ORANGE GROVE MANAGEMENT PROJECT

MANAGEMENT SCIENCE SENIOR DESIGN

MAY 9, 2008

RAFAEL ALVAREZ, SYLIA GALLEGOS, JUAN CARLOS GONZALEZ, SKY NOYD Table of Contents Page I. Management Summary 4 II. Background and Description of the Problem Situation 5 III. Analysis of the Situation 7 IV. Technical Description of the Model 9 V. Analysis and Managerial Interpretation 11 VI. Conclusions and Critique 15 VII. Appendix a. Tardias Model without Irrigation 16 b. Tardias Model with Irrigation 17 c. Tardias Data 18 d. Mayeras Model without Irrigation 19 e. Mayeras Model with Irrigation 20 f. Mayeras Data 21 g. Navels Model without Irrigation 22 h. Navels Model with Irrigation 23 i. Navels Data 24 j. Nova Model without Irrigation 25 k. Nova Model with Irrigation 26 l. Nova Data 27 m. Zarzuma Model without Irrigation 28 n. Zarzuma Model with Irrigation 29 o. Zarzuma Data 30 p. Delicias Model without Irrigation 31 q. Delicias Model with Irrigation 32 r. Delicias Data 33 s. Mars Model without Irrigation 34 t. Mars Model with Irrigation 35 u. Mars Data 36 v. Results: Tardias Model without irrigation 37 w. Results: Tardias Model with Irrigation 38 x. Results: Mayeras Model without Irrigation 39 y. Results: Mayeras Model with Irrigation 40 z. Results: Navelas Model without Irrigation 41 aa. Results: Navelas Model with Irrigation 42 bb. Results: Nova Model without Irrigation 43 cc. Results: Nova Model with Irrigation 44 dd. Results: Zarzuma Model without Irrigation 45 ee. Results: Zarzuma Model with irrigation 46 ff. Results: Delicias Model without Irrigation 47 gg. Results: Delicias Model with Irrigation 48 hh. Results: Mars Model without Irrigation 49 ii. Results: Mars Model with Irrigation 50

2 | P a g e The Vision

“My family has owned an orange grove in Veracruz, Mexico for a number of years.

Since I was a boy, I dreamed of managing the family orange grove. My dream has finally become a reality. Now I want to transform the orange grove into a family business of unparalleled management where not only the passion exists, but the operation is the most efficient it can be.”

– JUAN CARLOS GONZALEZ

3 | P a g e EXECUTIVE SUMMARY For this project we will be representing both the client and the consultant. This is possible because Juan is the owner of an orange grove in Veracruz, Mexico and he is interested in maximizing his profit by analyzing the revenues and costs of production for his orange grove. The orange grove consists of nearly 4,000 trees of which there are 6 different kinds of trees and 7 different kinds of oranges. In making the model, we will consider each type of tree, and consequently fruit, as independent of one another. This will help the model maximize the profit for each type of tree. The profit from each tree will then be added to determine the total profit. The model will indicate the best choice to follow in order to have a maximum profit. A maximum profit includes minimal cost at the end of each month and provides the best choice of selling the fruit (on the tree, weight station or wholesaler). The maximum profit was determined by a linear programming model. The accuracy of the solution depends on the accuracy of the data being used. For the problem, we had limited information that may not necessarily be accurate. It is the quality and size of the orange that will determine the selling price. To maximize the quality and size of the oranges different cost factors were considered in determining the optimal solution. The factors included fertilizer, pesticide, maintenance, and tilling. Also, the model will determine whether or not an irrigation system should be installed. Since an irrigation system will benefit all the trees, a decision analysis was performed. To do this we developed 2 different sets of models; one without the irrigation system and one with the irrigation system. After running each model and compiling the results, we concluded that the best option was to install the irrigation system and use the following activities for each type of tree (see table below). Following this schedule would yield an estimated $132,791 in total profit. This amount includes the estimated $60,000 cost for the irrigation system. Decision Analysis Results with Irrigation Tree Type Fertilizer Pesticide Maintenance Trimming Tardias YES YES YES YES Mayeras YES YES YES YES Navels NO YES YES NO Nova NO YES YES NO Zarzuma NO YES YES YES Delicias NO YES YES NO Mars YES YES YES YES

BACKGROUND AND DESCRIPTION OF THE PROBLEM

4 | P a g e The purpose of the model is to find a real world situation where we can apply our engineering management science skills. The most important reason for the applications of management science models is to improve the organization and management of the Orange Grove. At the same time, to find the best opportunity to maximize profit where we analyze revenue and costs, and with the combination of these factors select the best profit decision by implementing a linear programming model. Furthermore, we want to find a better alternative system to organize the Orange Grove by recording expenses, revenues, number of employees, agricultural implements, labor costs etc. in order to have the best efficiently management impact.

The types of decisions involved in the Orange Grove are according to costs, lack of inventory, poor management control. First, we are going to organize all the information regarding costs, for example, labor cost, fertilize implementation cost, pesticide cost, tilling cost, maintenance cost, etc. For these costs we are deciding to implement excel and begin recording the costs, first, seasonally and then annually. Also, we are going to try to go in detail for each tree-cost according to the farm. Second, there will be the application of a linear program once the fruit is ready to sell. The reason for this will be to find the best way to sell it according to the option that we have which can be on the tree, weight station or wholesaler, including the costs of it. Finally, a forecasting model to identify the total yield in the farm and in addition the total orange quality A yield and quality B yield in order to document accurate profit.

The questions to be answered are:  How much is the annual cost?  What kind of cost are we having?  What is the best way to go for a better profit, once the fruit is ready for sale?  What is going to be the annual yield?  If there is an irrigation system how these numbers are going to change?

People that are going to be involved and affected positively by the problem being studied could be the owner and all the farmers since it is a program that can be implemented in each one in order to have a better inventory management system. Also, it can be employed by the state or government to support any study in the zone.

The orange grove is a 28.9 acre property, including 3,932 new orange trees made up of six different varieties: Tardias, Navels, Zarsuma, Delicias, Mars, and Nova. The orange grove is located in the city of Alamo Veracruz, Mexico. The farm was previously planted with orange trees that were 30 years old. Also, there are no records of revenues, costs, schedules, inventory or any type of management plan to manage the farm to obtain the maximum profit, hence the name “Orange Grove Management Project”.

5 | P a g e Project Location:

6 | P a g e ANALYSIS OF THE SITUATION

The problem presented to us was maximizing the profit while considering cost and revenue variables. Given the data, we decided to optimize the solution by using an integer programming model. To make the problem more complex, we decided to include a decision analysis on whether or not it would be best to install an irrigation system.

Cost factors we will be analyzing are:

 Fertilizer – used to increase the quality and size of the oranges o If used, fertilizer will result in a 70% probability of producing grade A oranges  Pesticide – used to prevent infections and increase the quality of the oranges o If used, pesticide will result in a 90% probability of producing grade A oranges *** note: if both fertilizer and pesticide are used, there is a 100% probability of producing grade A oranges  Maintenance – used to improve the quality of the tree, indirectly increases the quality and size of the oranges  Worker Salary (Picking) – depends on the time of year and the type of tree  Tilling – used to improve the quality of the tree, indirectly increase the quality of the fruit  Transportation (Shipping) – used in determining the optimal method of selling the oranges

Each of these factors will have to be applied to six types of orange trees (each with different sales value). The sales value is determined by the quality of the orange. So, to maximize profit, we need to maximize the quality of the oranges. Each of the factors has a positive influence on the quality of the oranges and each has a different effect and price. Furthermore, none of the cost factors are required to be used. Therefore, the number of options to maximize the profit is exponentially greater. The questions we aim to answer from our analysis are:

 How much profit should we expect in the future? – We will forecast five (5) years into the future.  Which cost factors should be implemented for each type of tree?  Should an irrigation system be installed?

In designing the model, certain assumptions were made. These included how much each cost factor would improve the quality and size of the oranges. Juan helped make these assumptions from his experience on the farm. We also used a couple of other sources to determine sales prices and selling costs.

1. Engineer from the area: Victor Marín 2. CONCITVER.com in their section for fruit CEDER FRUT / FOCIVER

7 | P a g e a. This website provides oranges prices that are updated daily. For example, the price for one (1) ton of oranges (tardias) today is $100 USD per ton.

Other assumptions that were made include the idea that the longer the fruit stays on the tree, the bigger it will be when it is picked. Also, if an orange is sold on the tree, it does not require any picking or shipping costs. Another important assumption is that we are not considering theft as an option. According to Juan, it is possible for oranges, especially ones of high quality, to be stolen off the tree by local farmers.

In designing the model, there are several factors that are considered:

 There will be production movement from month to month  Each node represents a month  There will be three basic costs associated with sales  All types of oranges will follow the Figure 1 generic production and sales model

FIGURE 1. Production and Sales Model

8 | P a g e TECHNICAL DESCRIPTION OF THE MODEL

To program the model, we used AMPL.

The model begins by defining sets of nodes for the months and the amount sold each month: set Month; set Sell;

Next, we define the parameters:

param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0;

Next, we define the variables:

var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0;

This ensures that the capacity being sent over the arcs between nodes will be greater than or equal to zero, but cannot exceed the given capacity for that arc.

Next, we define the variables that represent the costs of production. Each model will determine which of cost factors will be used. To do this we defined the cost factors as binary variables:

var P binary; var F binary; var T binary; var M binary;

Next, we define the objective function. To do this, we defined a simple formula to calculate the revenue and cost and subtracted them. maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M;

Next, we defined the constraints. We started by defining the total amount of oranges that would be produced for each type of tree. Next, we constrained each node so that the input equaled the output. Then we made sure all of the oranges were ‘picked’ and that no oranges were ‘lost’.

subject to Start: X[1] = (# of Trees)*(5 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i];

9 | P a g e The data files will include revenue and cost information in matrix form. The matrix is a mathematical representation of Figure 1. If the oranges do not sell in a given month, they can always carry over to the next month (rightward motion along the matrix) or simply be sold the month it is harvested (downward motion along the matrix). The revenue for each scenario is as follows:

param Revenue: 1 2 3 := 1 150 180 200 2 210 240 270 3 300 340 380 ;

Likewise, depending on what month you are selling the oranges, different costs will apply. For example:

param Cost: 1 2 3 := 1 0 0 0 2 60 60 70 3 150 160 180 ;

The program will search the matrix and determine which month and selling method is most profitable.

10 | P a g e ANALYSIS AND MANAGERIAL INTERPRETATION

Our analysis, without irrigation, concluded the following results. The total projected profit is estimated at $597,974 for the year. The decision analysis revealed the activities that would cover their costs. For example, all activities are recommended for the Tardias, Mayeras, and Mars trees. For the Navels, Nova, and Delicias tree, the model recommends only using Pesticide and Maintenance for the return on fertilizer and pesticide would not cover their costs.

Table 1. Decision Analysis Results without Irrigation Tree Type Fertilizer Pesticide Maintenance Trimming Tardias YES YES YES YES Mayeras YES YES YES YES Navels NO YES YES NO Nova NO YES YES NO Zarzuma NO YES YES YES Delicias NO YES YES NO Mars YES YES YES YES

Table 2. Financial Analysis Results and Projection without Irrigation Year 1 2 3 4 5 Revenue Tardias $14,490 $18,837 $24,488 $31,835 $41,385 Mayeras $48,300 $62,790 $81,627 $106,115 $137,950 Navels $1,258 $1,635 $2,126 $2,763 $3,592 Nova $1,303 $1,693 $2,201 $2,862 $3,720 Zarzuma $2,256 $2,933 $3,813 $4,956 $6,443 Delicias $1,034 $1,344 $1,747 $2,271 $2,952 Mars $5,490 $7,137 $9,278 $12,062 $15,680 Cost Tardias $1,715 $2,230 $2,898 $3,768 $4,898 Mayeras $1,715 $2,230 $2,898 $3,768 $4,898 Navels $615 $800 $1,039 $1,351 $1,757 Nova $615 $800 $1,039 $1,351 $1,757 Zarzuma $1,015 $1,320 $1,715 $2,230 $2,899 Delicias $615 $800 $1,039 $1,351 $1,757 Mars $1,715 $2,230 $2,898 $3,768 $4,898 Profit Tardias $12,775 $16,608 $21,590 $28,067 $36,487 Mayeras $46,585 $60,561 $78,729 $102,347 $133,051 Navels $643 $836 $1,086 $1,412 $1,836 Nova $688 $894 $1,162 $1,511 $1,964 Zarzuma $1,241 $1,613 $2,097 $2,726 $3,544

11 | P a g e Delicias $419 $544 $707 $919 $1,195 Mars $3,775 $4,908 $6,380 $8,294 $10,782 Projected Income $66,125 $85,962 $111,751 $145,276 $188,859 Total $597,974

Table 3. Projected Profit by Grade GRADE PROFIT A B A B Tardias 100% 0% $12,775 $0 Mayeras 100% 0% $46,585 $0 Navels 90% 10% $578 $64 Nova 90% 10% $619 $69 Zarzuma 90% 10% $1,117 $124 Delicias 90% 10% $377 $42 Mars 100% 0% $3,775 $0 Income $65,826 $299 Total $66,125

Our analysis, with considering irrigation, concluded the following results. The total projected profit is estimated at $881,506 for the year. Once again, the decision analysis revealed the activities that would cover their costs. For example, all activities are recommended for the Tardias, Mayeras, and Mars trees. For the Navels, Nova, and Delicias trees, the model recommends only using Pesticide and Maintenance for the return on fertilizer and pesticide would not cover their costs.

Table 4. Decision Analysis Results with Irrigation Tree Type Fertilizer Pesticide Maintenance Trimming Tardias YES YES YES YES Mayeras YES YES YES YES Navels NO YES YES NO Nova NO YES YES NO Zarzuma NO YES YES YES Delicias NO YES YES NO Mars YES YES YES YES

Table 5. Financial Analysis Results and Projection with Irrigation With Irrigation System Cost = $60,000 Year 1 2 3 4 5 Revenue

12 | P a g e Tardias $20,286 $26,372 $34,283 $44,568 $57,939 Mayeras $67,620 $87,906 $114,278 $148,561 $193,129 Navels $2,032 $2,641 $3,434 $4,464 $5,803 Nova $2,806 $3,648 $4,742 $6,165 $8,014 Zarzuma $3,384 $4,399 $5,719 $7,435 $9,665 Delicias $1,670 $2,170 $2,821 $3,668 $4,768 Mars $7,686 $9,992 $12,989 $16,886 $21,952 Cost Tardias $1,715 $2,230 $2,898 $3,768 $4,898 Mayeras $1,715 $2,230 $2,898 $3,768 $4,898 Navels $615 $800 $1,039 $1,351 $1,757 Nova $615 $800 $1,039 $1,351 $1,757 Zarzuma $1,015 $1,320 $1,715 $2,230 $2,899 Delicias $615 $800 $1,039 $1,351 $1,757 Mars $1,715 $2,230 $2,898 $3,768 $4,898 Profit Tardias $18,571 $24,142 $31,385 $40,800 $53,041 Mayeras $65,905 $85,677 $111,379 $144,793 $188,231 Navels $1,417 $1,842 $2,394 $3,113 $4,046 Nova $2,191 $2,848 $3,703 $4,814 $6,258 Zarzuma $2,369 $3,080 $4,004 $5,205 $6,766 Delicias $1,055 $1,371 $1,782 $2,317 $3,012 Mars $5,971 $7,762 $10,091 $13,118 $17,054 Projected Income $97,478 $126,722 $164,738 $214,160 $278,408 Total $881,506

Additional Saving $31,353 $40,759 $52,987 $68,883 $89,548 NPV $192,791

Net gain/loss $132,791

Table 6. Projected Profit by Grade GRADE PROFIT A B A B Tardias 100% 0% $18,571 $0 Mayeras 100% 0% $65,905 $0 Navels 90% 10% $1,275 $142 Nova 90% 10% $1,972 $219 Zarzuma 90% 10% $2,132 $237 Delicias 90% 10% $949 $105

13 | P a g e Mars 100% 0% $5,971 $0 Income $96,775 $703 Total $97,478

The net present value of the projected profit to be gained over the next five years is $192,791. When considering whether or not an irrigation system should be installed, at a price of $60,000 the estimated gain is still $132,791.

The assumptions made in the model are given by the owner of the property. The costs of picking, transporting, and the effects on the tree of using each activity are determined by the owner from past results. It is because of these assumptions that the model may over or under estimate the results. Another assumption was the estimated 30% increase in production from year to year in the projections. Given the growth rate of the trees and amount of fruit produced by each tree, 30% seemed like a reasonable yearly increase.

14 | P a g e CONCLUSIONS AND CRITIQUE

The model we used to solve the problem was an integer programming model. To solve the model we used AMPL. We decided that because each tree has different costs and revenues that we should solve the problem by viewing each tree as independent of another. This would ensure that our solution is optimal for each tree and would maximize the total profit of the entire orchard. As expected, the recommended different activities for different trees. Additionally, another activity (irrigation) was included in the model. However, an irrigation decision for each type of tree was no longer an option because we could only consider irrigation for the entire orchard or not at all.

After running each model and compiling the results, we concluded that the best option was to install the irrigation system and use the following activities for each type of tree.

Table 5. Decision Analysis Results with Irrigation Tree Type Fertilizer Pesticide Maintenance Trimming Tardias YES YES YES YES Mayeras YES YES YES YES Navels NO YES YES NO Nova NO YES YES NO Zarzuma NO YES YES YES Delicias NO YES YES NO Mars YES YES YES YES

Following this schedule would yield an estimated $132,791 in total profit. This amount includes the estimated $60,000 cost for the irrigation system.

Some of the limitations we faced were in our assumptions in costs and estimated revenues. Since profit is dependent on both of these factors, our estimated profit will change with any change in cost or selling prices. Also, our optimal solution from the models may change with different cost and price numbers. We believe that with more compiled data from previous years, we could more accurately project future earnings and make better decisions.

15 | P a g e APPENDIX a. Tardias Model without Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 1932*(5 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..3}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[4] = Y[4]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Tardias_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

16 | P a g e b. Tardias Model with Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 1932*(25 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..3}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[4] = Y[4]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Tardias_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

17 | P a g e c. Tardias Data set Month := 1, 2, 3, 4; set Sell := 1, 2, 3; param Revenue: 1 2 3 4 := 1 90 90 120 150 2 110 110 160 230 3 145 145 220 310 ; param Cost: 1 2 3 4 := 1 0 0 0 0 2 20 20 40 80 3 55 55 100 170 ;

18 | P a g e d. Mayeras Model without Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 1932*(5 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..3}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[4] = Y[4]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Mayeras_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

19 | P a g e e. Mayeras Model with Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 1932*(25 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..3}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[4] = Y[4]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Mayeras_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

20 | P a g e f. Mayeras Data set Month := 1, 2, 3, 4; set Sell := 1, 2, 3; param Revenue: 1 2 3 4 := 1 400 400 450 500 2 520 520 570 620 3 680 680 760 860 ; param Cost: 1 2 3 4 := 1 0 0 0 0 2 120 120 170 220 3 260 260 360 460 ;

21 | P a g e g. Navels Model without Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 387*(5 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Navels_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

22 | P a g e h. Navels Model with Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 387*(25 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Navels_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

23 | P a g e i. Navels Data set Month := 1, 2, 3; set Sell := 1, 2, 3; param Revenue: 1 2 3 := 1 80 90 100 2 100 110 140 3 135 155 195 ; param Cost: 1 2 3 := 1 0 0 0 2 20 30 40 3 55 75 95 ;

24 | P a g e j. Nova Model without Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 334*(5 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Nova_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

25 | P a g e k. Nova Model with Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 334*(25 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Nova_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

26 | P a g e l. Nova Data set Month := 1, 2, 3; set Sell := 1, 2, 3; param Revenue: 1 2 3 := 1 90 90 120 2 110 110 160 3 145 145 220 ; param Cost: 1 2 3 := 1 0 0 0 2 20 20 40 3 55 55 100 ;

27 | P a g e m. Zarzuma Model without Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 282*(5 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Zarzuma_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

28 | P a g e n. Zarzuma Model with Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 282*(25 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Zarzuma_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

29 | P a g e o. Zarzuma Data set Month := 1, 2, 3; set Sell := 1, 2, 3; param Revenue: 1 2 3 := 1 150 180 200 2 210 240 270 3 300 340 380 ; param Cost: 1 2 3 := 1 0 0 0 2 60 60 70 3 150 160 180 ;

30 | P a g e p. Delicias Model without Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 265*(5 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Delicias_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

31 | P a g e q. Delicias Model with Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 265*(25 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Delicias_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

32 | P a g e r. Delicias Data set Month := 1, 2, 3; set Sell := 1, 2, 3; param Revenue: 1 2 3 := 1 90 90 120 2 110 110 160 3 145 145 220 ; param Cost: 1 2 3 := 1 0 0 0 2 20 20 40 3 55 55 100 ;

33 | P a g e s. Mars Model without Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 732*(5 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Mars_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

34 | P a g e t. Mars Model with Irrigation set Month; set Sell; param Revenue{Sell, Month} >= 0; param Cost{Sell, Month} >= 0; var X{Month} >= 0; var Y{Month} >= 0; var Z{Sell, Month} >= 0; var P binary; var F binary; var T binary; var M binary; maximize Profit: sum{i in Sell, j in Month} (Revenue[i,j] - Cost[i,j])*Z[i,j] - 390*F - 700*P - 225*T - 400*M; subject to Start: X[1] = 732*(25 + 20*F + 10*P + 7.5*T + 7.5*M)/1000; subject to Balance{i in 1..2}: X[i] = Y[i] + X[i+1]; subject to Balance4: X[3] = Y[3]; subject to Selling{i in Month}: Y[i] = sum{j in Sell} Z[j,i]; data Mars_Data.txt; option solver cplex; solve; display X, Y, Z, P, F, T, M;

35 | P a g e u. Mars Data set Month := 1, 2, 3; set Sell := 1, 2, 3; param Revenue: 1 2 3 := 1 120 135 150 2 160 190 220 3 220 265 300 ; param Cost: 1 2 3 := 1 0 0 0 2 40 55 70 3 100 130 160 ;

36 | P a g e v. Results: Tardias Model without Irrigation ampl: model Tardias_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 12775 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 96.6 0 2 96.6 0 3 96.6 0 4 96.6 96.6 ;

Z := 1 1 0 1 2 0 1 3 0 1 4 96.6 2 1 0 2 2 0 2 3 0 2 4 0 3 1 0 3 2 0 3 3 0 3 4 0 ;

P = 1 F = 1 T = 1 M = 1

37 | P a g e w. Results: Tardias Model with Irrigation ampl: model Tardias_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 18571 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 135.24 0 2 135.24 0 3 135.24 0 4 135.24 135.24 ;

Z := 1 1 0 1 2 0 1 3 0 1 4 135.24 2 1 0 2 2 0 2 3 0 2 4 0 3 1 0 3 2 0 3 3 0 3 4 0 ;

P = 1 F = 1 T = 1 M = 1

38 | P a g e x. Results: Mayeras Model without Irrigation ampl: model Mayeras_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 46585 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 96.6 0 2 96.6 0 3 96.6 0 4 96.6 96.6 ;

Z := 1 1 0 1 2 0 1 3 0 1 4 96.6 2 1 0 2 2 0 2 3 0 2 4 0 3 1 0 3 2 0 3 3 0 3 4 0 ;

P = 1 F = 1 T = 1 M = 1

39 | P a g e y. Results: Mayeras Model with Irrigation ampl: model Mayeras_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 65905 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 135.24 0 2 135.24 0 3 135.24 0 4 135.24 135.24 ;

Z := 1 1 0 1 2 0 1 3 0 1 4 135.24 2 1 0 2 2 0 2 3 0 2 4 0 3 1 0 3 2 0 3 3 0 3 4 0 ;

P = 1 F = 1 T = 1 M = 1

40 | P a g e z. Results: Navels Model without Irrigation ampl: model Navels_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 642.75 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 12.5775 0 2 12.5775 0 3 12.5775 12.5775 ;

Z := 1 1 0 1 2 0 1 3 12.5775 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 0 F = 1 T = 1 M = 0

41 | P a g e aa. Results: Navels Model with Irrigation ampl: model Navels_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 1416.75 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 20.3175 0 2 20.3175 0 3 20.3175 20.3175 ;

Z := 1 1 0 1 2 0 1 3 20.3175 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 0 F = 1 T = 1 M = 0

42 | P a g e bb. Results: Nova Model without Irrigation ampl: model Nova_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 687.6 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 10.855 0 2 10.855 0 3 10.855 10.855 ;

Z := 1 1 0 1 2 0 1 3 10.855 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 0 F = 1 T = 1 M = 0

43 | P a g e cc. Results: Nova Model with Irrigation ampl: model Nova_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 1489.2 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 17.535 0 2 17.535 0 3 17.535 17.535 ;

Z := 1 1 0 1 2 0 1 3 17.535 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 0 F = 1 T = 1 M = 0

44 | P a g e dd. Results: Zarzuma Model without Irrigation ampl: model Zarzuma_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 1241 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 11.28 0 2 11.28 0 3 11.28 11.28 ;

Z := 1 1 0 1 2 0 1 3 11.28 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 0 F = 1 T = 1 M = 1

45 | P a g e ee. Results: Zarzuma Model with Irrigation ampl: model Zarzuma_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 2369 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 16.92 0 2 16.92 0 3 16.92 16.92 ;

Z := 1 1 0 1 2 0 1 3 16.92 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 0 F = 1 T = 1 M = 1

46 | P a g e ff. Results: Delicias Model without Irrigation ampl: model Delicias_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 418.5 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 8.6125 0 2 8.6125 0 3 8.6125 8.6125 ;

Z := 1 1 0 1 2 0 1 3 8.6125 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 0 F = 1 T = 1 M = 0

47 | P a g e gg. Results: Delicias Model with Irrigation ampl: model Delicias_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 1054.5 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 13.9125 0 2 13.9125 0 3 13.9125 13.9125 ;

Z := 1 1 0 1 2 0 1 3 13.9125 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 0 F = 1 T = 1 M = 0

48 | P a g e hh. Results: Mars Model without Irrigation ampl: model Mars_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 3775 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 36.6 0 2 36.6 0 3 36.6 36.6 ;

Z := 1 1 0 1 2 0 1 3 36.6 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 1 F = 1 T = 1 M = 1

49 | P a g e ii. Results: Mars Model with Irrigation ampl: model Mars_Model.txt; CPLEX 11.0.1: optimal integer solution; objective 5971 0 MIP simplex iterations 0 branch-and-bound nodes : X Y := 1 51.24 0 2 51.24 0 3 51.24 51.24 ;

Z := 1 1 0 1 2 0 1 3 51.24 2 1 0 2 2 0 2 3 0 3 1 0 3 2 0 3 3 0 ; P = 1 F = 1 T = 1 M = 1

50 | P a g e