1、P94，例3.4 選址問題目錄題目1第一步，舊址基礎上只求運量的程序1第二步，舊址基礎上選擇新址的NLP程序2題目6個工地的地址(坐標表示，距離單位KM)及水泥用量(單位：噸)如下表，而在P(5,1)及Q(2,7)處有兩個臨時料場，日儲量各有20t，如何安排運輸，可使總的噸公里數最小?新料場應選何處？能節約多少噸公里數？123456A1.258.750.55.7537.25B1.250.754.7556.57.75D3547611第一步，舊址基礎上只求運量的程序MODEL:Title Location Problem;sets: demand/1.6/:a,b,d; supply/1.2/:x

2、,y,e; link(demand,supply):c;endsetsdata:!locations for the demand(需求點的位置);a=1.25,8.75,0.5,5.75,3,7.25;b=1.25,0.75,4.75,5,6.5,7.75;!quantities of the demand and supply（供需量）;d=3,5,4,7,6,11; e=20,20;x,y=5,1,2,7;enddatainit:!initial locations for the supply（初始點）;endinit!Objective function（目標）;OBJ min=su

3、m(link(i,j): c(i,j)*(x(j)-a(i)2+(y(j)-b(i)2)(1/2) );!demand constraints（需求約束）;for(demand(i):DEMAND_CON sum(supply(j):c(i,j) =d(i););!supply constraints（供應約束）;for(supply(i):SUPPLY_CON sum(demand(j):c(j,i) <=e(i); );!for(supply: free(x); !free(Y); !);for(supply: bnd(0.5,X,8.75); bnd(0.75,Y,7.75); )

4、;END運行可得到全局最優解 Global optimal solution found. Objective value: 136.2275 Total solver iterations: 1 Model Title: Location Problem Variable Value Reduced Cost X( 1) 5.000000 0.000000 X( 2) 2.000000 0.000000 Y( 1) 1.000000 0.000000 Y( 2) 7.000000 0.000000 E( 1) 20.00000 0.000000 E( 2) 20.00000 0.000000

5、第二步，舊址基礎上選擇新址的NLP程序!選新址的程序;MODEL:Title Location Problem;sets: demand/1.6/:a,b,d; supply/1.2/:x,y,e; link(demand,supply):c;endsetsdata:!locations for the demand(需求點的位置);a=1.25,8.75,0.5,5.75,3,7.25;b=1.25,0.75,4.75,5,6.5,7.75;!quantities of the demand and supply（供需量）;d=3,5,4,7,6,11; e=20,20;enddataini

6、t:!initial locations for the supply（初始點）;!x,y=5,1,2,7;endinit!Objective function（目標）;OBJ min=sum(link(i,j): c(i,j)*(x(j)-a(i)2+(y(j)-b(i)2)(1/2) );!demand constraints（需求約束）;for(demand(i):DEMAND_CON sum(supply(j):c(i,j) =d(i););!supply constraints（供應約束）;for(supply(i):SUPPLY_CON sum(demand(j):c(j,i) &

7、lt;=e(i); );!for(supply: free(x); !free(Y); !);for(supply: bnd(0.5,X,8.75); bnd(0.75,Y,7.75); );END求解結果只得到局部最優解 Local optimal solution found. Objective value: 89.88347 Total solver iterations: 67 Model Title: Location Problem Variable Value Reduced Cost X( 1) 5.695966 0.000000 X( 2) 7.250000 -0.3212

8、138E-05 Y( 1) 4.928558 0.000000 Y( 2) 7.750000 -0.1009767E-05如果不要初始數據，可能計算時間更長，本例的結果更優：Local optimal solution found. Objective value: 85.26604 Total solver iterations: 29 Model Title: Location Problem Variable Value Reduced Cost X( 1) 3.254883 0.000000 X( 2) 7.250000 -0.2958858E-05 Y( 1) 5.652332 0.000000 Y( 2) 7.750000 -0.1114154E-05如果想求全局最優解，結果將會出現如下錯誤版本限制，但會得到一個的局部最優解，結果與不要初始數據時算出的結果一樣。 Local optimal solution found. Objective value: 85.26604 Extended solver steps: 18 Total solver iterations: 16121 Model Title: Location Pro