
best and worst slope, the final result will vary only 0.25 !!!
Contribution of slope criterion to total if slope = 10 -> 10 * 0.125
Contribution of slope criterion to total if slope = 8-> 8 * 0.125 =
You can try to change the weights as much as you want, we will always
find a mismatch of results with human reasoning. Simply because we do not
weight criteria in our heads. What makes
matters more confusing is the fact that we give different importance to
criteria and assign them a ciphered weight.
We will come back later on. First, let’s finish the exercise with
Weighted Overlay with Spatial Analyst
Open the Weighted Overlay tool. (Spatial Analyst
Tools -> Overlay -> Weighted Overlay)
- Type 1,
10, and 1 in the From, To, and By text boxes . The
default rating scale is 1 at 9 in steps of 1. A scale of 1
at 10 was used when reclassifying the datasets, before adding input
rasters to the Weighted Overlay tool, so you must set the rating scales of 1
at 10 in steps of 1. This means that you avoid having to update
the scale values after adding your input datasets.
- Click on
- Add the SlopeReclass
layer to the Weighted Overlay tool .
- Click the
Add Raster Line button .
- For the input
Raster , select SlopeReclass in the
drop-down list and leave the input field set to Value .
- Click OK
- Click the
The raster is added to the weighted Overlay
Table. The Field column displays the values of the SlopeReclass
data . The Scale Value column replicates the Field column because
the Evaluation Scale has been set to include the range of values for each
input raster. At this point, you can change the Scale Values for each
class, but for this entry the values have already been properly weighted
during the reclassification.
Repeat the previous step for each of the reclassified datasets: LanduseReclass
, DistSchoolsReclass, and DistRecSitesReclass .
For the SlopeeReclass entry , in the Scale Value column , click
the cell that has the value 1.
Click the drop-down arrow, scroll down the view, and then click Restricted
In this way, whatever the values of the other layers, the resultant
value will be 0.
Repeat this for values 2 and 3.
For the LandUseReclass layer , set the Scale Value that
represents Water and Wetlands (0) to Restricted .
Now, you have to assign a weighting percentage to each raster, based on
the importance (or weighting) of each in the final aptitude map.
In the % Influence column , enter the percentages for each of the
DistSchoolsReclass out of 25
DistRecSitesReclass out of 50
SlopeReclass of 13
LlandUseReclass of 12
Click OK to run the tool.
The final result, our aptitude map, appears as follows:
The blue areas are the excluded areas (from the weighted overlay
Green areas are the most suitable areas.
In the next article we will perform these operations using the flexible
aggregation command, applying fuzzy logic instead of the weights used in