Optimal points of operational parameters of sulphuric acid, phosphic acid, and ammonia production process

PT Petrokimia Gresik is the most complete fertilizer producer in Indonesia and produces various types of fertilizers. NPK is a fertilizer that contains nitrogen (N) to make plants greener, phosphorus (P) to stimulate root growth, and potassium (K) to make plants more upright and sturdy. This study aims to determine the optimum value of using sulfuric acid, phosphoric acid, and ammonia so as to obtain the best value in producing NPK fertilizer. This research method uses quantitative descriptive research methods by collecting secondary data, namely the use of sulfuric acid, phosphoric acid, and ammonia for a period of 1 year. Researchers use multiple regression analysis to determine whether or not there is an influence between the three variables to be studied, then the data is processed and analyzed to determine whether the regression model used is feasible to use before finally calculating the optimal point of NPK fertilizer production. The stages of data optimization in this study are the data collection stage, the data processing stage (descriptive data, parameter estimation, regression model), the data analysis and interpretation stage (F test, t test, IIDN assumption test for determining the optimum point), and the conclusion and suggestion stage. This study concluded that obtaining the optimum point for the amount of NPK fertilizer production can be done by regulating the use of phosphoric acid at 211.91 tons, Sulphuric acid at 844,099 tons, and ammonia at 308,055 tons. Total fertilizer production achieved with the use of optimum parameters amounted to 2085.38 tons. This is an open access article under the CC BY-SA license.

(WPPA) as much as 4870.42kg, ammonia as much as 283.01 kg, sulfuric acid as much as 227.69 kg will produce NPK superphosphate-based fertilizer as much as 931.59 kg and NPK-based fertilizer Ammophosphate amounted to 1329.33 kg.According to research conducted by Mukhtorovna et. al., (2021) to produce 1000 kg of NPK fertilizer, it is necessary to use 360 kg of nitric acid (HNO3), 264.88 kg of water (H2O), 205.8 kg of potassium chloride (KCl), and the use of phosphoric acid (H3PO4) of 650 kg.Another study conducted by Allamuratova et al., (2021) to produce NPK fertilizer with the formula 16.7:16.7:16.7 requires the amount of nutritional components as much as 50.1%, the consumption coefficient of Kyzyl Kum phosphorite is 1.51, nitric acid consumption coefficient is 0.58, sulfuric acid consumption coefficient is 1.33,The consumption coefficient of potassium chloride is 0.55, the ammonia consumption coefficient is 0.24.Nomozov et al., (2020) explained that in his research with the use of phosphoric acid of 1103.63 kg, ammonia of 86.90 kg, urea of 124.24 kg, and potassium chloride of 170.31 kg, NPK fertilizer of 1000 kg would be obtained.Rasulov et al., (2020) explained that to obtain fertilizer with nitrogen-phosphate (NPK) content with the use of wet processing phosphate acid (WPA) of 415.14 kg, sulfuric acid of 120.06 kg, ammonia of 61.45 kg, phosphorite powder as much as 119.84 kg will produce a total of 406.34 kg of fertilizer with an N content of 11.55% and P2O5 of 24.61%.Research conducted by Numonov et al., (2019) stated that the use of sulfuric acid as much as 985 kg, consumption of 1000 kg WBPC (Washed And Burned Phosphoconcentrate), the use of 215 kg of ammonia, nitric acid (HNO3) as much as 550.4 kg, potassium chloride of 402 kg will produce NPK fertilizer of 1743 kg with a Nitrogen content of 14.34%, P2O5 13.91%, and K2O 13.81%.Pagaleshkin and Kolpakov (2019) stated that there is a technology developed for NPK fertilizers with formula types 16-16-16 and 17-17-17.The process is carried out under conditions of producing nitroammophonskafertilizer with a high molar ratio of ammonia (NH3) and phosphoric acid (H3PO4) in fertilizer with a ratio of 1.65 : 1.8 mol / mol.According to the final project research conducted by Tania and Machdalia (2017) to produce NPK fertilizer with a capacity of 500 tons per day requires ammonia raw materials of 68784.11kg, the use of sulfuric acid as much as 72807.86kg, phosphoric acid of 251832.50kg, urea of 29855.34kg, ammonium nitrate of 65135.57kg, KCl of 131341.21kg, coating powder of 1100 kg, and coating oil of 1200 kg.Zhang et al., (2017) explained that to obtain fertilizer with an amount of 1 ton of product requires sulfuric acid as much as 1.53 tons, sulfur (sulfur) as much as 0.50 tons, phosphoric acid as much as 2.27 tons, water as much as 3.22 m 3 , ammonia as much as 0.14 tons.
Based on the existing explanation, an optimization step is needed from the use of plant operations which are parameters in the production process at the NPK Phonska Reaction plant.The optimization problem is a problem of finding the optimum value of a function.The necessary condition that must be met in order for the optimal fungisional is a partial derivative of the goal function against all its variables to zero (Khoiriyah, 2020).A partial differential equation is a differential equation involving a partial derivative containing two or more variables (Raming et al., 2023).
This study aims to know, identify the optimal value of using sulphuric acid, phosphoric acid, and ammonia for use in the production process at the NPK Phonska Reaction fertilizer plant, and determine the optimum point of operational parameters at the NPK Phonska Reaction plant in producing the amount of fertilizer with the use of sulphuric acid, phosphoric acid, ammonia.The results of this study provide recommendations for optimizing the operational parameters of the NPK Phonska IV (Phonska Reaction) plant at PT Petrokimia Gresik as a reference in order to carry out the production process effectively and efficiently in the long term.

METHOD
This research method uses quantitative descriptive research methods, while the analysis uses multiple regression models.This research methodology is divided into four stages, namely the data collection stage, data processing stage, data analysis and interpretation stage, and conclusion and suggestion stage.The flow diagram of the research stages is shown in the figure 1 below.These stages are carried out using Minitab and Microsoft Excel software.The data collection stage is a way to obtain data and needs that support the research conducted.The data collected by this study is secondary data sourced from the NPK Phonska Reaction fertilizer production process at PT Petrokimia Gresik from January 1, 2022 to December 31, 2022, namely phosphoric acid usage data, sulphuric acid usage data, ammonia usage data, downtime data, and the amount of fertilizer production.The data processing stage consists of descriptive data, parameter estimation, regression models.Descriptive statistics are a suite of statistics that summarize the characteristics and distribution of a set of data values (Lee, 2020).Parameter estimation is an estimation of population characteristic values based on sample characteristics (Aster, 2019).Meanwhile, regression models are used to determine the relationship between variables (Mohr, 2022).
The analysis phase includes testing the feasibility of the model and determining the optimum point.Model feasibility tests include F test, t test, correlation test, IIDN assumption test.Test F is performed to determine whether all independent variables entered into the model have a joint influence on the dependent variable (Siegel and Wagner, 2022).The t test is used to determine the effect of each independent variable on the dependent variable (Thukral and Paturu, 2023).The correlation test is used to describe how strong the relationship of a variable with other variables (Rizk, 2023).Testing identical residual assumptions is used to determine the homogeneity of the regression model's residual variance (Rositawati, 2020).Independent residual assumption testing is used to determine the presence of autocorrelation in (Salsabila, 2022).Testing of normally distributed residual assumptions can be done by looking at graphs, namely histogram graphs and normal probability plot graphs and by Kolmogorov-Smirnov testing (Dzikir, 2020).

Figure 1. Research Flowchart
The results of the univariate analysis in this study will provide results in the form of average values, standard deviations, minimum values, and maximum values from the use of phosphoric acid, sulfuric acid, ammonia, downtime, and the amount of fertilizer production in the NPK Phonska Reaction.While the results of the bivariate analysis will show the relationship between each variable and the amount of NPK fertilizer production.These results were obtained using Minitab software.

RESULTS AND DISCUSSION
Researchers conduct model feasibility tests to determine whether the regression model is feasible to use in this study before determining the optimal point of operational parameters.Model feasibility tests include F test, t test, correlation test, identical assumption test, independent assumption test, and normal distribution assumption test. 1) The hypotheses in this test are: a) H0 : β1 = β2 = 0, meaning that there is no significant effect together or simultaneously from variable X1 on variable Y. b) H1: β1 ≠ β2 ≠ 0, meaning that at least one of the variables X1 has an effect on variable Y.
3) The calculated F test value is 229.34.4) Statistical test criteria: a) If the value of Fcount < Ftable then H0 is accepted (no effect) b) If the value of Fcount > Ftable then H0 is rejected (effect) 5) Decision Making: From the test results, it is found that Fcount > Ftable or p-value value < 0.05, then a decision can be made to reject H0 which means that individually there is a significant influence of the variables studied.
Total Fertilizer Production Against the Use of Sulphuric Acid (X2) 1) The hypotheses in this test are: a) H0 : β1 = β2 = 0, meaning that there is no significant effect together or simultaneously from variable X2 on variable Y. b) H1: β1 ≠ β2 ≠ 0, meaning that at least one of the variables X2 has an effect on variable Y.
3) The calculated F test value is 474.03.4) Statistical test criteria: a) If the value of Fcount < Ftable then H0 is accepted (no effect) b) If the value of Fcount > Ftable then H0 is rejected (effect) 5) Decision Making: From the test results, it is found that Fcount > Ftable or p-value value < 0.05, then a decision can be made to reject H0 which means that individually there is a significant influence of the variables studied. 1) The hypotheses in this test are: a) H0 : β1 = β2 = 0, meaning that there is no significant effect together or simultaneously from variable X3 on variable Y. b) H1: β1 ≠ β2 ≠ 0, meaning that at least one of the variables X3 has an effect on variable Y.
3) The calculated F test value is 469.77.4) Statistical test criteria: a) If the value of Fcount < Ftable then H0 is accepted (no effect) b) If the value of Fcount > Ftable then H0 is rejected (effect) 5) Decision Making: From the test results, it is found that Fcount > Ftable and the p-value < 0.05 then a decision can be made to reject H0 which means that individually there is a significant influence from the variables studied. 1) The hypotheses in this test are: a) H0 : β1 = β2 = 0, meaning that there is no significant effect together or simultaneously from variable X3 on variable Y. b) H1: β1 ≠ β2 ≠ 0, meaning that at least one of the variables X3 has an effect on variable Y.

Amount of Fertilizer Production Against Plant Downtime
2) The F test in this test uses a significance level of α = 5%.Degree of freedom (df1) = k -1 = 3 -1 = 2, and df2 = nk = 312 -3 = 309, so the F value of the table for significance 0.05 is 3.025.
3) The calculated F test value is 130592.4) Statistical test criteria: a) If the value of Fcount < Ftable then H0 is accepted (no effect) b) If the value of Fcount > Ftable then H0 is rejected (effect) 5) Decision Making: From the test results, it is found that Fcount > Ftable and the p-value < 0.05 then a decision can be made to reject H0 which means that individually there is a significant influence from the variables studied.

Partial Regression Coefficient Test (t Test)
Total Fertilizer Production Against Phosphoric Acid Use (X1) -0,01279 0,00159 -8,05 0,000 1) The hypotheses in this test are: a) H0: the independent variable has no significant effect on the yield of the amount of fertilizer production in the NPK Phonska Reaction plant unit.b) H1: the independent variable has a significant effect on the yield of the amount of fertilizer production in the NPK Phonska Reaction plant unit.
2) The t test in this test uses a significance level of α = 5%.Degree of freedom (df1) = n -2 = 312 -2 = 310, so the table t value for significance of 0.05 is 1.9676.3) Statistical test criteria: a) If the significance value of the t-test < 0.05 then H0 is rejected and H1 is accepted (effect).b) If the significance value of the t-test > 0.05 then H0 is accepted and H1 is rejected (no effect).4) Decision Making: From the test results, it was found that the significance value was 0.008 or pvalue < 0.05, then a decision could be made to reject H0 which means that the independent variable has a significant effect on the results of the amount of fertilizer production in the NPK Phonska Reaction plant unit.-0,003430 0,000700 -4,90 0,000

Total Fertilizer Production Against the Use of Sulphuric Acid (X2)
1) The hypotheses in this test are: a) H0: the independent variable has no significant effect on the yield of the amount of fertilizer production in the NPK Phonska Reaction plant unit.b) H1: the independent variable has a significant effect on the yield of the amount of fertilizer production in the NPK Phonska Reaction plant unit.
2) The t test in this test uses a significance level of α = 5%.Degree of freedom (df1) = n -2 = 312 -2 = 310, so the table t value for significance of 0.05 is 1.9676.3) Statistical test criteria: a) If the significance value of the t-test < 0.05 then H0 is rejected and H1 is accepted (effect).b) If the significance value of the t-test > 0.05 then H0 is accepted and H1 is rejected (no effect).
4) Decision Making: From the test results, it was found that the significance value was 0.006 or pvalue < 0.05, then a decision could be made to reject H0 which means that the independent variable has a significant effect on the results of the amount of fertilizer production in the NPK Phonska Reaction plant unit.-0,02049 0,00341 -6,01 0,000

Total Fertilizer Production Against Ammonia Use (X3)
1) The hypotheses in this test are: a) H0: the independent variable has no significant effect on the yield of the amount of fertilizer production in the NPK Phonska Reaction plant unit.b) H1: the independent variable has a significant effect on the yield of the amount of fertilizer production in the NPK Phonska Reaction plant unit.
2) The t test in this test uses a significance level of α = 5%.Degree of freedom (df1) = n -2 = 312 -2 = 310, so the table t value for significance of 0.05 is 1.9676.3) Statistical test criteria: a) If the significance value of the t-test < 0.05 then H0 is rejected and H1 is accepted (effect).b) If the significance value of the t-test > 0.05 then H0 is accepted and H1 is rejected (no effect).4) Decision Making: From the test results, it was found that the significance value was 0.002 or pvalue < 0.05, then a decision could be made to reject H0 which means that the independent variable has a significant effect on the results of the amount of fertilizer production in the NPK Phonska Reaction plant unit.1) The hypotheses in this test are: a) H0: the independent variable has no significant effect on the yield of the amount of fertilizer production in the NPK Phonska Reaction plant unit.b) H1: the independent variable has a significant effect on the yield of the amount of fertilizer production in the NPK Phonska Reaction plant unit.

Amount of Fertilizer Production Against Plant Downtime
2) The t test in this test uses a significance level of α = 5%.Degree of freedom (df1) = n -2 = 312 -2 = 310, so the table t value for significance of 0.05 is 1.9676.3) Statistical test criteria: a) If the significance value of the t-test < 0.05 then H0 is rejected and H1 is accepted (effect).b) If the significance value of the t-test > 0.05 then H0 is accepted and H1 is rejected (no effect).4) Decision Making: From the test results, it is found that the significance value is 0.000 or p-value < 0.05, then a decision can be made to reject H0 which means that the independent variable has a significant effect on the results of the amount of fertilizer production in the NPK Phonska Reaction plant unit.From the results of the correlation test, a value (r) of 0.716 was obtained.This value indicates that there is a strong correlation with a positive direction between the amount of fertilizer production and the use of phosphoric acid.From the results of the correlation test, a value (r) of 0.857 was obtained.This value indicates that there is a very strong correlation with a positive direction between the amount of fertilizer production and the use of Sulphuric acid.From the results of the correlation test, a value (r) of 0.851 was obtained.This value indicates that there is a very strong correlation with a positive direction between the amount of fertilizer production and ammonia use.From the results of the correlation test, a value (r) of -0.999 was obtained.This value indicates that there is a very strong correlation with the negative direction between the amount of fertilizer production and plant downtime.1) The hypotheses in this test are: a) H0: Residual data on the amount of fertilizer production against the use of phosphoric acid in the NPK Phonska plant unit The reaction is identical.b) H1: Residual data on the amount of fertilizer production against the use of phosphoric acid in the NPK Phonska plant unit The reaction is not identical.2) The t test in this test uses a significance level of α = 5%.Degree of freedom (df) = n -2 = 312 -2 = 310, so the F value of the table for significance of 0.05 is 3.024.3) Statistical test criteria: a) If the significance value < 0.05 or Fcount > F table then reject H0 or not identical.b) If the significance value > 0.05 or Fcount < Ftable then H0 reject fails or is identical.4) Decision Making: From the test results, it was found that the significance value was 0.751 or Pvalue > 0.05, then a decision could be made to fail to reject H0 which means that residual data on the amount of fertilizer production on the use of phosphoric acid in the NPK Phonska plant unit The reaction is identical.1) The hypotheses in this test are: a) H0: Residual data on the amount of fertilizer production against the use of Sulphuric acid in the NPK Phonska plant unit The reaction is identical.b) H1: Residual data on the amount of fertilizer production against the use of Sulphuric acid in the NPK Phonska plant unit The reaction is not identical.

The Amount of Fertilizer Production Against the Use of Sulphuric Acid
2) The t test in this test uses a significance level of α = 5%.Degree of freedom (df) = n -2 = 312 -2 = 310, so the F value of the table for significance of 0.05 is 3.024.3) Statistical test criteria: a) If the significance value < 0.05 or Fcount > Ftable then reject H0 or not identical.b) If the significance value > 0.05 or Fcount < Ftable then H0 reject fails or is identical.4) Decision Making: From the test results, it was found that the significance value was 0.599 or Pvalue > 0.05, so a decision could be made to fail to reject H0, which means that the residual data on the amount of fertilizer production on the use of sulphuric acid in the NPK Phonska plant unit The reaction is identical.1) The hypotheses in this test are: a) H0: Residual data on the amount of fertilizer production against ammonia use in the NPK Phonska plant unit The reaction is identical.b) H1: Residual data on the amount of fertilizer production against ammonia use in the NPK Phonska plant unit The reaction is not identical.2) The t test in this test uses a significance level of α = 5%.Degree of freedom (df) = n -2 = 312 -2 = 310, so the F value of the table for significance of 0.05 is 3.024.3) Statistical test criteria: a) If the significance value < 0.05 or F count > F of the table then reject H0 or not identical.b) If the significance value > 0.05 or F count < F table then H0 reject fails or is identical.4) Decision Making: From the test results, it was found that the significance value was 0.599 or Pvalue > 0.05, then a decision could be made to fail to reject H0, which means that residual data on the amount of fertilizer production on ammonia use in the NPK Phonska plant unit The reaction is identical.1) The hypotheses in this test are: a) H0: Residual data on the amount of fertilizer production against plant downtime in the NPK Phonska plant unit The reaction is identical.

Amount of Fertilizer Production Against Plant Downtime
b) H1: Residual data on the amount of fertilizer production against plant downtime in the NPK Phonska plant unit The reaction is not identical.2) The t test in this test uses a significance level of α = 5%.Degree of freedom (df) = n -2 = 312 -2 = 310, so the F value of the table for significance of 0.05 is 3.024.3) Statistical test criteria: a) If the significance value < 0.05 or Fcount > Ftable then reject H0 is rejected or not identical.b) If the significance value > 0.05 or Fcount < Ftable then H0 reject fails or is identical.4) Decision Making: From the test results, it was found that the significance value was 0.213 or Pvalue > 0.05, then a decision could be made to fail to reject H0, which means that the residual data on the amount of fertilizer production on plant downtime in the NPK Phonska plant unit The reaction is identical.

Independent Residual Assumption Testing
The Amount of Fertilizer Production Against the Use of Phosphoric Acid 1) The hypotheses in this test are: a) H0 : Residual amount of fertilizer production against the use of phosphoric acid in the NPK Phonska plant unit The reaction is independent.b) H1 : Residual amount of fertilizer production against the use of phosphoric acid in the NPK Phonska plant unit The reaction is dependent.
2) The Durbin-Watson test in this test uses a significance level of α = 5% with n = 2 and a lot of data as much as 200, so that from the Durbin-Watson table obtained a dL value of 1.7483 and a dU value of 1.7887.3) Statistical test criteria: a) If the value of dcount < dL then reject H0 or residual is not independent.b) If the value of dU ≤ d calculate ≤ 4 -dU then it fails to reject H0 or the residual is independent.4) Decision Making: From the results of the Durbin-Watson test, a calculated d value of 1.8532 is obtained, so that 1.7887 ≤ 1.8532 ≤ 2.2113, a decision can be made to fail to reject H0 which means that residual data is independent.1) The hypotheses in this test are: a) H0: Residual amount of fertilizer production against the use of Sulphuric acid in the NPK Phonska plant unit The reaction is independent.b) H1: Residual amount of fertilizer production against the use of Sulphuric acid in the NPK Phonska plant unit The reaction is dependent.

The Amount of Fertilizer Production Against the Use of Sulphuric Acid
2) The Durbin-Watson test in this test uses a significance level of α = 5% with n = 2 and a lot of data as much as 200, so that from the Durbin-Watson table obtained a dL value of 1.7483 and a dU value of 1.7887.3) Statistical test criteria: a) If the value of dcount < dL then reject H0 or residual is not independent.b) If the value of dU ≤ dcount ≤ 4 -dU then it fails to reject H0 or the residual is independent.4) Decision Making: From the results of the Durbin-Watson test, a calculated value of 1.9575 is obtained, so that 1.7887 ≤ 1.9575 ≤ 2.2113, a decision can be made to fail to reject H0 which means that residual data is independent.a) If the value of dcount < dL then reject H0 or residual is not independent.b) If the value of dU ≤ dcount ≤ 4 -dU then it fails to reject H0 or the residual is independent.4) Decision Making: From the results of the Durbin-Watson test, a calculated value of 2.1751 was obtained, so that 1.7887 ≤ 2.1751 ≤ 2.2113, a decision can be made to fail to reject H0 which means that residual data is independent.

Normal Distribution Residual Assumption Testing
The Amount of Fertilizer Production Against the Use of Phosphoric Acid 2) The Kolmogorov-Smirnov test in this test uses a significance level of α = 5%.
3) Statistical test criteria: a) If the significance value (P-value) < 0.05 then reject H0 or residuals are not normally distributed.b) If the significance value (P-value) > 0.05 then it fails to reject H0 or the residual is normally distributed.4) Decision Making: From the picture above, a P-value of > 0.150 or a P-value of > 0.05 can be made, so a decision can be made to fail to reject H0, which means that residual data on the amount of fertilizer production against the use of Sulphuric acid is normally distributed.So that the optimum value of fertilizer production is 2085.38 tons.From the calculation above, the optimum value of operational parameters of phosphoric acid (X1) is 211,191 tons, Sulphuric acid (X2) is 844,099 tons, ammonia (X3) is 308,055 tons, and the amount of fertilizer production in the NPK Phonska Reaction unit is 2085.38 tons.

CONCLUSION
The optimum value of using phosphoric acid in the production process at the NPK Phonska Reaction plant unit is 211,191 tons, the optimum value of Sulphuric acid is 844,099 tons, and the optimum value of ammonia is 308,055 tons.The minimum factory downtime value is 0 days and the maximum is 0.94 days.The higher the factory downtime value, the amount of fertilizer production will decrease and the lower the factory downtime value, the amount of fertilizer production will increase.The optimum value of the fertilizer production process in the NPK Phonska Reaction plant unit is 2085.38 tons with the use of phosphoric acid of 211.91 tons, Sulphuric acid of 844,099 tons, and ammonia of 308,055 tons.

Figure 2 .Figure 3 .
Figure 2. Residual Graph of the Amount of Fertilizer Production Against Phosphoric Acid Use 1) The hypotheses in this test are: a) H0: Residual amount of fertilizer production against the use of phosphoric acid in the NPK Phonska plant unit The reaction is normally distributed.b) H1: Residual amount of fertilizer production against the use of phosphoric acid in the NPK Phonska plant unit The reaction is not normally distributed.2) The Kolmogorov-Smirnov test in this test uses a significance level of α = 5%.3) Statistical test criteria: a) If the significance value (P-value) < 0.05 then reject H0 or residuals are not normally distributed.b) If the significance value (P-value) > 0.05 then it fails to reject H0 or the residual is normallydistributed. 4) Decision Making: From the picture above, a P-value of 0.076 or a P-value of > 0.05 can be made, so a decision can be made to fail to reject H0, which means that residual data on the amount of fertilizer production against the use of phosphoric acid is normally distributed.The Amount of Fertilizer Production Against the Use of Sulphuric acid

Figure 4 .
Figure 4. Residual Graph of the Amount of Fertilizer Production Against the Use of Sulphuric Acid 1) The hypotheses in this test are: a) H0 : Residual amount of fertilizer production against ammonia use in NPK Phonska plant unit Normal distributed reaction.b) H1 : Residual amount of fertilizer production against ammonia use in NPK Phonska plant unit The reaction is not normally distributed.2) The Kolmogorov-Smirnov test in this test uses a significance level of α = 5%.3) Statistical test criteria: a) If the significance value (P-value) < 0.05 then reject H0 or residuals are not normally distributed.b) If the significance value (P-value) > 0.05 then it fails to reject H0 or the residual is normally distributed.4) Decision Making: From figure 4.19 above, a P-value of > 0.150 or a P-value of > 0.05 can be made, so a decision can be made to fail to reject H0, which means that residual data on the amount of fertilizer production against the use of ammonia is normally distributed.Optimum Point DeterminationWays to determine the optimum values of operational parameters of phosphoric acid, Sulphuric acid, ammonia in the plant units of NPK Phonska Reaction using partial derivatives of regression model equations.The equation of the regression model of the amount of fertilizer production against operational parameters is as follows:  = −245,9 − 1,17 1 + 0,00277 1 2 + 2,718 2 − 0,00161 2 2 + 8,49 3 − 0,01378 3 2 Where: Y : Total Fertilizer Production X1 : Phosphoric Acid X2 : Sulphuric Acid X3 :AmmoniaTo find the optimum value, it is necessary to find a value where the partial derivative of each variable X1, X2, and X3 is equal to zero (0).Partial derivative of X1:  1 = −245,9 − 1,17 1 + 0,00277 1 2 + 2,718 2 − 0,00161 2 2 + 8,49 3 − 0,01378 3 2

Table 1 .
Test F Amount of Fertilizer Production Against Phosphoric Acid Use

Table 2 .
Test F Amount of Fertilizer Production Against the Use of Sulphuric Acid

Table 3 .
Test F Amount of Fertilizer Production Against Ammonia Use

Table 4 .
Test F Amount of Fertilizer Production Against Plant Downtime

Table 5 .
Test t Amount of Fertilizer Production Against the Use of Phosphoric Acid

Table 6 .
Test t Amount of Fertilizer Production Against the Use of Sulphuric Acid

Table 7 .
Test t Amount of Fertilizer Production Against Ammonia Use

Table 8 .
Test t Amount of Fertilizer Production Against Plant Downtime

Table 9 .
The Value of the Correlation of the Amount of Fertilizer Production to the Use of Phosphoric Acid

Table 10 .
The Value of the Correlation of the Amount of Fertilizer Production to the Use of Sulphuric Acid

Table 11 .
The Value of the Correlation of the Amount of Fertilizer Production to the Use of Ammonia

Table 12 .
The Correlation Value of the Amount of Fertilizer Production to Factory Downtime

Table 13 .
Testing of Identical Residual Assumptions of Phosphoric Acid

Table 14 .
Testing of Identical Residual Assumptions of Sulphuric Acid

Table 15 .
Testing of Ammonia's Identical Residual Assumption

Table 16 .
Testing Identical Residual Assumptions Factory Downtime

Table 17 .
Durbin-Watson Test CalculationTable Total Production of Fertilizer-

Table 18 .
Durbin-Watson Test Calculation Total Production of Fertilizer-

Table 19 .
Durbin-Watson Test CalculationTable Total Fertilizer-Ammonia Production Residual amount of fertilizer production against ammonia use in NPK Phonska plant unit The reaction is dependent.2) The Durbin-Watson test in this test uses a significance level of α = 5% with n = 2 and a lot of data as much as 200, so that from the Durbin-Watson table obtained a dL value of 1.7483 and a dU value of 1.7887.3) Statistical test criteria: a) If the value of dcount < dL then reject H0 or residual is not independent.b) If the value of dU ≤ dcount ≤ 4 -dU then it fails to reject H0 or the residual is independent.4) Decision Making: From the results of the Durbin-Watson test, a calculated value of 1.9596 is obtained, that 1.7887 ≤ 1.9596 ≤ 2.2113, a decision can be made to fail to reject H0 which means that residual data is independent.

Table 20 .
Durbin-Watson Test Calculation Table Total Fertilizer Production-