Appendix: Finding the optimal trajectory using LQR method
First of all, the mathematical model should be determined. It should be mentioned that in this paper, subscript 1 stands for well 1, 2 for well 2, and so on.
State variables:
The state variable is as the following form:
$$ {\mathrm{x}}_1\left(\mathrm{t}\right),{\mathrm{x}}_2\left(\mathrm{t}\right),\dots .,{\mathrm{x}}_{\mathrm{n}}\left(\mathrm{t}\right) $$
(1)
In which:
$$ {\displaystyle \begin{array}{c}{\mathrm{x}}_1\left(\mathrm{t}\right)-{\mathrm{x}}_6\left(\mathrm{t}\right):{\mathrm{q}}_{\mathrm{o}1}\left(\mathrm{t}\right)-{\mathrm{q}}_{\mathrm{o}6}\left(\mathrm{t}\right)\\ {}{\mathrm{x}}_7\left(\mathrm{t}\right)-{\mathrm{x}}_{12}\left(\mathrm{t}\right):{\mathrm{q}}_{\mathrm{g}1}\left(\mathrm{t}\right)-{\mathrm{q}}_{\mathrm{g}6}\left(\mathrm{t}\right)\\ {}{\mathrm{x}}_{13}\left(\mathrm{t}\right)-{\mathrm{x}}_{18}\left(\mathrm{t}\right):{\mathrm{q}}_{\mathrm{w}1}\left(\mathrm{t}\right)-{\mathrm{q}}_{\mathrm{w}6}\left(\mathrm{t}\right)\end{array}} $$
Thus:
$$ {\displaystyle \begin{array}{c}x(t)=\left[{\mathrm{x}}_1\left(\mathrm{t}\right),{\mathrm{x}}_2\left(\mathrm{t}\right),\dots .,{\mathrm{x}}_{18}\left(\mathrm{t}\right)\right]\\ {}=\left[{\mathrm{q}}_{\mathrm{o}1}\left(\mathrm{t}\right),\dots, {\mathrm{q}}_{\mathrm{o}6}\left(\mathrm{t}\right),{\mathrm{q}}_{\mathrm{g}1}\left(\mathrm{t}\right),\dots, {\mathrm{q}}_{\mathrm{g}6}\left(\mathrm{t}\right),{\mathrm{q}}_{\mathrm{w}1}\left(\mathrm{t}\right),\dots .,{\mathrm{q}}_{\mathrm{w}6}\left(\mathrm{t}\right)\right]\end{array}} $$
(2)
Control inputs:
As mentioned earlier, by a sensitivity analysis, the lift gas injection rates were determined as the control inputs:
$$ \mathrm{u}\left(\mathrm{t}\right)=\left[{\mathrm{u}}_1\left(\mathrm{t}\right),{\mathrm{u}}_2\left(\mathrm{t}\right),\dots {\mathrm{u}}_6\left(\mathrm{t}\right)\right]=\left[{\mathrm{q}}_{\mathrm{inj}1}\left(\mathrm{t}\right),\dots, {\mathrm{q}}_{\mathrm{inj}6}\left(\mathrm{t}\right)\right] $$
(3)
Now, the system should be described in the following form:
$$ \dot{\mathrm{x}}\left(\mathrm{t}\right)=\mathrm{a}\left(\mathrm{x}\left(\mathrm{t}\right),\mathrm{u}\left(\mathrm{t}\right),\mathrm{t}\right) $$
(4)
The general for of proxy models is:
$$ \dot{{\mathrm{x}}_{\mathrm{i}}}\left(\mathrm{t}\right)={\mathrm{b}}_1.{\mathrm{q}}_{\mathrm{g}\mathrm{inj}1}+\dots +{\mathrm{b}}_6.{\mathrm{q}}_{\mathrm{g}\mathrm{inj}6}+{\mathrm{b}}_7.{\mathrm{q}}_{\mathrm{o}1}+\dots +{\mathrm{b}}_{12}.{\mathrm{q}}_{\mathrm{o}6}+{\mathrm{b}}_{13}{\mathrm{q}}_{\mathrm{g}1}+\dots +{\mathrm{b}}_{18}{\mathrm{q}}_{\mathrm{g}6}+{\mathrm{b}}_{19}{\mathrm{q}}_{\mathrm{w}1}+\dots +{\mathrm{b}}_{24}{\mathrm{q}}_{\mathrm{w}6} $$
(5)
In which, \( \dot{\mathrm{x}}\left(\mathrm{t}\right) \) is a representative for all factors of states and coefficients are constants with respect to the time. The value of these parameters is listed in Tables 5, 6, and 7. Based on these tables, four new variables are defined, bginj, bo, bg, and bw. Each variable is an 18 × 6 matrix in which each row is the coefficient of the injection lift gas, oil rates, gas rate, and water rates, respectively, in proxy models and there are 18 rows in which each row is related to one of the proxy models (first 6 for \( {\dot{\mathrm{q}}}_{\mathrm{o}} \), second 6 for \( {\dot{\mathrm{q}}}_{\mathrm{g}} \) and the third 6 for \( {\dot{\mathrm{q}}}_{\mathrm{w}} \)). These data are extracted from Tables 5, 6, and 7.
$$ {\mathrm{b}}_{\mathrm{ginj}}=\left[\begin{array}{cc}\begin{array}{c}\begin{array}{c}19.51869\\ {}77.90273\\ {}-128.864\end{array}\kern0.5em \begin{array}{c}63.99556\\ {}133.3043\\ {}-432.835\end{array}\kern0.5em \begin{array}{c}-43.6708\\ {}52.13013\\ {}-49.3166\end{array}\\ {}\begin{array}{c}109.7704\\ {}-44.6052\\ {}18.63184\end{array}\kern0.5em \begin{array}{c}68.12849\\ {}-27.6772\\ {}-38.2472\end{array}\kern0.5em \begin{array}{c}-157.979\\ {}-168.19\\ {}-47.0328\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}5475.533\\ {}24189.97\\ {}-40945.5\end{array}& \begin{array}{c}44134.73\\ {}32682.59\\ {}-137149\end{array}& \begin{array}{c}-46529.7\\ {}25692.67\\ {}-14317.7\end{array}\end{array}\end{array}& \begin{array}{c}\begin{array}{ccc}\begin{array}{c}-5.74883\\ {}98.12983\\ {}-312.155\end{array}& \begin{array}{c}35.95253\\ {}-54.9185\\ {}56.22917\end{array}& \begin{array}{c}9.092735\\ {}69.70845\\ {}-104.394\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}144.0481\\ {}7.59055\\ {}-54.1682\end{array}& \begin{array}{c}22.23096\\ {}127.9731\\ {}10.14346\end{array}& \begin{array}{c}141.8283\\ {}-79.2866\\ {}-26.1613\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-24590.6\\ {}22380.04\\ {}-99272.1\end{array}& \begin{array}{c}43149.59\\ {}-24794.2\\ {}17038.02\end{array}& \begin{array}{c}-4371.01\\ {}21343.64\\ {}-33375.7\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{ccc}\begin{array}{c}44214.27\\ {}-20817.3\\ {}21421.14\end{array}& \begin{array}{c}56303.33\\ {}-12549.7\\ {}-12178.5\end{array}& \begin{array}{c}-47109.7\\ {}-70534.9\\ {}-25986\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.75099\\ {}3945.394\\ {}-1430.34\end{array}& \begin{array}{c}2.737366\\ {}6905.261\\ {}-4924.41\end{array}& \begin{array}{c}-0.45077\\ {}2246.148\\ {}-620.212\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}4685.601\\ {}-822.303\\ {}0.377439\end{array}& \begin{array}{c}5191.368\\ {}-754.013\\ {}-1.1793\end{array}& \begin{array}{c}-6819.2\\ {}-3183.02\\ {}-1.56708\end{array}\end{array}\end{array}& \begin{array}{c}\begin{array}{ccc}\begin{array}{c}76072.82\\ {}-1076.59\\ {}-27916.4\end{array}& \begin{array}{c}4566.836\\ {}55973.33\\ {}-1610.56\end{array}& \begin{array}{c}58981.14\\ {}-35597.5\\ {}-10708.9\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.219743\\ {}5028.765\\ {}-3533.97\end{array}& \begin{array}{c}0.471764\\ {}-2447.13\\ {}649.9386\end{array}& \begin{array}{c}0.170363\\ {}3409.162\\ {}-1161.85\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}7364.745\\ {}-1106.43\\ {}-1.81484\end{array}& \begin{array}{c}996.7621\\ {}2350.329\\ {}0.575917\end{array}& \begin{array}{c}6902.153\\ {}-2800.46\\ {}-0.89859\end{array}\end{array}\end{array}\end{array}\right] $$
$$ {\mathrm{b}}_{\mathrm{o}}=\left[\begin{array}{cc}\begin{array}{c}\begin{array}{c}-3.91722\\ {}12.89984\\ {}-8.53318\end{array}\kern0.5em \begin{array}{c}5.943169\\ {}-26.5249\\ {}30.12196\end{array}\kern0.5em \begin{array}{c}-1.91383\\ {}24.08847\\ {}-9.33007\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-11.5793\\ {}-14.9257\\ {}-1.53236\end{array}& \begin{array}{c}3.409462\\ {}28.79974\\ {}6.915694\end{array}& \begin{array}{c}9.707503\\ {}-20.9951\\ {}-5.04067\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-5280.58\\ {}4441.473\\ {}-2584.63\end{array}& \begin{array}{c}7701.111\\ {}-9096.09\\ {}9570.217\end{array}& \begin{array}{c}-4404.73\\ {}8922.27\\ {}-2800.31\end{array}\end{array}\end{array}& \begin{array}{c}\begin{array}{ccc}\begin{array}{c}0.398172\\ {}-1.93508\\ {}2.499365\end{array}& \begin{array}{c}1.201417\\ {}-1.55252\\ {}4.598442\end{array}& \begin{array}{c}0.188561\\ {}1.906188\\ {}-2.25239\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-3.27551\\ {}2.577603\\ {}0.697694\end{array}& \begin{array}{c}3.481796\\ {}3.05501\\ {}0.610075\end{array}& \begin{array}{c}2.157504\\ {}-2.07794\\ {}-1.96687\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}446.1933\\ {}-712.447\\ {}816.4239\end{array}& \begin{array}{c}1235.091\\ {}-355.831\\ {}1486.094\end{array}& \begin{array}{c}-347.146\\ {}667.5498\\ {}-723.234\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{ccc}\begin{array}{c}1882.871\\ {}-6493.98\\ {}1141.214\end{array}& \begin{array}{c}-2285.82\\ {}\mathrm{12,219.46}\\ {}2626.637\end{array}& \begin{array}{c}9969.335\\ {}-9097.29\\ {}44.77462\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-0.06946\\ {}635.477\\ {}-114.678\end{array}& \begin{array}{c}0.107933\\ {}-1328.3\\ {}328.6332\end{array}& \begin{array}{c}-0.03175\\ {}1177.418\\ {}-126.795\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}150.2845\\ {}-434.718\\ {}-0.0294\end{array}& \begin{array}{c}-168.21\\ {}756.1789\\ {}0.22992\end{array}& \begin{array}{c}1080.819\\ {}-95.5565\\ {}-0.15267\end{array}\end{array}\end{array}& \begin{array}{c}\begin{array}{ccc}\begin{array}{c}-839.542\\ {}1108.605\\ {}274.198\end{array}& \begin{array}{c}1114.976\\ {}1245.054\\ {}272.4091\end{array}& \begin{array}{c}109.2575\\ {}-757.668\\ {}-1823.2\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.007628\\ {}-94.0684\\ {}24.3523\end{array}& \begin{array}{c}0.025871\\ {}-83.4397\\ {}45.6129\end{array}& \begin{array}{c}-0.00051\\ {}93.61422\\ {}-22.0002\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-90.886\\ {}46.5982\\ {}0.025594\end{array}& \begin{array}{c}129.87\\ {}178.6252\\ {}0.015262\end{array}& \begin{array}{c}8.654326\\ {}39.51658\\ {}-0.08096\end{array}\end{array}\end{array}\end{array}\right] $$
$$ {\mathrm{b}}_{\mathrm{g}}=\left[\begin{array}{cc}\begin{array}{c}\begin{array}{c}0.00295\\ {}-0.01194\\ {}0.011461\end{array}\kern0.5em \begin{array}{c}-0.01516\\ {}0.064522\\ {}-0.07434\end{array}\kern0.5em \begin{array}{c}0.00842\\ {}-0.08361\\ {}0.040148\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.014634\\ {}0.013661\\ {}0.002549\end{array}& \begin{array}{c}-0.00969\\ {}-0.07384\\ {}-0.01678\end{array}& \begin{array}{c}-0.02937\\ {}0.08051\\ {}0.018769\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}4.185925\\ {}-3.9599\\ {}3.501068\end{array}& \begin{array}{c}-19.5151\\ {}22.11203\\ {}-23.6131\end{array}& \begin{array}{c}17.2177\\ {}-30.739\\ {}12.23742\end{array}\end{array}\end{array}& \begin{array}{c}\begin{array}{ccc}\begin{array}{c}-0.00067\\ {}0.006711\\ {}-0.00966\end{array}& \begin{array}{c}-0.00233\\ {}0.005558\\ {}-0.01664\end{array}& \begin{array}{c}-0.00031\\ {}-0.00237\\ {}5.17\mathrm{E}-05\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.007456\\ {}-0.00921\\ {}-0.00179\end{array}& \begin{array}{c}-0.00603\\ {}-0.01085\\ {}-0.0021\end{array}& \begin{array}{c}-0.00661\\ {}0.002334\\ {}0.000682\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-0.87069\\ {}2.448072\\ {}-3.18104\end{array}& \begin{array}{c}-2.75093\\ {}1.310461\\ {}-5.34408\end{array}& \begin{array}{c}0.483808\\ {}-0.89485\\ {}0.039186\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{ccc}\begin{array}{c}-0.8383\\ {}5.89965\\ {}-0.30403\end{array}& \begin{array}{c}5.24383\\ {}-31.3031\\ {}-6.16007\end{array}& \begin{array}{c}-32.4893\\ {}34.74476\\ {}0.830269\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}3.45\mathrm{E}-05\\ {}-0.59202\\ {}0.147895\end{array}& \begin{array}{c}-0.00028\\ {}3.225792\\ {}-8.11\mathrm{E}-01\end{array}& \begin{array}{c}0.000131\\ {}-4.08033\\ {}0.519035\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-0.01804\\ {}0.40027\\ {}5.77\mathrm{E}-05\end{array}& \begin{array}{c}0.369355\\ {}-1.91864\\ {}-0.00056\end{array}& \begin{array}{c}-3.47246\\ {}0.677646\\ {}0.000582\end{array}\end{array}\end{array}& \begin{array}{c}\begin{array}{ccc}\begin{array}{c}1.01575\\ {}-3.93974\\ {}-0.58507\end{array}& \begin{array}{c}-1.71787\\ {}-4.46108\\ {}-0.94445\end{array}& \begin{array}{c}-1.30057\\ {}0.88306\\ {}1.040257\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-9.62\mathrm{E}-06\\ {}0.326058\\ {}-9.12\mathrm{E}-02\end{array}& \begin{array}{c}-3.73\mathrm{E}-05\\ {}0.293292\\ {}-1.71\mathrm{E}-01\end{array}& \begin{array}{c}1.34\mathrm{E}-05\\ {}-0.11558\\ {}-3.98\mathrm{E}-03\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.097345\\ {}-0.16528\\ {}-7.45\mathrm{E}-05\end{array}& \begin{array}{c}-0.21098\\ {}-0.55329\\ {}-6.10\mathrm{E}-05\end{array}& \begin{array}{c}-0.16377\\ {}-0.06059\\ {}4.96\mathrm{E}-05\end{array}\end{array}\end{array}\end{array}\right] $$
$$ {\mathrm{b}}_{\mathrm{w}}=\left[\begin{array}{cc}\begin{array}{c}\begin{array}{c}-2.993365431\\ {}-2.964251078\\ {}-17.90377957\end{array}\kern0.5em \begin{array}{c}0.702129\\ {}0.010334\\ {}0.545039\end{array}\kern0.5em \begin{array}{c}0.605118\\ {}0.310069\\ {}0.781682\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-25.6804916\\ {}-13.61059976\\ {}0.051890422\end{array}& \begin{array}{c}0.176114\\ {}0.762349\\ {}0.92728\end{array}& \begin{array}{c}0.860919\\ {}0.185616\\ {}0.259876\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-802.0512528\\ {}-1610.738998\\ {}-5779.778909\end{array}& \begin{array}{c}0.773891\\ {}0.430653\\ {}0.907082\end{array}& \begin{array}{c}0.285287\\ {}0.875233\\ {}0.492885\end{array}\end{array}\end{array}& \begin{array}{c}\begin{array}{ccc}\begin{array}{c}0.038267\\ {}0.155207\\ {}0.3274\end{array}& \begin{array}{c}0.19515\\ {}0.994083\\ {}0.095263\end{array}& \begin{array}{c}0.330682\\ {}0.450239\\ {}0.441683\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.655948\\ {}0.51677\\ {}0.832475\end{array}& \begin{array}{c}0.856538\\ {}0.640152\\ {}0.175126\end{array}& \begin{array}{c}0.27094\\ {}0.868951\\ {}0.879842\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.824397\\ {}0.067641\\ {}0.20714\end{array}& \begin{array}{c}0.173061\\ {}0.339907\\ {}0.025819\end{array}& \begin{array}{c}0.863643\\ {}0.177062\\ {}0.170209\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{ccc}\begin{array}{c}-\mathrm{10,300.81067}\\ {}-5297.440964\\ {}-415.5435951\end{array}& \begin{array}{c}0.784127\\ {}0.076086\\ {}0.417079\end{array}& \begin{array}{c}0.170393\\ {}0.40626\\ {}0.554477\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-0.128342286\\ {}-117.9729736\\ {}-1.77\mathrm{E}+02\end{array}& \begin{array}{c}0.041599\\ {}0.764898\\ {}0.113998\end{array}& \begin{array}{c}0.75928\\ {}0.996463\\ {}0.84042\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}-1259.767197\\ {}-795.6930059\\ {}0.025446028\end{array}& \begin{array}{c}0.288928\\ {}0.309191\\ {}0.23777\end{array}& \begin{array}{c}0.124354\\ {}0.747136\\ {}0.127723\end{array}\end{array}\end{array}& \begin{array}{c}\begin{array}{ccc}\begin{array}{c}0.731294\\ {}0.430213\\ {}0.029752\end{array}& \begin{array}{c}0.246622\\ {}0.659312\\ {}0.866706\end{array}& \begin{array}{c}0.081841\\ {}0.218391\\ {}0.894414\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.324874\\ {}0.265018\\ {}0.337497\end{array}& \begin{array}{c}0.521822\\ {}0.792067\\ {}0.429436\end{array}& \begin{array}{c}0.951888\\ {}0.134609\\ {}0.793779\end{array}\end{array}\\ {}\begin{array}{ccc}\begin{array}{c}0.514081\\ {}0.336661\\ {}0.53412\end{array}& \begin{array}{c}0.387512\\ {}0.883354\\ {}0.056697\end{array}& \begin{array}{c}0.97861\\ {}0.333656\\ {}0.468575\end{array}\end{array}\end{array}\end{array}\right] $$
For that, the proxy models which were developed earlier are used. As all proxy models are linear, for finding the optimal control trajectory, it is better to describe the problem as a linear regulator problem in the following form:
$$ \dot{\mathrm{x}}\left(\mathrm{t}\right)=\mathrm{A}\left(\mathrm{t}\right)\mathrm{x}\left(\mathrm{t}\right)+\mathrm{B}\left(\mathrm{t}\right)\mathrm{u}\left(\mathrm{t}\right) $$
(6)
In the proxy models of this study, A(t) and B(t) are not time-variant, we have:
$$ \mathrm{A}\left(\mathrm{t}\right)=\left[{\mathrm{b}}_{\mathrm{o}}\ {\mathrm{b}}_{\mathrm{g}}\ {\mathrm{b}}_{\mathrm{w}}\right] $$
(7)
In which “A(t)” or “A” is an 18 × 18 matrix.
$$ \mathrm{B}\left(\mathrm{t}\right)={\mathrm{b}}_{\mathrm{ginj}} $$
(8)
Also, “B(t)” or “B” is a 6 × 18 matrix. As/while/where “x(t)” is a 18 × 1 and “u(t)” is a 6 × 1 so, the equation is consistent. In the problem of this paper, x(t0) and x(tf) are unknown. In addition, the total amount of lift gas is limited and equal to 10 MSCF and no negative injection rate is meaningful, so:
Constraints:
$$ \mathrm{u}\left(\mathrm{t}\right).{\left[1\ 1\ 1\ 1\ 1\ 1\right]}^{\prime }<10 $$
(9)
$$ 0\le \mathrm{u}\left(\mathrm{t}\right)\le 10 $$
(10)
The objective of controlling the injection gas rate is to maximize the total NPV. The NPV function is defined as follows:
$$ {\mathrm{J}}_0={\int}_{{\mathrm{t}}_0}^{{\mathrm{t}}_{\mathrm{f}}}\mathrm{x}\left(\mathrm{t}\right).\mathrm{pr}\left(\mathrm{t}\right)-\mathrm{OPEX}\left(\mathrm{t}\right)\mathrm{dt} $$
(11)
In which pr(t) is the price of oil, gas, and the cost of water treatment, while it is an 18 × 1 matrix of 6 time repetition of oil price, gas price, and water treatment cost. Here, the oil price is assumed 65 USD/BBL, gas price is 3 USD/MSCF, and water treatment cost is 5 USD/BBL. Thus pr(t0) is:
$$ \mathrm{pr}\left({\mathrm{t}}_0\right)=\left[\mathrm{65,65,65,65,65,65,3},3,3,3,3,3,-5,-5,-5,-5,-5,-5\right] $$
(12)
Similarly, the OPEX in the start time is assumed 10,000 USD per month. Both OPEX and price are money and the money value discount during the time. Here, for this purpose, the following equations are defined:
$$ \mathrm{pr}\left(\mathrm{t}\right)=\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{pr}\left({\mathrm{t}}_0\right) $$
(13)
$$ \mathrm{OPEX}\left(\mathrm{t}\right)=\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{OPEX}\left({\mathrm{t}}_0\right) $$
(14)
In which r is the discount rate and its value is assumed 0.01 per month. In addition, the problem is solved in time zero to month 20, so t0 = 0 and tf is 20.
Thus, the performance function is as follows:
$$ {\mathrm{J}}_0={\int}_0^{20}\mathrm{x}\left(\mathrm{t}\right).\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{pr}\left({\mathrm{t}}_0\right)-\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{OPEX}\left({\mathrm{t}}_0\right)\ \mathrm{dt} $$
(15)
As the amount of lift gas injection rate and the compressor capacity is constant during the simulation time, the injection rate is not involved in the performance function. The equation can be separated into two other equations as follows:
$$ {\mathrm{J}}_0={\int}_0^{20}\mathrm{x}\left(\mathrm{t}\right).\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{pr}\left({\mathrm{t}}_0\right)\mathrm{dt}-{\int}_0^{20}\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{OPEX}\left({\mathrm{t}}_0\right)\ \mathrm{dt} $$
(16)
$$ {\int}_0^{20}\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{OPEX}\left({\mathrm{t}}_0\right)\ \mathrm{dt}={\int}_0^{20}\frac{1}{{\left(1+0.01\right)}^{\mathrm{t}}}\ 100\ 000\ \mathrm{dt}=1813600\kern0.5em $$
(17)
$$ {\mathrm{J}}_0={\int}_0^{20}\mathrm{x}\left(\mathrm{t}\right).\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{pr}\left({\mathrm{t}}_0\right)\mathrm{dt}-1813600 $$
(18)
As the purpose of the optimal control is to maximize J, thus, a constant value does not affect, and it should be emphasized that this is because the value of J is not important; its extreme is important. So J can be reduced as follows: (J0 is equivalent to J and introduced to prevent the mistakes. Also, a negative sign is given to the equation because the calculation minimizes the equation.
$$ \mathrm{J}={\int}_0^{20}-\mathrm{x}\left(\mathrm{t}\right).\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{pr}\left({\mathrm{t}}_0\right)\mathrm{dt} $$
(19)
In addition, it should be emphasized that the control trajectory of this study has some limitations which will be explained in the following equations.
$$ \mathrm{u}\left(\mathrm{t}\right).{\left[1\ 1\ 1\ 1\ 1\ 1\right]}^{\prime }<10 $$
(20)
$$ 0\le \mathrm{u}\left(\mathrm{t}\right)\le 10 $$
(21)
As mentioned, the total amount of lift gas is limited and this can be applied in performance function by adding a new term to that.
$$ \mathrm{J}={\int}_0^{20}\left(-\mathrm{x}\left(\mathrm{t}\right).\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{pr}\left({\mathrm{t}}_0\right)+\mathrm{f}\left(\mathrm{u}\left(\mathrm{t}\right)\right)\right)\mathrm{dt} $$
(22)
f(u(t)) is a function of u(t), in which as the summation of injection gas or u is smaller than the maximum available lift gas; it has a small and almost fixed value, but as the total summation of u exceeds the maximum available lift gas, and its value becomes huge, and thus, it falls off the optimum point. For this purpose, exponential functions can be useful. Here, f is defined below:
$$ \mathrm{f}\left(\mathrm{u}\left(\mathrm{t}\right)\right)={\mathrm{e}}^{10^4\ \left(\mathrm{u}\left(\mathrm{t}\right).{\left[1\ 1\ 1\ 1\ 1\ 1\right]}^{\mathrm{T}}-10\right)} $$
(23)
The value of 104 is found by the trial and error method. The value of f(u(t)) for 9.99 is 3.72 × 10−44 which is small enough and has an ignorable effect on the performance function value and for the values smaller than 9.99, the value of f(u(t)) decreases. In addition, for 10.01, the value of f(u(t)) is 2.69 × 1043 which is large enough to increase the value of the performance function as large as it cannot be accepted as the optimum point.
In which:
$$ \mathrm{g}\left(\mathrm{x}\left(\mathrm{t}\right)\right)=-\mathrm{x}\left(\mathrm{t}\right).\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{pr}\left({\mathrm{t}}_0\right)+\mathrm{f}\left(\mathrm{u}\left(\mathrm{t}\right)\right) $$
(24)
Now, the Hamiltonian equation should be developed. The general form of the Hamiltonian is as follows.
Hamiltonian:
$$ \mathrm{H}\left(\mathrm{x}\left(\mathrm{t}\right),\mathrm{u}\left(\mathrm{t}\right),\mathrm{p}\left(\mathrm{t}\right),\mathrm{t}\right)=\mathrm{g}\left(\mathrm{x}\left(\mathrm{t}\right),\mathrm{u}\left(\mathrm{t}\right),\mathrm{t}\right)+{\mathrm{p}}^{\mathrm{T}}\left(\mathrm{t}\right)\left[\mathrm{a}\left(\mathrm{x}\left(\mathrm{t}\right),\mathrm{u}\left(\mathrm{t}\right),\mathrm{t}\right)\right]+\mathrm{f}\left(\mathrm{u}\left(\mathrm{t}\right)\right) $$
(25)
By substituting equations and in equation:
$$ \mathrm{H}\left(\mathrm{x}\left(\mathrm{t}\right),\mathrm{u}\left(\mathrm{t}\right),\mathrm{p}\left(\mathrm{t}\right),\mathrm{t}\right)=-\mathrm{x}\left(\mathrm{t}\right).\frac{1}{{\left(1+\mathrm{r}\right)}^{\mathrm{t}}}\ \mathrm{pr}\left({\mathrm{t}}_0\right)+{\mathrm{p}}^{\mathrm{T}}\left(\mathrm{t}\right)\left[\mathrm{A}.\mathrm{x}\left(\mathrm{t}\right)+\mathrm{B}.\mathrm{u}\left(\mathrm{t}\right)\right]+{\mathrm{e}}^{10^4\ \left(\mathrm{u}\left(\mathrm{t}\right).{\left[1\ 1\ 1\ 1\ 1\ 1\right]}^{\mathrm{T}}-10\right)} $$
(26)
In which r, pr(t0), A, and B are known.
By simplifying and inserting equations in it, it will result in:
$$ {\displaystyle \begin{array}{c}\begin{array}{c}\mathrm{H}=\frac{-1}{(1.01)^{\mathrm{t}}}\left(65{\mathrm{q}}_{\mathrm{o}1}+\cdots +65{\mathrm{q}}_{\mathrm{o}6}+3{\mathrm{q}}_{\mathrm{g}1}+\cdots +3{\mathrm{q}}_{\mathrm{g}6}-5{\mathrm{q}}_{\mathrm{w}1}-{}_{\cdots }-5{\mathrm{q}}_{\mathrm{w}6}\right)\\ {}+\sum \limits_{\mathrm{i}=1}^{18}\left[{\mathrm{p}}_{\mathrm{i}}\cdot \left({\mathrm{b}}_{\mathrm{o}\mathrm{i}1}{\mathrm{q}}_{\mathrm{o}1}+\cdots +{\mathrm{b}}_{\mathrm{o}\mathrm{i}6}{\mathrm{q}}_{\mathrm{o}6}+{\mathrm{b}}_{\mathrm{g}\mathrm{i}1}{\mathrm{q}}_{\mathrm{g}1}+\cdots {\mathrm{b}}_{\mathrm{g}\mathrm{i}6}{\mathrm{q}}_{\mathrm{g}6}+{\mathrm{b}}_{\mathrm{w}\mathrm{i}1}{\mathrm{q}}_{\mathrm{w}1}\right.\right.\end{array}\\ {}\left.\left.+\cdots +{\mathrm{b}}_{\mathrm{w}\mathrm{i}6}{\mathrm{q}}_{\mathrm{w}\mathrm{i}6}+{\mathrm{b}}_{\mathrm{g}\mathrm{i}\mathrm{nj}\mathrm{i}1}{\mathrm{q}}_{\mathrm{g}\mathrm{i}\mathrm{nj}1}+\cdots +{\mathrm{b}}_{\mathrm{g}\mathrm{i}\mathrm{nj}\mathrm{i}6}{\mathrm{q}}_{\mathrm{g}\mathrm{i}\mathrm{nj}6}\right)\right]\\ {}+{\mathrm{e}}^{10^4\left(\mathrm{u}\left(\mathrm{t}\right).{\left[111111\right]}^{\mathrm{T}}-10\right)}\end{array}} $$
(27)
Now, the state, co-state, and Pontryagin equations should be formed and then they will be solved. First, the condition is ignored.
State, co-state, and Pontryagin equations are as follows:
$$ {\dot{\mathrm{x}}}^{\ast}\left(\mathrm{t}\right)=\frac{\mathrm{\partial H}}{\mathrm{\partial p}}\left({\mathrm{x}}^{\ast}\left(\mathrm{t}\right),{\mathrm{u}}^{\ast}\left(\mathrm{t}\right),{\mathrm{p}}^{\ast}\left(\mathrm{t}\right),\mathrm{t}\right) $$
(28)
$$ {\dot{\mathrm{p}}}^{\ast}\left(\mathrm{t}\right)=-\frac{\mathrm{\partial H}}{\mathrm{\partial x}}\left({\mathrm{x}}^{\ast}\left(\mathrm{t}\right),{\mathrm{u}}^{\ast}\left(\mathrm{t}\right),{\mathrm{p}}^{\ast}\left(\mathrm{t}\right),\mathrm{t}\right) $$
(29)
$$ \frac{\mathrm{\partial H}}{\mathrm{\partial u}}\left({\mathrm{x}}^{\ast}\left(\mathrm{t}\right),{\mathrm{u}}^{\ast}\left(\mathrm{t}\right),{\mathrm{p}}^{\ast}\left(\mathrm{t}\right),\mathrm{t}\right)=0 $$
(30)
State equations:
State equation results in a matrix form of all first 18 state equations.
Co-state equations:
For finding the equation of these forms, differentiation is required and finally, a set of equations in the following form are developed. For example, for p1, it is:
$$ {\dot{\mathrm{p}}}_1=65+\left({\mathrm{b}}_{\mathrm{oi}1}+{\mathrm{b}}_{\mathrm{oi}2}+\dots +{\mathrm{b}}_{\mathrm{oi}18}\right)\ {\mathrm{p}}_1 $$
(31)
For other p, there are similar equations in which the whole set of equations can be summarized in the following form:
$$ \dot{\mathrm{p}}=\mathrm{C}.\mathrm{D}-\mathrm{Ep} $$
(32)
In which:
$$ \dot{\mathrm{p}}={\left[{\dot{\mathrm{p}}}_1,\dots, {\dot{\mathrm{p}}}_{18}\right]}^{\mathrm{T}} $$
(33)
$$ C=\left[\begin{array}{cccccccccccccccccc}-65& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& -65& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& -65& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& -65& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& -65& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& -65& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& -3& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& -3& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& -3& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& -3& 0& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& 0& -3& 0& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& -3& 0& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 5& 0& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 5& 0& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 5& 0& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 5& 0& 0\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 5& \begin{array}{c}0\\ {}0\end{array}\\ {}0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 5\end{array}\right] $$
(34)
$$ \mathrm{D}={\left[1,\dots, 1\right]}^{\mathrm{T}}\kern1.5em \mathrm{size}:1\times 18 $$
(35)
$$ \mathrm{E}=\left[\begin{array}{ccc}{\mathrm{b}}_{1,7}& \dots & {\mathrm{b}}_{1,24}\\ {}:& \dots & :\\ {}{\mathrm{b}}_{18,1}& \dots & {\mathrm{b}}_{18,24}\end{array}\right]\kern2.75em \mathrm{size}:18\times 18 $$
(36)
$$ \mathrm{p}={\left[{\mathrm{p}}_1,\dots, {\mathrm{p}}_{18}\right]}^{\mathrm{T}}\kern1.25em \mathrm{size}:1\times 18 $$
(37)
By substituting values in E, there is:
$$ \mathrm{E}=\left[{\mathrm{E}}_1,{\mathrm{E}}_2,{\mathrm{E}}_3\right] $$
(38)
In which E1 to E3 are:
$$ {E}_1=\left[\begin{array}{cccccc}-3.91722& 12.89984& -8.53318& -11.5793& -14.9257& -1.53236\\ {}5.943169& -26.5249& 30.12196& 3.409462& 28.79974& 6.915694\\ {}-1.91383& 24.08847& -9.33007& 9.707503& -20.9951& -5.04067\\ {}0.398172& -1.93508& 2.499365& -3.27551& 2.577603& 0.697694\\ {}1.201417& -1.55252& 4.598442& 3.481796& 3.05501& 0.610075\\ {}0.188561& 1.906188& -2.25239& 2.157504& -2.07794& -1.96687\\ {}0.00295& -0.01194& 0.011461& 0.014634& 0.013661& 0.002549\\ {}-0.01516& 0.064522& -0.07434& -0.00969& -0.07384& -0.01678\\ {}0.00842& -0.08361& 0.040148& -0.02937& 0.08051& 0.018769\\ {}-0.00067& 0.006711& -0.00966& 0.007456& -0.00921& -0.00179\\ {}-0.00233& 0.005558& -0.01664& -0.00603& -0.01085& -0.0021\\ {}-0.00031& -0.00237& 5.17\mathrm{E}-05& -0.00661& 0.002334& 0.000682\\ {}-2.99337& -2.96425& -17.9038& -25.6805& -13.6106& 0.05189\\ {}0.702129& 0.010334& 0.545039& 0.176114& 0.762349& 0.92728\\ {}0.605118& 0.310069& 0.781682& 0.860919& 0.185616& 0.259876\\ {}0.038267& 0.155207& 0.3274& 0.655948& 0.51677& 0.832475\\ {}0.19515& 0.994083& 0.095263& 0.856538& 0.640152& 0.175126\\ {}0.330682& 0.450239& 0.441683& 0.27094& 0.868951& 0.879842\end{array}\right] $$
(39)
$$ {E}_2=\left[\begin{array}{cccccc}-5280.58& 4441.473& -2584.63& 1882.871& -6493.98& 1141.214\\ {}7701.111& -9096.09& 9570.217& -2285.82& 12219.46& 2626.637\\ {}-4404.73& 8922.27& -2800.31& 9969.335& -9097.29& 44.77462\\ {}446.1933& -712.447& 816.4239& -839.542& 1108.605& 274.198\\ {}1235.091& -355.831& 1486.094& 1114.976& 1245.054& 272.4091\\ {}-347.146& 667.5498& -723.234& 109.2575& -757.668& -1823.2\\ {}4.185925& -3.9599& 3.501068& -0.8383& 5.89965& -0.30403\\ {}-19.5151& 22.11203& -23.6131& 5.24383& -31.3031& -6.16007\\ {}17.2177& -30.739& 12.23742& -32.4893& 34.74476& 0.830269\\ {}-0.87069& 2.448072& -3.18104& 1.01575& -3.93974& -0.58507\\ {}-2.75093& 1.310461& -5.34408& -1.71787& -4.46108& -0.94445\\ {}0.483808& -0.89485& 0.039186& -1.30057& 0.88306& 1.040257\\ {}-802.051& -1610.74& -5779.78& -10300.8& -5297.44& -415.544\\ {}0.773891& 0.430653& 0.907082& 0.784127& 0.076086& 0.417079\\ {}0.285287& 0.875233& 0.492885& 0.170393& 0.40626& 0.554477\\ {}0.824397& 0.067641& 0.20714& 0.731294& 0.430213& 0.029752\\ {}0.173061& 0.339907& 0.025819& 0.246622& 0.659312& 0.866706\\ {}0.863643& 0.177062& 0.170209& 0.081841& 0.218391& 0.894414\end{array}\right] $$
(40)
$$ {E}_3=(0.1)\left[\begin{array}{cccccc}-0.06946& 635.477& -114.678& 150.2845& -434.718& -0.0294\\ {}0.107933& -1328.3& 328.6332& -168.21& 756.1789& 0.22992\\ {}-0.03175& 1177.418& -126.795& 1080.819& -95.5565& -0.15267\\ {}0.007628& -94.0684& 24.3523& -90.886& 46.5982& 0.025594\\ {}0.025871& -83.4397& 45.6129& 129.87& 178.6252& 0.015262\\ {}-0.00051& 93.61422& -22.0002& 8.654326& 39.51658& -0.08096\\ {}3.45\mathrm{E}-05& -0.59202& 0.147895& -0.01804& 0.40027& 5.77\mathrm{E}-05\\ {}-0.00028& 3.225792& -8.11\mathrm{E}-01& 0.369355& -1.91864& -0.00056\\ {}0.000131& -4.08033& 0.519035& -3.47246& 0.677646& 0.000582\\ {}-9.62\mathrm{E}-06& 0.326058& -9.12\mathrm{E}-02& 0.097345& -0.16528& -7.45\mathrm{E}-05\\ {}-3.73\mathrm{E}-05& 0.293292& -1.71\mathrm{E}-01& -0.21098& -0.55329& -6.10\mathrm{E}-05\\ {}1.34\mathrm{E}-05& -0.11558& -3.98\mathrm{E}-03& -0.16377& -0.06059& 4.96\mathrm{E}-05\\ {}-0.12834& -117.973& -1.77\mathrm{E}+02& -1259.77& -795.693& 0.025446\\ {}0.041599& 0.764898& 0.113998& 0.288928& 0.309191& 0.23777\\ {}0.75928& 0.996463& 0.84042& 0.124354& 0.747136& 0.127723\\ {}0.324874& 0.265018& 0.337497& 0.514081& 0.336661& 0.53412\\ {}0.521822& 0.792067& 0.429436& 0.387512& 0.883354& 0.056697\\ {}0.951888& 0.134609& 0.793779& 0.97861& 0.333656& 0.468575\end{array}\right] $$
(41)
Algebraic equation (Pontryagin method):
$$ \frac{\mathrm{H}}{\mathrm{\partial u}}\left({\mathrm{x}}^{\ast}\left(\mathrm{t}\right),{\mathrm{u}}^{\ast}\left(\mathrm{t}\right),{\mathrm{p}}^{\ast}\left(\mathrm{t}\right),\mathrm{t}\right)=0 $$
(42)
In which it will result in:
$$ \mathrm{F}.\mathrm{p}+{10}^4{\mathrm{e}}^{10^4\ \left(\mathrm{u}\left(\mathrm{t}\right).{\left[1\ 1\ 1\ 1\ 1\ 1\right]}^{\mathrm{T}}-10\right)}=0 $$
(43)
And F is:
$$ \mathrm{F}=\left[\begin{array}{ccc}{\mathrm{b}}_{\mathrm{ginj}1,1}& \dots & {\mathrm{b}}_{\mathrm{ginj}18,1}\\ {}:& \dots & :\\ {}{\mathrm{b}}_{\mathrm{ginj}1,6}& \dots & {\mathrm{b}}_{\mathrm{ginj}18,6}\end{array}\right] $$
(44)
By substituting the values of binj, it will result in :
$$ \mathrm{F}=\left[{\mathrm{F}}_1,{\mathrm{F}}_2{\mathrm{F}}_3\right] $$
(45)
$$ {F}_1=\left[\begin{array}{cccccc}19.51869& 77.90273& -128.864& 109.7704& -44.6052& 18.63184\\ {}63.99556& 133.3043& -432.835& 68.12849& -27.6772& -38.2472\\ {}-43.6708& 52.13013& -49.3166& -157.979& -168.19& -47.0328\\ {}-5.74883& 98.12983& -312.155& 144.0481& 7.59055& -54.1682\\ {}35.95253& -54.9185& 56.22917& 22.23096& 127.9731& 10.14346\\ {}9.092735& 69.70845& -104.394& 141.8283& -79.2866& -26.1613\end{array}\right] $$
(46)
$$ {F}_2=\left[\begin{array}{cccccc}5475.533& 24189.97& -40945.5& 44214.27& -20817.3& 21421.14\\ {}44134.73& 32682.59& -137149& 56303.33& -12549.7& -12178.5\\ {}-46529.7& 25692.67& -14317.7& -47109.7& -70534.9& -25986\\ {}-24590.6& 22380.04& -99272.1& 76072.82& -1076.59& -27916.4\\ {}43149.59& -24794.2& 17038.02& 4566.836& 55973.33& -1610.56\\ {}-4371.01& 21343.64& -33375.7& 58981.14& -35597.5& -10708.9\end{array}\right] $$
(47)
$$ {F}_3=\left[\begin{array}{cccccc}0.75099& 3945.394& -1430.34& 4685.601& -822.303& 0.377439\\ {}2.737366& 6905.261& -4924.41& 5191.368& -754.013& -1.1793\\ {}-0.45077& 2246.148& -620.212& -6819.2& -3183.02& -1.56708\\ {}0.219743& 5028.765& -3533.97& 7364.745& -1106.43& -1.81484\\ {}0.471764& -2447.13& 649.9386& 996.7621& 2350.329& 0.575917\\ {}0.170363& 3409.162& -1161.85& 6902.153& -2800.46& -0.89859\end{array}\right] $$
(48)
Now, there are 42 linear equations, 18 from equation, 18 from equation and 6 from equation.
From equation, it results in:
$$ \mathrm{F}.\mathrm{p}=\mathrm{G} $$
(49)
In which G is:
$$ {\left[-{10}^4{\mathrm{e}}^{10^4\ \left(\mathrm{u}\left(\mathrm{t}\right).{\left[1\ 1\ 1\ 1\ 1\ 1\right]}^{\mathrm{T}}-10\right)},\dots, -{10}^4{\mathrm{e}}^{10^4\ \left(\mathrm{u}\left(\mathrm{t}\right).{\left[1\ 1\ 1\ 1\ 1\ 1\right]}^{\mathrm{T}}-10\right)}\right]}^{\mathrm{T}}18\times 1 $$
(50)
If the total amount of lift gas is less than 10 (assumed total amount of lift gas), each term of the G is very near to 0. For example, if it is 9.99, each term of G will be 3.7E−40. If it is a little more than 10, each term of G will be very negative, for example, 10.01 each term of G is − 2.7E−47. Thus, the algebraic equation is not sensitive to the value of a specific u but the summation of all terms of u.
