Originally posted by: shatha

hello, the model below is a transportation model using lingo 9.0, but when solving it the lingo keeps giving me "unexpected jacobian overflow, can anyone help?
MODEL:
! A 3 Warehouse, 4 Customer
Transportation Problem;

SETS:
country / morroco, poland, ksa/ :available_veh1, sc ;
warehouse / marka, layan/ :available_veh2;
CUSTOMER / carrefoure, safeway, cozmo/;
item / cheese, cornd_beef, milk/;
time/ 1,2/;
transfer1 ( country, warehouse, item, time) : quantity1;
transfer2 (warehouse, customer, item, time) : quantity2;
pair1 (country, item,time) : supply, price;
pair2 (customer, item, time) : demand;
pair3 (warehouse, item) ;
pair4 (customer, item);
link1 (country,warehouse, time) :cost1, travel_time1, arrival_time1;
link2 (warehouse,customer, time) :cost2, travel_time2, arrival_time2;
link3 (country, time) :required_veh1;
link4 (warehouse, time) :required_veh2;

ENDSETS

! The objective;
OBJ MIN = 0.75*(@SUM( transfer1(i,j,r,t):(sc(i))+ ((COST1(i,j,t)+price(i,r,t)) * quantity1(i,j,r,t))+ (fixed1 * required_veh1(i,t)))
+ @SUM( transfer2(j,k,r,t): (COST2(j,k,t) * quantity2(j,k,r,t))+ (fixed2 * required_veh2(j,t))))
+ 0.25*(@SUM( link1(i,j,t): arrival_time1(i,j,t)) + @SUM ( link2(j,k,t):arrival_time2(j,k,t)));

! The demand constraints;
@FOR( pair2(k,r,t): DEM
@SUM( warehouse(j): quantity2( j,k,r,t)) >=
DEMAND( k,r,t));

! The supply constraints;
@FOR( pair1(i,r,t): SUP
@SUM( warehouse( j): quantity1( i,j,r,t)) <=
supply( i,r,t));

!The flow conservation;
@sum (transfer1(i,j,r,t): quantity1(i,j,r,t)) = @sum (transfer2(j,k,r,t): quantity2(j,k,r,t));

! The vehicles requirement;
@for (link3(i,t):vehicles1
required_veh1= @if ((@sum (pair3(j,r): quantity1(i,j,r,t))/capacity1) #LE# 1, 1,
@sum (pair3(j,r): quantity1(i,j,r,t))/capacity1));

@for ( link3(i,t):
required_veh1 <= available_veh1);
@for (link4(j,t):vehicles2
required_veh2= @if ((@sum (pair4(k,r): quantity2(j,k,r,t))/capacity2) #LE# 1, 1,
@sum (pair4(k,r): quantity2(j,k,r,t))/capacity2));

@for ( link4(j,t):
required_veh2 <= available_veh2);

!arrival time;

@for (link1(i,j,t): arvl1
arrival_time1= @if(@sum(item(r): quantity1(i,j,r,t)) #GT# 0, travel_time1(i,j,t), 0));

@for (link2(j,k,t): arvl2
arrival_time2= @if(@sum(item(r): quantity2(j,k,r,t)) #GT# 0, travel_time2(j,k,t), 0));

!integer values;
@for (link3(i,t): @GIN (required_veh1(i,t)));
@for (link4(j,t):@GIN (required_veh2(j,t)));

@for (transfer1(i,j,r,t): @GIN (quantity1(i,j,r,t)));
@for (transfer2(j,k,r,t):@GIN (quantity2(j,k,r,t)));

! Here are the parameters;
DATA:
sc = 2000 4000 2000;

travel_time1 = 115 227
763 236
353 165;

travel_time2= 2.5 4 1.5
4 1.5 6;
supply = 4000 2000 4000
4000 2000 4000
2000 1000 3000
4000 2000 4000
4000 2000 4000
2000 1000 3000
;

DEMAND = 2000 1000 2500
1700 850 2125
1530 522 1915
2000 1000 2500
1700 850 2125
1530 522 1915
;

COST1 = 0.49 0.5
0.49 0.5
0.4 0.25;

cost2 = 0.5 0.85 0.5
0.7 1 1.2 ;
price= 1.189 0.55 4.3
1.56 0.50 4.1
1.2 0.47 4.06;

Fixed1=100;
Fixed2=70;

available_veh1=60;
available_veh2=25;

capacity1=10000;
capacity2=1000;

ENDDATA
END
#DecisionOptimization
#MathematicalProgramming-General