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

facilities: 50% - 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

your study.

**Weighting analysis**

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

decision.

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

occupancy).

By applying these weights we will (we believe) respect the decision maker

reasoning.

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

= 8.75

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

three criteria:

0.5 x 10 + 0.25 x 10 + 0.125 x 10

= 8.75

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

maker?

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

= 1.25

Contribution of slope criterion to total if slope = 8-> 8 * 0.125 =

1.00

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

Spatial Analyst.

**Weighted Overlay with Spatial Analyst**

**Steps** **:**

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

Apply - Add the
*SlopeReclass*

layer to thetool .*Weighted Overlay*- Click the

button .*Add Raster Line*

- For the
, select*input*

Raster*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

input rasters:

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

Restricted command).

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

Spatial Analyst.

Thank you for this