I declare a macro which returns a float bounded by [-2.0,2.0].
I multiply the value of this macro by  0.05 and add it to 0.91.
What is the upper limit of the sum?

The variable of interest below is "brakepct"

//CODE BELOW HELPS YOU UNDERSTAND MY PREDICAMENT

#declare years=0;
#while(years<1)          //

#declare RRR=seed(years+99+0) ;             //note years+99 is base
#macro rdr()
(rand(RRR)-0.5)+(rand(RRR)-0.5)+(rand(RRR)-0.5)+(rand(RRR)-0.5)
#end

#declare wallbraketurns=24.5+18.0*rdr();
#declare wallaccelturns=61.0+30.0*rdr();
#declare carbraketurns= 14.0+8.00*rdr();
#declare caraccelturns= 38.0+18.0*rdr();
#declare steerfactor=   1.20+0.29*rdr();
#declare brakepct=      0.91+0.05*rdr();
#declare accelboost=    0.93+0.41*rdr();
#declare VMAX=          6.60+2.80*rdr();
#declare otherang=      8.00+5.00*rdr();

#debug "\n The value for brakepct is "
#debug str(brakepct,7,4)
#debug "  \n"

#debug "\n brakepct is calculated by 0.91+0.05*rdr()"

// now here's a routine to investigate the upper limit of rdr();

#declare counter=0;
#declare maxrdr=0;
#while(counter<10000)
        #declare n=rdr();
        #if(maxrdr<n)
                #declare maxrdr=n;
        #end
#declare counter=counter+1;
#end

#debug "\n 10000 iterations of rdr meanwhile gives a max of  only :"
#debug str(maxrdr,6,4)
#debug "\n   "
#debug "This value approaches 2.0, the expected value."


#declare brakepct2=      0.91+0.05*maxrdr;
#debug "\n \n So the max value of brakepct should be slightly higher
than:"
#debug str(brakepct2,6,4)
#debug "\n   "

#declare years=years+1;
#end

camera {
  location <0, -65,-0>
  look_at  <0, 0, 0>
  angle 30
}
light_source{<10,-200,50>rgb .2}

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