IE-301_OptionalProject_Group2_Report

IE-301 Integer Linear Programming Modelling and Implementation
pg. 1
IE-301
Optional Project
Integer Linear Programming Modeling and
Implementation
Fall 2016
Group #3
Project Group Members:
Sarp Uzel 18184
Barışcan Savaş16590
Utku Suer 16531
Supervised By:Kerem Bülbül
Date:08/12/2016

pg. 2
Table of Contents
Abstract................................................................................................................................................... 3
Introduction............................................................................................................................................ 3
Part A....................................................................................................................................................... 6
Part B..................................................................................................................................................... 11
Part C..................................................................................................................................................... 15
APPENDIX.............................................................................................................................................. 20
Code for connecting the data file in part A:.......................................................................................... 20
Code for connecting the data file in part B and C:................................................................................ 20
References: ........................................................................................................................................... 20

pg. 3
Abstract
The Linear Integer Programming is a widely used method in optimization of warehouse
selection problems(or p-median problems or location-allocation) by many Industrial
Engineers.This project focuses on a p-median problem, where the project members were
assumed to imagine Turkey’s well known charity organzation require them to find the optimal
values for disturbing relief supplies from depots to districts.The members of the project group
were asked to minimize the total cost of disturbing relief items, which are supplied by Kızılay,
to 13 different disctricts located in the Asia part of Istanbul by selecting 5 depot points out of
13 districts and determining the number of supplies sent from 5 depots to 13 districts, in case
of a disaster. The core important parts which separate this project’s problem from classical p-
median problems can be listed as: the fact that the multi-sourcing is allowed, the fact that
uncertainity is concerned.The project group, first structured the mathematical model of the
Kızılay’s minimum cost warehouse selection problems in different scenerios and then
implemented the model by using IBM ILOG CPLEX Optimization software, in order to find the
solution of the Linear Integer Program.At the very end, the project group found the objective
values for different demands of Kızılay as well as the optimum solution for decision variables.
Introduction
In this project, project team assumed that Kızılay is aware of the importance of the natural
disasters, so they want to be ready to disturibute relief supplies that they have with respect
to the different demands of different districts, under different types of circumstances, in case
of a disaster. Not only do Kızılay require to distribute these items according to the demands

pg. 4
of the districts, but also they want to minimize the cost of disturbing these items. The problem
states that there are 13 districts that have be supplied by 5 depots whom locations will be
selected amoung the same 13 districts. All of the depot capacities for stocking relief supplies
are the same for each depot candidate. The demands of the districts are given.The distance
between two different districts is also provided as a 13x13 matrix.The depots can only serve
to districts that are at most 20 km far away. The cost of carrying one item for 1 km is 1 TL.
Every district must be assigned to atleast one depot. There is no fixed cost of opening a new
depot. Below you can find the description of the original problem that is assigned to the
project members:

pg. 5
The p-median problem is stated as:” Given a graph or a network G = (V, E), find Vp ⊆ V such that
|Vp| = p, where p may either be variable or fixed (see Section 2.3), and that the sum of the shortest
distances from the vertices in {V Vp} to their nearest vertex in Vp is minimized. In this section we
provide an extended problem definition and a unified notation scheme.”( J. Reese,2005).

pg. 6
The reason why the project group state that this problem is also a p-median problem is the fact that
it satisfies the requirements of p-median problems.
Here are the criteria of p-median problem: (Jamshidi,2009)
1-The cost is directly proportional to the distance.
2-There is no time limit.
3-Total number of warehouses is known.
4-There is no fixed cost for opening a warehouse.
5-All warehouses have the same properties.
6-The candidates for warehouses are also known.
Part A
The data for distance,capacity and demands were all provided by the supervisor.
Here is the data for distance between two districts:
Here is the capacity data:

pg. 7
Here is the data for demand:
First thing that the project team did was analyzing the data. By using the =MAX() funtion in excel,
project team members determined the possible maximum relief item sent from district i to district
j(i=1,2..,13; j=1,2..,13).
Thus, analyze resulted in 6880.

pg. 8
The second step was to construct the MILP(Mixed Integer Linear Program)
Parameters:
M=6880 –Possible maximum item can be send from one district the another
𝑑𝑖𝑠𝑡𝑖𝑗=distance difference between district i and j—provided in the excel sheet(i=1,2..,13; j=1,2..,13).
𝑐𝑖=capacity of candidate i for stocking relief supplies. —provided in the excel sheet(j=1,2..,13).
𝑑𝑒𝑚𝑗=demand of district j-—provided in the excel sheet(j=1,2..,13).
Decision Variables:
𝑌𝑖=1 if candidate i is a depot , 0 else. (j=1,2..,13).
𝑋𝑖𝑗=1 if district i can carry relief items to district j, 0 else. (i=1,2..,13; j=1,2..,13).
𝑅𝑖𝑗=# of supplies carried from district i to district j(i=1,2..,13; j=1,2..,13).
Objective Function:
Minimize:Z=∑ ∑ (𝑑𝑖𝑠𝑡𝑖𝑗 ∗ 𝑅𝑖𝑗
13
𝑗=0
13
𝑖=1 *1)
Constraints:
∑ 𝑌𝑖
13
𝑖=1 =5 –Number of depot constraint.
𝑋İ𝐽 ≤ 𝑌İ (i=1,2..,13; j=1,2..,13)---i can only carry items to j if and only if there is a depot in i
∑ 𝑋İ𝐽
13
𝑖=1 ≥ 1 (i=1,2..,13; j=1,2..,13)---j must be supplied by at least one depot
∑ 𝑅İ𝐽
13
𝑗=1 ≤ 𝑌İ ∗ 𝑐𝑖 (i=1,2..,13; j=1,2..,13)---i is capable of suppliying items with respect to its
capacity
𝑋İ𝐽 ≤ 20 ∗ 𝑑𝑖𝑠𝑡𝑖𝑗(i=1,2..,13; j=1,2..,13)---if the distance is more than 20 km i can’t serve j.
∑ 𝑅İ𝐽
13
𝑖=1 =𝑑𝑒𝑚𝑗 (i=1,2..,13; j=1,2..,13)---#number of items sent to j should be equal to demand of j
𝑅𝑖𝑗 ≤ 𝑀 ∗ 𝑋𝑖𝑗 –at most M number of items can be sent from i to j
𝑅𝑖𝑗 ≥0
𝑋İ𝐽 , 𝑌İ 𝜖(0,1)
Here is the CPLEX code:
int M=6880;//Maximum demand of one district
int NDistricts=13;//There are 13 districts that will be supplied
range I=1..NDistricts;//Set of candidates for depot locations
range J=1..NDistricts;//Set of districts
//Note that there are in total 13 districts and also 13 canditates
//Also note that "..." meaning that data is going to be chosen from the excel file
//in the following lines
int dist[i in I,j in J]=...;//Distance from location i to location j

pg. 9
int dem[j in J]=...;//Demand of location i for relief supplies
int c[i in I]=...;//Capacity of each candidate distribution points
dvar boolean y[i in I];//if candidate i is assigned as a depot it takes value "1"
dvar boolean X[i in I,j in J];//if location j is assigned to depot location i x
takes value "1"
dvar int+ R[i in I,j in J];//ammount of supply coming from dist point i to j
dvar int Z;//The objective function
minimize Z;
subject to{
Z==sum(i in I,j in J) dist[i,j]*R[i,j] ;//since the price of one relief supply
for per km travelled is one we can directly multiply demand by distance
sum(i in I)y[i]==5;//The project concerns to locate 5 depots
forall(i in I,j in J)
X[i,j]<=y[i];// i can only carry items to j if and only if there is a depot in i
forall(j in J)
sum(i in I)X[i,j]>=1;//Every district must be attained to a distribution
point(atleast)
forall(i in I)
sum(j in J)R[i,j]<=y[i]*c[i];//capacity constraint
forall(j in J)
sum(i in I)R[i,j]==dem[j];//demand of j should be satisfied
R[i,j]<=M*X[i,j];//maximum number of supply carried from i to j
X[i,j]*dist[i,j]<=20;//distance constraint
}

pg. 10
Here are the optimal solutions provided by IBM ILOG :
// solution (optimal) with objective 274980
// Quality Incumbent solution:
// MILP objective 2.7498000000e+005
// MILP solution norm |x| (Total, Max) 3.26606e+005 2.74980e+005
// MILP solution error (Ax=b) (Total, Max) 0.00000e+000 0.00000e+000
// MILP x bound error (Total, Max) 0.00000e+000 0.00000e+000
// MILP x integrality error (Total, Max) 0.00000e+000 0.00000e+000
// MILP slack bound error (Total, Max) 0.00000e+000 0.00000e+000
//
Z = 274980;
R = [[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[4190 0 220 4650 0 0 0 0 0 0 0 0 5400]
[0 0 0 0 4570 4870 0 0 0 0 2340 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 6810 0 3210 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 330 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 2490 2090 0 0 0 0 3540 0 0 0 6880 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]];
y = [0 0 0 1 1 0 1 0 0 1 0 1 0];
X = [[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 0 1 1 1 1 0 1 0 0 0 1 1]
[0 0 0 1 1 1 1 0 1 0 1 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 0 0 0 1 1 1 1 1 0 1 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 1 1 1 0 1 0 1 0 0 0 1 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0]];
Optimal objective function value:274980 TL
Above one can see that all of the demands given in the excel sheet was satisfied from the objective
value of R[i,j]( 𝑅𝑖𝑗).
Computational Time:01:44

pg. 11
Part B
In part B,there are different scenarios and each scenario must be satisfied.Kızılay do not want to risk
anything, hence our program is to determine the specific number of items sent from i to j which
satisfies all the scenarios.There are 20 different scenarios in this part; in addition , all must be
satisfied.Here are the demand data for different scenarios:
The maximum number of demand again is calculated, in order to decide the possible maximum
number of item distribution.

pg. 12
Parameters:
MS=7377 –Possible maximum item can be send from one district the another
𝑢𝑐𝑑𝑒𝑚𝑗𝑠= uncertain demand of district j with respect to scenario s-—provided in the excel
sheet(j=1,2..,13,s=1,2,3,..20).
Decision Variables:
Objective Function:
13
𝑗=0
13
𝑖=1 *1)
Constraints:
∑ 𝑌𝑖
13
𝑖=1 =5 –Number of depot constraint.
∑ 𝑋İ𝐽
13
∑ 𝑅İ𝐽
13
capacity
∑ 𝑅İ𝐽
13
𝑖=1 ≥ 𝑢𝑐𝑑𝑒𝑚𝑗𝑠 (i=1,2..,13; j=1,2..,13; s=1,2,3...20)---in order to satisfy all demands under every
scenario
𝑅𝑖𝑗 ≤ 𝑀𝑆 ∗ 𝑋𝑖𝑗 –at most MS number of items can be sent from i to j
𝑅𝑖𝑗 ≥0
𝑋İ𝐽 , 𝑌İ 𝜖(0,1)

pg. 13
int Nscenarios=20;//Number of Scenarios
range S=1..Nscenarios;//Set of Scenarios
int ucdemand[j in J,s in S]=...;
int MS=7377;
takes value "1"
minimize Z;
subject to{
X[i,j]<=y[i];
forall(j in J)
sum(i in I)X[i,j]>=1;//Every district must be attained to a distribution
point(atleast)
forall(i in I)
sum(j in J)R[i,j]<=y[i]*c[i]; //capacity constraint
forall(j in J,s in S)
sum(i in I)R[i,j]>=ucdemand[j,s]; //demand constraint
R[i,j]<=MS*X[i,j];//possible maximum number of supply sent

pg. 14
}
Here is the report:
//
Z = 311658;
R = [[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[4347 0 0 4962 0 0 0 0 0 0 0 0 5691]
[0 0 0 0 4899 5191 0 0 0 0 2711 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[240 0 0 0 0 0 7228 1665 3553 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 732 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 2680 2754 0 0 0 0 2189 0 0 0 7377 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]];
y = [0 0 0 1 1 0 1 0 0 1 0 1 0];
X = [[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 0 1 1 1 1 0 1 0 0 0 1 1]
[0 0 0 1 1 1 1 0 1 0 1 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 0 0 0 1 1 1 1 1 0 1 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 1 1 1 0 1 0 1 0 0 0 1 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0]];
Optimal objective value is=311658
Computational Time: 01:70
It is no surprise that summation of the elements in the columns of R matrix is equal to the maximum
demand for the corresponding district under among all scenarios.This is the only way to satisfy all the
scenarios. Part B will give the maximum optimal cost for this problem, since the program collects the
maximum demand for each district out of 20 scenarios, in order to satisfy all the scenarios.This is why
the optimal objective value is higher than the part a. Moreover; computational time increased when
part b is compared to the part a which means that the software spent more computational effort.This is

pg. 15
just because adding new constraint **∑ 𝑅İ𝐽
13
𝑖=1 ≥ 𝑢𝑐𝑑𝑒𝑚𝑗𝑠**. This constraint definetely increased the
complexity of the problem.
For(j=1;j<=13;i++)
For(s=1;s<=10;s++)
For(i=1;i<=13;i++)
The complexity of this psedue-code which is the representation of the **∑ 𝑅İ𝐽
13
𝑖=1 ≥ 𝑢𝑐𝑑𝑒𝑚𝑗𝑠**
constraint is basically j*s*i.
The interesting part is optimal depot locations are the same as part a. And also optimal values for 𝑋İ𝐽’s
Are the same.This explains why the cost is higher than part a better.Since, matrix R is the multiplied
version of matrix X. Since X matrices are same for part a and b, also the summation of the columns
are higher than part a the optimal objective function value is also greater than part a. Obviously
number of items sent from i to j in part b is greater than the items sent from i to j in part a, hence the
optimal solution of part b’s objective function is greater,since they have the same depot locations
and same assignments for depots.
Part C
In this part, the uncertain data did not change.However, in the first phase of the question, it is stated
that Kızılay is optimistic and it is enough to satisfy 80 percent of the scenarios.In the second phase we
will try the same problem with 60 percent satisfaction of scenarios.
Parameters:
MS=7377 –Possible maximum item can be send from one district the another
𝑢𝑐𝑑𝑒𝑚𝑗𝑠= uncertain demand of district j with respect to scenario s-—provided in the excel
sheet(j=1,2..,13,s=1,2,3,..20).

pg. 16
Decision Variables:
𝑆𝑐𝑛 𝑠=If scenario s is satisfied it returns 1; 0 otherwise (s=1,2,3...20)
Objective Function:
13
𝑗=0
13
𝑖=1 *1)
Constraints:
∑ 𝑌𝑖
13
𝑖=1 =5 --Number of depot constraint.
∑ 𝑋İ𝐽
13
∑ 𝑅İ𝐽
13
capacity
∑ 𝑅İ𝐽
13
𝑖=1 ≥ 𝑢𝑐𝑑𝑒𝑚𝑗𝑠 *𝑆𝑐𝑛 𝑠 (i=1,2..,13; j=1,2..,13; s=1,2,3...20)---in order to satisfy all demands
under every scenario
𝑅𝑖𝑗 ≤ 𝑀𝑆 ∗ 𝑋𝑖𝑗 –at most MS number of items can be sent from i to j
∑13
𝑖=1 𝑆𝑐𝑛 𝑠=16 (s=1,2,3...20)
𝑅𝑖𝑗 ≥0 ,𝑋İ𝐽 , 𝑌İ 𝑆𝑐𝑛 𝑠 𝜖(0,1) (i=1,2..,13; j=1,2..,13; s=1,2,3...20)
int Nscenarios=20;//Number of Scenarios
range S=1..Nscenarios;//Set of Scenarios
int ucdemand[j in J,s in S]=...;
int MS=7377;
takes value "1"

pg. 17
dvar boolean Scn[s in S];//If scene s is satisfied then it returns 1
minimize Z;
subject to{
X[i,j]<=y[i];// i can serve j if and only if there is a depot in i
forall(j in J)
sum(i in I)X[i,j]>=1;//Every district must be assigned to a distribution
point(atleast)
forall(i in I)
sum(j in J)R[i,j]<=y[i]*c[i];//capacity constraint
forall(j in J,s in S)
sum(i in I)R[i,j]>=Scn[s]*ucdemand[j,s];//demand contraint of the satisfied
scenarios
sum(s in S)Scn[s]==16;//80 percent satisfaction of scenarios
R[i,j]<=MS*X[i,j];//possible maximum number of carrying items from i to j
}

pg. 18
Here is the report:
//
Z = 303513;
R = [[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[4347 0 0 4962 0 0 0 0 0 0 0 0 5691]
[0 0 0 0 4899 5191 0 0 0 0 2711 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[24 0 0 0 0 0 7228 1440 3553 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 7377 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 2680 2754 0 0 0 0 2189 0 0 0 7377 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]];
y = [0 0 0 1 1 0 1 0 0 1 0 1 0];
X = [[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 0 1 1 1 1 0 1 0 0 0 1 1]
[0 0 0 1 1 1 1 0 1 0 1 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 0 0 0 1 1 1 1 1 0 1 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 1 1 1 0 1 0 1 0 0 0 1 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0]];
Scn = [0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0];
Optimal objective value when p=0.8 : 303513 TL
***Note that R[10,10] is 7377 ; however, in demand/scenario data there is no demand as 7377 for
Şile under any scenario. But, this is not going to change anything, since the distance between Şile and
Şile is 0.Moreover; Şile is more than 20 km far away from any district, so it cannot send any items to
any district and also there must be a depot in Şile to satisfy the demand of Şile. Therefore, Şile can

pg. 19
use all of the capacity it has just for itself with 0 cost. Also note that, scenarios 1,3,19,20 can not be
satisfied.
For the second part of the part c, all we need to modify line 59:
** sum(s in S)Scn[s]==12;//60 percent satisfaction of scenarios **(Make the
summation of scenarios satisfied equal to 12 since it is enough)
Here are the results:
// solution (integer optimal, tolerance) with objective 296732
//
Z = 296732;
R = [[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[4311 0 0 4962 0 0 0 0 0 0 0 0 5691]
[0 0 0 0 4899 5098 0 0 0 0 2602 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 7228 955 3553 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 7377 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 2680 2623 0 0 0 0 2674 0 0 0 7023 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]];
y = [0 0 0 1 1 0 1 0 0 1 0 1 0];
X = [[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 0 1 1 1 1 0 1 0 0 0 1 1]
[0 0 0 1 1 1 1 0 1 0 1 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 0 0 0 1 1 1 1 1 0 1 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0]
[1 1 1 1 0 1 0 1 0 0 0 1 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0]];
Scn = [0 1 0 0 1 1 1 0 1 1 1 1 0 1 0 1 1 1 0 0];
Optimal value for objective function :296732TL

pg. 20
Note that, scenarios 1,3,4,8,13,15,19,20 can not be satisfied.
The optimal solutions for p=0.8 p=0.6 and p=1 are totally different. p=1>p=0.8>p=0.6. The main
reason is when the all scenarios do not necessarily have to be satisfied, Cplex eliminates the
scenarios which cost more ,since the objective funtion tries to minimize the cost.Number of items
sent from i to j decreased for almost every element of the R matrix as p gets closer to 0. For instance
when p=0.8 ammount of supply sent from 4 to 1 was 4371 but when p was P=0.6 of supply sent
from 4 to 1 became 4311. This is what Cplex handles, as we decrese the size of the set of the feasible
region for a minimization problem by relaxing the constraints, we are more prone to observe lower
values for optimal objective funtion solution. The depot locations and their assignments in part c is
again the same as part a and b.
APPENDIX
Code for connecting the data file in part A:
SheetConnection sheet("KizilayData.xlsx");
dist from SheetRead(sheet,"Distances!B2:N14");
dem from SheetRead(sheet,"Demands_CaseA!B2:B14");
c from SheetRead(sheet,"Capacities!B2:B14");
Code for connecting the data file in part B and C:
SheetConnection sheet("KizilayData.xlsx");
dist from SheetRead(sheet,"Distances!B2:N14");
c from SheetRead(sheet,"Capacities!B2:B14");
ucdemand from SheetRead(sheet,"Demands_CaseB_C!B2:U14");
References:
Jamshidi, M. (2009). “Median Location Problem”, Facility Location: Concepts, Models, Algorithms and
Case Studies, Ed. by. R.Z. Farahani and M. Hekmatfar, PhysicaVerlag Heidelberg, pp.177-191

pg. 21
J. Reese(2005) ” Methods for Solving the p-Median Problem: An Annotated Bibliography” pp.2-3

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