Comparative analysis of Jacobi's and game theory models for wealth creation from solid wastes in Enugu State, Nigeria

 Corresponding author: Anthony Nnamdi Ezemerihe Copyright © 2025 Author(s) retain the copyright of this article. This article is published under the terms of the Creative Commons Attribution Liscense 4.0. Comparative analysis of Jacobi’s and game theory models for wealth creation from solid wastes in Enugu State, Nigeria Anthony Nnamdi Ezemerihe 1, * and Shedrack Agafenachukwu Ume 2 1 Department of Building, Faculty of Environmental Sciences, Enugu State University of Science and Technology, Enugu, Enugu State, Nigeria. 2 Managing Director/Chief Executive, Everwinners Construction Company (Nigeria) Limited, Enugu, Enugu State, Nigeria. World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 Publication history: Received on 08 April 2025; revised on 16 May 2025; accepted on 19 May 2025 Article DOI: https://doi.org/10.30574/wjarr.2025.26.3.1900 Abstract The study is aimed at carrying out comparative analysis of Jacobi’s iteration Optimization and Game theory models for wealth creation from Solid Wastes generation in Enugu. The objectives are to quantify solid wastes volume and characteristics generated, develop the optimization modeling cost for cost effective solid wastes management and planning, apply Jacobi’s iteration model and compare the results with Game theory models to optimize wealth creation in Enugu. There is progressive increase on solid wastes generation due to urbanization and industrialization, unhygienic disposal and associated problems that pose serious threat to public health and environment. Lack of public enlightenment on environmental control measures, protection of environment and human lives, and other pollution menace made the current Solid waste planning and management unsustainable. There is lack of capacity on the part of Enugu State Waste Management Authority (ESWAMA) to live up to the challenges. The methodology involve the use of Jacobi’s iteration Optimization and Game theory optimization models to optimize the Solid Wastes generated in Enugu in order to create wealth for the inhabitants of Enugu. The results show that Jacobi’s iteration optimization model objective function maximized profit at Z = N21,072,853.00 (N21.07million) per ton per day based on fifty years projection from 2006 census projection of the population of Enugu municipal. The benefit of cost recovery for optimal solution using Jacobi’s iteration was N7.67billion per ton per annum with total revenue of N18.409trillion. The result obtained from Game theory optimization model was N18.284trillion which compares favourably with Jacobi’s iteration with difference of N125.00billion. The work concludes that proper management of solid waste generated can create wealth through source reduction, re-use, recycling, recovery with treatment and disposal. Effective legislation based on the outcome of these optimization techniques will market solid waste as essential commodity, which will in turn improve the general sanitation of Enugu urban. Appropriate measures must be put in place on the implementation for the economic benefit of the inhabitants. Keywords: Game theory; Jacobi’s iteration; Optimization; Solid waste planning and management; Wealth creation 1. Introduction The analysis of Jacobi’s iteration and Game theory optimization for wealth creation in the planning and management of municipal Solid wastes in Enugu Nigeria is imperative due to; •Progressive increase in Solid waste generation rates as a result of urbanization and industrialization with corresponding increase in disposal costs, environmental pollution, health hazards, limited landfill, composting and unkept dump sites that deface the environment. World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 640 • There are insufficient dumpsters and dump sites, and substantial qualities of municipal solid wastes are disposed off unhygienic in open dumps which create problems to public health and environment. • There are delays in collection of solid wastes; inadequate plant, machinery and equipment to face the challenges and lack of reliable data. • Lack of public enlightenment to educate the inhabitants on environmental control measures to protect environment, human lives and reduce pollution menace for effective waste management. • Lack of solid waste treatment plants and waste recycling plants which would assist in creating wealth and employment. • Inconsistent government policies by successive governments on Solid waste planning and management have adversely affected the development of a master plan for solid waste in Enugu. Solid waste generation planning and management has been an issue of concern to successive government in Enugu both at state and local government areas. Various strategies to tackling the problem enunciated by successive governments are inconsistent with the global standards. Solid waste problem in Enugu contributes to the contamination of the streams, river, land and the atmosphere. Waste disposal operations are becoming increasingly sophisticated with specialist companies and facilities, leaving the Enugu State with great responsibility to rise up to the challenges. Aramabi (1998) states that the possession of corrected and adequate information on the rate of generation and composition of wastes generated will make it easy to propose and implement an effective method of management. Therefore, the generation rate and composition of the wastes generated in Enugu must be first identified in order to use the best management options to manage the waste generated. Sincero and Sincere (2006) opined that Solid waste survey and characterization are special tools in bringing to light the generation rate and composition of solid waste. It is too common to have solid waste disposed of in Enugu without attempting to explore the wealth creation options of solid waste management. Management methods such as recovery, recycling, and reuse are very important tools for creating wealth from waste. Oyinlola (1999) stated that it is not essentially every composition of solid waste that can be further utilized as resources for wealth creation. Therefore, solid waste components in Enugu must be well classified to make the compositions readily differentiable to intended stakeholders. Enugu State Waste Management Authority (ESWAMA) charged with management of solid waste in Enugu urban is still grappling with the challenges of articulating an effective and efficient solid waste management programme. Most of the inhabitants in the area do not know what is expected of them as regards solid waste management. Out of the need to get rid of the solid waste from their domains and relieve themselves of its nuisance, some of the inhabitants dump their refuse behind their houses and indiscriminately in nearby open dumps. Some of the solid waste is dumped in the water channels, gullies, river side and any available spaces. In most cases, the refuse accumulates, encroaching on roads and streets. Other inhabitants opt for open burning as their main method of reducing the volume of solid waste leading to air pollution in Enugu. In fact, solid waste collection and disposal is one of the highest environmental problems because, there is no effective existing solid waste management disposal technology for the area. The integrated Solid Wastes Management (ISWM) option refers to the selection and use of appropriate management programs, technologies, and techniques to achieve particular waste management goals and objectives. The U.S. Environmental Protection Agency (EPA) states that ISWM is composed of waste source reduction, recycling, waste combustion, and landfills. These activities can be done in either an interactive or hierarchical way (Clarke and Meantay, 2006). It is important to stress that better solid waste management programs are urgently needed in some countries. Only about half of the waste generated in cities and one-quarter of what is produced in rural areas is collected. Internationally, the World Bank (1984) warns that global waste could increase by 70% by 2050 in a business-as-usual scenario, if ongoing efforts to improve the waste management system are an important part of preserving a healthy human and ecological future. It is against this back drop that the study articulated the use of Jacobis iteration and Game theory optimization models as a basis for wealth creation from solid waste generated in Enugu state. 2. Literature Review The literature review was based on concept of Jacobi’s iteration model and linear programming techniques of Game theory optimization model. World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 641 2.1. Jacobi’s Method The method is named after Carl Gustav Jacob Jacobi. The Jacobi’s method is an iterative algorithm for determining the solutions of a strictly diagonally dominant system of linear equations in numerical linear algebra. It consists of solving for each diagonal element before an approximate value is plugged in. The process is then iterated until it converges. This algorithm is a stripped-down version of the Jacobi transformation method of matrix diagonalization as explained by the following procedure; Description Let,AX = b be a square system of n linear equations, where: …….. (1) 𝐴 = [𝑎 11 𝑎 12 … 𝑎 1𝑛 𝑎 21 𝑎 22 … 𝑎 2𝑛 ⋮ ⋮ ⋱ ⋮ 𝑎 𝑛1 𝑎 𝑛2 … 𝑎 𝑛𝑛 ],X= [𝑥 1 𝑥 2 ⋮ 𝑥 𝑛 ],𝒃=[𝑏 1 𝑏 2 ⋮𝑏 1 ],...........................................(2) Then A can be decomposed into a diagonal component D, and the remainder R: 𝐴 =𝐷+𝑅 𝑤ℎ𝑒𝑟𝑒 𝐷= [𝑎 11 0 … 0 0 𝑎 22 … 0 ⋮ ⋮ ⋱ ⋮ 0 0 … 𝑎 𝑛𝑛 ],and R= [0 𝑎 22 … 𝑎 1𝑛 𝑎 11 0 … 𝑎 2𝑛 ⋮ ⋮ ⋱ ⋮ 𝑎 𝑛1 𝑎 𝑛2 … 0],.......(3) The solution is then obtained iteratively via 𝑥(𝑘+1)= 𝐷−1 (𝑏−𝑅𝑥(𝑘)),..................................................(4) wherex(k)is the kth approximation or iteration of x and x(k+1)is the next or k + 1 iteration of x. The element-based formula is thus: xi(k+1)= 1 aii(bi− ∑aijxj(k) j≠i ),i=1,2,…,n................................(5) The computation of xi(k+1)requires each element in x(k) except itself. Unlike the Gauss–Seidel method, we can't overwrite xi(k) with xi(k+1), as that value will be needed by the rest of the computation. The minimum amount of storage is two vectors of size n. Convergence: The standard convergence condition (for any iterative method) is when the spectral radius of the iteration matrix is less than 1: 𝜌(𝐷−1𝑅)<1. …………… (6) A sufficient (but not necessary) condition for the method to converge is that the matrix A is strictly or irreducibly diagonally dominant. Strict row diagonal dominance means that for each row, the absolute value of the diagonal term is greater than the sum of absolute values of other terms: |𝑎𝑖𝑖|>∑|𝑎𝑖𝑗| 𝑗≠𝑖 .................................. (7) The Jacobi method sometimes converges even if these conditions are not satisfied. Note that the Jacobi method does not converge for every symmetric positive-definite matrix. For example 𝐴= (29 2 1 2 6 1 1 1 15 )→𝐷 −1 𝑅=(0 2 29 1 29 13 0 16 5 5 0) ⇒𝜌(𝐷 −1 𝑅)≈1.0661 ………. (8) World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 642 Example: A linear system of the form Ax = b with initial estimate 𝑥(0) 𝐴= [2 1 5 7],𝑏= [11 13]𝑎𝑛𝑑𝑥(0) = [11] We use the equation,𝑥(𝑘+1)= 𝐷−1(𝑏−𝑅𝑥(𝑘)), described above, to estimate x. First, we rewrite the equation in a more convenient form D − 1 (b − R 𝑥(k)) = T 𝑥(k) + C Where T= 𝐷−1𝑅 𝑎𝑛𝑑 C = D− 1 b. Note that 𝑅=𝐿+𝑈 where L and U are the strictly lower and upper parts of A. From the known values 𝐷−1= [12 ⁄1 017 ⁄],𝐿= [0 0 5 0]𝑎𝑛𝑑 𝑈=[0 1 0 0]. 𝑤𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑇=−𝐷−1(𝐿+𝑈) 𝑎𝑠 𝑇= [12 ⁄1 017 ⁄]{[0 0 −5 0]+[0 −1 0 0]}=[ 0−12 ⁄ −57 ⁄0] Further, C is found as C=[12 ⁄0 0 17 ⁄][11 13]= [112 ⁄ 137 ⁄]. With𝑇 and 𝐶 calculated, we estimate as 𝑥(1) and T𝑥(0)+𝐶: 𝑥(1)= [ 0−12 ⁄ −57 ⁄0][11]+ [112 ⁄ 137 ⁄]=[5.0 8/7]≈[ 5 1.143]. The next iteration yields 𝑥(2)= [ 0−12 ⁄ −57 ⁄0][5.0 87 ⁄]=[112 ⁄ 137 ⁄]= [6914 ⁄ −127 ⁄]≈[4.929 −1.714] This process is repeated until convergence (i.e.,until ‖𝐴𝑥(𝑛)−𝑏‖ is small). The solution after 25 iterations is 𝑥= [7.111 −3.222] Another Example: Suppose we are given the following linear system: 10𝑥1−𝑥2+2x3=6 , −𝑥1+11x2−𝑥3 + 3𝑥4=25 2𝑥1−𝑥2+10x3−𝑥4 =−11 , 3𝑥2−𝑥3+8x4= 15 If we choose (0, 0, 0, 0) as the initial approximation, then the first approximate solution is given by 𝑥1=(6+0−(2∗0))10 ⁄ =0.6, 𝑥2=(25+0+0−(3∗0))11 ⁄ =2511 ⁄=2.2727, World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 643 𝑥3=(−11−( 2∗0)+0+0 )10= −1.1, ⁄ 𝑥4 =(15−( 3∗0 )+0)8=1.875 ⁄ Using the approximations obtained, the iterative procedure is repeated until the desired accuracy has been reached. The following are the approximated solutions after five iterations as shown in Table 1. Table 1 Approximated Solution after five iterations in Jacobi’s Method 𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 0.6 2.27272 -1.1 1.875 1.04727 1.7159 -0.80522 0.88522 0.93263 2.05330 -1.0493 1.13088 1.01519 1.95369 -0.9681 0.97384 0.98899 2.0114 -1.0102 1.02135 The exact solution of the system is (1, 2, -1, 1) Weighted Jacobi method: The weighted Jacobi iteration uses a parameter𝜔to compute the iteration as X(𝑘−1)= 𝜔𝐷−1(𝑏−𝑅𝑥(𝑘))+(1−𝜔)𝑥(𝑘).................. (9) With ω = 2 / 3 being the usual choice. Convergence in the Symmetric Positive Definite Case: In case that the system matrix A is of symmetric positive-definite type one can show convergence. Let 𝐶 =𝐶𝜔 = I − ωD−1𝐴be the iteration matrix. Then, convergence is guaranteed for 𝜌(𝐶𝑤)<1 0<𝜔< 2 𝜆𝑚𝑎𝑥(𝐷−1 𝐴) ,....................................(10) where𝜆𝑚𝑎𝑥 is the maximal eigenvalue. The spectral radius can be minimized for a particular choice of ω = ωopt as follows min𝜌(𝐶𝑤)=(𝐶𝜔𝑜𝑝𝑡)=1− 2 𝑘(𝐷−1 𝐴)+1 f𝑜𝑟 𝜔𝑜𝑝𝑡∶= 2 𝜆𝑚𝑎𝑥(𝐷−1 𝐴)+𝜆𝑚𝑎𝑥(𝐷−1 𝐴) Where, 𝑘is the matrix condition number. 2.2. Linear programming method Game theory model There is some relationship between Game theory and linear programming. Two-person zero-sum games can also be solved by linear programming technique. It has an additional advantage of being able to solve mixed strategy games of larger dimension payroll matrix. To illustrate the transformation of a game problem to a Linear programming problem, consider a payroll matrix of m × n size. Let aij be the element in the ith row and jth column of game payroll matrix, and letting pi be the probabilities of m strategies (I = 1, 2, …, m) for player A. Then the expected gains for player A for each of B’s strategies will be ∑𝑝𝑖𝑎𝑖𝑗 𝑛 𝑖=1 ,𝑗=1,2,…𝑛..................................................(11) The aim of player A is to select asset of strategies with probability pi(I = 1, 2, …, m) on any play of game such that he can maximize his minimum expected gains. To obtain values of probability pi, the value of the game to player A for all strategies by player B must be at least equal to V. thus to maximize the minimum expected gains, it is necessary that 𝜔 𝜔 World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 644 Dividing both sides of the m inequalities and equation by V, the division is valid as long as V > 0. In case V < 0, the direction of the inequality constraints must be reserved. But if V = 0, division would be meaningless. In this case, a constant can be added to all entries of the matrix ensuring that the value of the game (V) for the revised matrix becomes more than zero. After optimal solution is obtained, the true value of the game is obtained by subtracting the same constant value. Let 𝑝𝑖 𝑉=𝑥𝑖,(≥0). Then we have where 𝑝1 𝑉+ 𝑝2 𝑉 + … + 𝑝𝑚 𝑉 = 1. Since the objective of player A is to maximize the value of the game, V which is equivalent to minimizing 1𝑉, the resulting linear programming problem can be stated as 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍𝑝(=1𝑉)=𝑥1+ 𝑥2+...+ 𝑥𝑛 Subject to the constraints: Similarly, player B has a similar problem with the inequalities of the constraints reversed, i.e. minimize the expected loss. Since minimizing of V is equivalent to maximizing 1𝑉, therefore, the resulting linear programming problem can be stated as: Maximize 𝑍𝑞(=1𝑉)= 𝑦1+𝑦2+ …+𝑦𝑛 World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 645 Subject to the constraints.............. It may be noted that the linear programming problem of player B is the dual of linear programming problem of player A and vice versa. Therefore, solution of the dual problem can be obtained from the primal simplex table. Since for both players Zp = Zq, the expected gain to player A in the game will be exactly equal to expected loss to player B. It should be noted that linear programming technique requires all variables to be non-negative and therefore to obtain a non-negative of value V of the game, the data of the problem, i.e. aij = 1 the payoff table should all be non-negative. If there are some negative elements in the payoff table, a constant to every element in the payoff table must be added so as to make the smallest element zero; the solution to this new game will give an optimal mixed strategy for the original game. The value of the original game then equals the value of the new game minus the constant. 2.3. Solid Waste Management Trends in Nigeria Ebikapade and Jim (2016); investigated the current trend of solid waste management and affirm that the challenges inhibiting the attainment of sustainable solid waste management, is a major concern in Nigeria. However, they identified inadequate environmental policies and Legislations, low level of environmental awareness, poor funding and inappropriate technology; corruption and unplanned development were some of the challenges facing solid waste management in the country. They opined that for waste management to work, various aspects of Government services such as engineering, urban planning, Geography, economics, public health and law among others must be brought into waste management system. They contended that effective waste management system could be achieved when there is development of a clear policies as well as adequate enforcement to serve as a major driver towards sustainability in waste management. Ogwueleka (2009) stated that “municipal solid waste management has emerged as one of the greatest challenges facing environmental protection agencies in developing countries especially Nigeria. Solid waste management is characterized by inefficient collection methods, insufficient coverage of the collection system and improper disposal. The waste density ranged from 280 to 370Kg/m3 and the waste generation ranged from 0.44 to 0.66Kg/capita/day”. He identified constraints faced by environmental agencies to include lack of institutional arrangement, insufficient financial resources, absence of bye-laws and standards, inflexible work schedules, insufficient information on quantity and composition of waste, and in appropriate technology; and suggested study of institutional, political, social, financial, economic and technical aspects of municipal solid waste management in order to achieve sustainable and effective solid waste management. The Federal Government of Nigeria has promulgated various laws and regulations to safeguard the environment. These include Federal Environmental Protection Agency (FEPA), which was created under the FEPA Act. Pursuant to the FEPA Act, each state and local government in the country set up its own environmental protection body for the protection and improvement within its jurisdiction. Municipal solid waste management is a major responsibility of state and local government environmental agencies. The agencies are charged with the responsibility of handling, employing and disposing of solid waste generated. The state agencies generate fund from subvention from state governments and internally generated revenue through sanitary levy and stringent regulations with heavy penalties for offenders of illegal dumping and littering of refuse along streets (Ogwueleka, 2003). World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 646 Municipal Solid Waste (MSW): is defined to include refuse from households, non-hazardous solid waste from industrial, commercial and institutional establishments (including hospitals), market waste, yard waste, and street sweepings. Municipal Solid Waste Management (MSWM) refers to collection, transfer, treatment, recycling, resources recovery and disposal of solid waste in urban areas. The goals of municipal solid waste management are to promote the quality of urban environment, generate employment and income, and protect environmental health and support the efficiency and productivity of the economy. Ogwueleka (2009) observed that the volume of solid waste being generated continues to increase at a faster rate than the ability of the agencies to improve on the financial and technical resources needed to parallel this growth. The quantity of solid waste generated in urban areas in industrialized countries is higher than in developing countries; still municipal solid waste management remains inadequate in our built urban environment. The distinction is in the areas of composition, density, political and economic framework, waste amount, access to waste for collection, awareness and attitude. The wastes are heavier, wetter and more corrosive in developing cities than developed cities. In developing countries, local authorities spend 77-95% of their revenue on collection and the balance on disposal (Ogwueleka, 2003), but can only collect almost 50-70% of municipal solid waste (MSW). In the past, the focus has been on the technical aspects of different means of collection and disposal (World Bank, 1992), but recently, attention has been on enhancing institutional arrangement to service delivery, with a special emphasis on privatization (Cointreau, 1994). Nigeria is presently experimenting with the privatization in this sector. The Federal government has instituted National Integrated Municipal Solid Waste Management Intervention Programme (NIMSWMIP) in seven (7) cities in Nigeria. The seven cities are Maiduguri, Kano, Kaduna, Onitsha, Uyo, Ota, and Lagos. Lagos state government established municipal solid waste management policy to encompass private sector participation in waste collection and transfer to designated Landfill sites. Chukwuemeka, Ugwu and Igwegbe (2012) stated that the environment of man lies at the mercy of both natural disaster and negligence on the part of man in the course of controlling the gifts of nature. The later, takes the form of dumping solid/industrial waste in an uncompromising, desert encroachment, erosion, depletion of ozone layer, depletion of natural resources, pollution of land, rivers, seas, the air and generally the environment. Kofoworola (2007) stated that recycling activities have increased all over the world during the last 10-15 years. The amount of resource separated from waste fractions being collected has also increased accordingly. However, the market has not been prepared to receive such vast amount of secondary raw materials, resulting in a large deviation between supply and demand. This situation leads to a low price for secondary raw materials, and in certain instance, even negative prices. At the same time, the collection and transportation costs related to all the waste fractions that are being sourced separately, have increased rapidly (Kofoworola, 2007). It is clear that the capacity of the end market should be investigated before any major recovery and recycling initiative is implemented. Recycling Programmed Implementation Risks: In any recycling programmed development, many decisions regarding financial planning and management of the programme involves risks that must be recognized and properly allocated to the programmed participants (Clarke and Philips, 1999). The risk of most concern to communities and private firms in recycling programmes is monetary loss. Thus, the allocation of risk is the assignment of monetary loss, if it occurs, to a specific party prior to the occurrence of the loss. It must be noted that monetary loss does not refer to a net programmed loss but rather a loss exceeding that budgeted and funded for using responsible assumptions. 3. Results and Discussion 3.1. Analysis of Total Solid Waste Generated in Enugu Urban Using Forecast for the Period Generated The total waste generated from the four (4) locations as at January, 2020 = 2,400,000 kg/day= 2,400 tons/day. The population projection based on 2006 Census of 900319 persons generates 2,400,000kg/day 900319 persons=2.666kg/person/day With the projected population of 1,584,403 in 2056 based on next 50 years from 2006 Census adjusted for each of the four locations we have 1055894 kg/day as stated above. The total waste generated as per projected population of 2056 is World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 647 1584403 900319 ×2,400,000kg/day= 4223577.6 𝑘𝑔/𝑑𝑎𝑦4223577.6 𝑘𝑔 4= 1055894 𝑘𝑔/𝑑𝑎𝑦 We use this to derive the Table 2 below for the quantity of options/day/tons. Table 2 Quantity of Options/Day/Ton Group E-waste (X) Plastics (X2) Ceramics (X3) Metal (X4) Total Available Wastes A 1055894 864 ×100.2 1000 = 122.46 tons 1055894 864 ×88.5 1000 = 108.16 tons 1055894 864 ×1 1000 = 2 tons 1055894 864 ×33.5 1000 = 40.94 tons 1055894 864 ×93.4 1000 = 1142 tons B 1055894 864 ×31.6 1000 = 38.62 tons 1055894 864 ×142.7 1000 = 174.4tons 1055894 864 ×142.5 1000 = 94.4 tons 1055894 864 ×29.2 1000 = 35.68 tons 1055894 864 ×1080 = 1320 tons C 1055894 864 ×90.6 1000 = 111 tons 190 128 864 ×87.6 1000 = 107.2 tons 190 128 864 ×96.4 1000 = 117.8 tons 190 128 864 ×48.5 1000 = 59.2 tons 190 128 864 ×1130 1000 = 1381 tons D 1055894 864 ×122.2 1000 = 149.4 tons 1055894 864 ×40.1 1000 = 49 tons 1055894 864 ×89.3 1000 = 109.2 tons 1055894 864 ×142.5 1000 = 174.2 tons 1055894 864 ×1200 1000 = 1467 tons Table 2 above shows the relationship with maximum available wastes and the various sample waste at location A, B, C, and D. Extracted from information in Appendix 2. These figures are derived based on the population projection for 2056 and will be used to formulate the constraints equation in the linear programming model. 3.2. Cost of the Recycling Options and Formulation of Linear Programming Model Table 3 Cost of the Recycling Options S/N Material type/Option Cost/kg (N) Cost per ton (N ton) 1 e-waste = 2350/kg1.5 (X1) = N3525 1055894 864 ×3525 = N 4308349.6 4308 2 Plastics =771.5 (X2) = N116 1055894 864 ×116 = N 141778.32 141.8 3 Ceramics = 8801.5 (X3) = N1320 1055894 864 ×1320 = N 1613339.44 1613.3 4 Metals = 8461.5 (X4) = N1269 1055894 864 ×1269 = N 1551005.88 1551 The Cost per tons as Stated in Table 3 above will be used to formulate objective function for Profit maximization. Using the forgoing data, the optimization problem can be written in the form World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 654 Discussion of results in Sixteenth (16th) iteration The result of the Sixteenth (16th) iteration shows that the values of the variables are; X1 = –1464.69, X2= –1498.30, X3 = –2627.70, and X4= –2256.26. These values will be used for the seventeenth (17th) iteration. It is worthy to note that values cannot converge on negative values hence, the need for the next iteration. Seventeenth (17th) iteration Substituting the new values of X1 = –1464.69, X2= –1498.30, X3 = –2627.70, and X4= –2256.26 in the RHS of equations (i) to (iv), we have the following result; X1=9.36−0.89(−1498.30)− 0.02(−2627.70)– 0.34 (−2256.26)=2162.03 X2=7.59−0.22 (−1464.69)− 0.54(−2627.70)– 0.21 (−2256.26)=2222.09 X3=11.70−0.94 (−1464.69)−0.91(−1498.30)– 0.5 (−2256.26)=3880.09 X4=8.43−0.86 (−1464.69)− 0.28(−1498.30) – 0.63(−2627.70) = 3343.04 Discussion on the results of Seventeenth (17th) iteration The result of the Seventeenth (17th) iteration shows that the values of the variables are; X1 = 2162.03, X2= 2222.09, X3 = 3880.09, and X4= 3343.04. The iteration shows that it has reached a convergence point where the respective values of X1, X2, X3, X4have attained the optimal solution. Therefore, the optimal solutions for the variables are actual values which are; X1 = 2162, X2= 2222, X3 = 3880, and X4= 3343. Table 4 Summary of Results of Jacobi’s Iteration Model S/N Iterations 0 1 2 3 4 5 6 7 8 9 10 1 X4=9.36–0.89X2 –0.02X3 – 0.34X4 0 9.36 – 0.50 14.90 – 8.54 24.80 – 25.47 47.92 – 60.68 99.34 – 136.61 2 X2=7.59–0.22X1 –0.34X3 – 0.21X4 0 7.59 – 2.56 14.05 – 9.08 25.87 – 25.29 50.22 – 61.21 102.99 – 139.25 3 X3=11.70–0.94X1–0.91X2– 0.50X4 0 11.70 – 8.22 19.06 – 22.47 38.15 – 51.39 80.75 - 113.94 173.30 – 250.17 4 X4=8.43–0.86X1 –0.28X2 – 0.63X3 0 8.43 – 9.12 14.76 – 20.33 32.47 – 44.18 69.79 – 97.72 149.54 – 215.01 S/N Iterations Continued 11 12 13 14 15 16 17 1 X4=9.36–0.89X2 –0.02X3 –0.34X4 211.40 – 301.70 454.99 – 675.17 992.89 – 1464.69 2162.03 2 X2=7.59–0.22X1 –0.34X3 –0.21X4 217.89 – 308.79 467.85 – 686.26 1025.97 – 1498.30 2222.09 3 X3=11.70–0.94X1–0.91X2–0.50X4 374.34 – 546.55 811.41 – 113.94 1212.19 – 2627.70 3880.09 4 X4=8.43–0.86X1 –0.28X2 –0.63X3 322.51 – 470.22 740.92 – 97.72 1025.05 – 2256.26 3343.04 These values of X1, X2, X3, and X4are now substituted in the objective function. Maximize Z = 4308X1 + 142X2 + 1613 X3 + 1551X4 Therefore, the optimal solution is World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 655 Z = 4308 2162 + 142 2222 + 1613 3880 + 1551 3343. = N 21,072,853 per ton per day, = N 21.07 million per ton per day, In a week, the waste generated will yield a revenue of N 21,072,853.00  7 days = N147,509,971.00 = N147.51 million per ton per week Then, the Annual cost will be N147,509,971.00  52 weeks = N7, 670,518,492 per ton per annum N7.67 billion per ton per annum However, The Total Revenue generated are as follows: For a day: Total revenue = N 21,072,853.00  2400 tons =N 50,574,847,200.00 =N 50.575 billion daily For a week: Total revenue = N 147,509,971.00  2400 tons = N 354,023,930,400.00 =N 354.024 billion weekly In a year: Total revenue = N7, 670,518,492  2400 tons = N18,409,244,380,000.00 = N 18.409 trillion annually. 3.4. Comparison of Jacobi’s Optimization Model to Game Theory Model In modeling Jacobi’s Model into Game theory model, we have to formulate the matrix for the computation of waste generation in Enugu urban. The difference from the characterized wastes was classified as organic wastes. The total waste from the option separated were deducted from the total waste generated from the projected period 2056 to determine the total organic waste generated at each location. Also some quantity of wastes generated in some wastes location at Enugu East and Enugu North were separated to create the fifth location referred to as Enugu East and the former Enugu East were renamed Enugu East Central in order to have a 5 × 5 matrix for the purpose of determining the Game theory model. This resulted to the information in Table 5 with the same quantity of tons of waste generated per day in the city. Table 5 Quantity of waste options generated at various locations in Enugu urban Waste locations Tons of available wastes generated E-waste (x1) Plastics (x2) Ceramics (x3) Metals (x4) Organic wastes (x5) Total wastes Enugu South (A1) 122 108 3 41 869 1142 Enugu East Central (A2) 31 142 82 28 752 1035 Enugu North East (A3) 111 107 118 59 986 1381 Enugu North Central (A4) 138 41 60 139 779 1157 World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 656 Enugu East (A5) 19 40 61 43 432 595 Total 421 438 323 310 3818 5310 The total revenue generated of N50.575 billion per day were used to determine the benefits of each of the tons of waste generated using pro-rata adjustment as shown below. First (1st) row: (𝑖).122 1142×50.575=5.40 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖).108 1142×50.575=4.78 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖𝑖).2 1142×50.575=0.09 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑣).41 1142×50.575=1.82 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑣).869 1142×50.575=38.48 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 Second (2nd) row: (𝑖).31 1035×50.575=1.51 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖).142 1035×50.575=6.94 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖𝑖).82 1035×50.575=4.01 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑣).28 1035×50.575=1.37 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑣).752 1035×50.575=36.75 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 Third (3rd) row: (𝑖).111 1381×50.575=4.07 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖).107 1381×50.575=3.92 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖𝑖).118 1381×50.575=4.32 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑣).59 1381×50.575=2.16 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑣).986 1381×50.575=36.11 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 Fourth (4th) row: (𝑖).138 1157×50.575=6.03 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖).41 1157×50.575=1.79 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖𝑖).60 1157×50.575=2.62 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑣).139 1157×50.575=6.08 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑣).779 1157×50.575=34.05 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 Fifth (5th) row: (𝑖).19 595×50.575=1.62 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖).40 595×50.575=3.40 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑖𝑖).61 595×50.575=5.19 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑖𝑣).43 595×50.575=3.66 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 (𝑣).432 595×50.575=36.72 𝑏𝑖𝑙𝑙𝑖𝑜𝑛 These costs are summarized in Table 6. Table 6 Summary of costs/benefits of various waste types generated Player A Player B Minimum B1 (X1) B2 (X2) B3 (X3) B4 (X4) B5 (X5) A1 5.40 4.78 0.09 1.82 38.48 0.09 A2 --1.51 6.94 4.01 1.37 36.75 1.37 A3 4.07 3.92 4.32 2.16 36.11 2.16 A4 6.03 1.79 2.62 6.08 34.05 1.79 A5 1.62 3.40 5.19 3.66 36.72 1.62 Maximum 6.03 6.94 5.19 6.08 38.48 Minimum = 2.16 World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 657 Maximum = 5.19 Since there is no saddle point, the value of the game will be calculated through iteration. Also no row or column is completely dominated by the other. So, we apply the linear programming of game theory. Let the value of the game = V and q1, q2, q3, q4, q5 be the probabilities of selecting the strategies B1, B2, B3, B4, B5 respectively. The objectives/benefits is to maximize value of the wastes which will be generated from Table 6. { 5.40𝑞1+4.78𝑞2+0.09𝑞3+1.82𝑞4+38.48𝑞5≤𝑉 1.51𝑞1+6.94𝑞2+4.01𝑞3+1.37𝑞4+36.75𝑞5≤𝑉 4.07𝑞1+3.92𝑞2+4.32𝑞3+2.16𝑞4+36.11𝑞5≤𝑉 6.03𝑞1+1.79𝑞2+2.62𝑞3+6.08𝑞4+34.05𝑞5≤𝑉 1.62𝑞1+3.4𝑞2+5.19𝑞3+3.66𝑞4+36.72𝑞5≤𝑉 𝑞1+𝑞2+𝑞3+𝑞4+𝑞5=1 (𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛) } ……… (32) Divide Equations 32 through by V, we have; { 5.40𝑞1/𝑉+4.78𝑞2/𝑉+0.09𝑞3/𝑉+1.82𝑞4/𝑉+38.48𝑞5/𝑉≤1 1.51𝑞1/𝑉+6.94𝑞2/𝑉+4.01𝑞3/𝑉+1.37𝑞4/𝑉+36.75𝑞5/𝑉≤1 4.07𝑞1/𝑉+3.92𝑞2/𝑉+4.32𝑞3/𝑉+2.16𝑞4/𝑉+36.11𝑞5/𝑉≤1 6.03𝑞1/𝑉+1.79𝑞2/𝑉+2.62𝑞3/𝑉+6.08𝑞4/𝑉+34.05𝑞5/𝑉≤1 1.62𝑞1/𝑉+3.4𝑞2/𝑉+5.19𝑞3/𝑉+3.66𝑞4/𝑉+36.72𝑞5/𝑉≤1 𝑞1∗/𝑉+𝑞2/𝑉+𝑞3/𝑉+𝑞4/𝑉+𝑞5/𝑉=1 } … (33) 𝑙𝑒𝑡𝑞1 𝑉=𝑥1,𝑞2 𝑉=𝑥2,𝑞3 𝑉=𝑥3,𝑞4 𝑉=𝑥4𝑎𝑛𝑑,𝑞5 𝑉=𝑥5 (34) The values in equations 33 and 34 are converted into a linear programming problem as; 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒𝑍𝑝=(1𝑉)=𝑥1+𝑥2+ 𝑥3+ 𝑥4+𝑥5 Subject to: 5.4𝑥1+4.78𝑥2+ 0.09𝑥3+ 1.82𝑥4+38.48𝑥5≤1 1.51𝑥1+6.94𝑥2+ 4.01𝑥3+ 1.37𝑥4+36.75𝑥5≤1 4.07𝑥1+3.92𝑥2+ 4.32𝑥3+ 2.16𝑥4+36.11𝑥5≤1....... 6.03𝑥1+1.97𝑥2+ 2.62𝑥3+ 6.08𝑥4+34.05𝑥5≤1 (35) 1.62𝑥1+3.40𝑥2+ 5.19𝑥3+ 3.66𝑥4+36.72𝑥5≤1 Since we have more than two variables, the simplex method of linear programming is used to solve the problem by introducing slack variables to convert the inequalities to equations which becomes; 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒𝑍𝑞=(1𝑉)=𝑥1+𝑥2+𝑥3+𝑥4+𝑥5+0𝑆1+0𝑆2+0𝑆3+0𝑆4+0𝑆5 Subject to the following constraints; 5.4𝑥1+4.78𝑥2+ 0.09𝑥+ 1.82𝑥4+38.48𝑥5+𝑆1+0𝑆2+0𝑆3+0𝑆4+0𝑆5=1 1.51𝑥1+6.94𝑥2+ 4.01𝑥3+ 1.37𝑥4+36.75𝑥5+0𝑆1+𝑆2+0𝑆3+0𝑆4+0𝑆5=1 World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 658 4.07𝑥1+3.92𝑥2+ 4.32𝑥3+ 2.16𝑥4+36.11𝑥5+0𝑆1+0𝑆2+𝑆3+0𝑆4 +0𝑆5=1 (36) 6.03𝑥1+1.97𝑥2+ 2.62𝑥3+ 6.08𝑥4+34.05𝑥5+0𝑆1+0𝑆2+0𝑆3+𝑆4+0𝑆5=1 1.62𝑥1+3.40𝑥2+ 5.19𝑥3+ 3.66𝑥4+36.72𝑥5+0𝑆1+0𝑆2+0𝑆3+0𝑆4+𝑆5=1 𝑥1,𝑥2,𝑥3,𝑥4,𝑥5,𝑆1,𝑆2,𝑆3,𝑆4,𝑆5≥0 The equations formulated are used to solve the Simplex method by forming the initial Simplex table. Table 7 Initial Simplex Table .......Variables x1 x2 x3 x4 x5 S1 S2 S3 S4 S5 Amount Basis Cj 1 1 1 1 1 0 0 0 0 0 Trade ratio S1 0 5.40 4.78 0.09 1.82 38.48 1 0 0 0 0 1 1 38.48=0.026 S2 0 1.51 6.94 4.01 1.37 36.75 0 1 0 0 0 1 1 36.75=0.027 S3 0 4.07 3.92 4.32 2.16 36.11 0 0 1 0 0 1 1 36.11=0.028 S4 0 6.03 1.79 2.62 6.08 34.05 0 0 0 1 0 1 1 34.05=0.029 S5 0 1.62 3.40 5.19 3.66 36.72 0 0 0 0 1 1 1 36.72=0.027 Zj 0 0 0 0 0 0 0 0 0 0 0 Cj – Zj 1 1 1 1 1 0 0 0 0 0 ....... Bring in World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 659 Table 8 Second (2nd) Simplex table Variables x1 x2 x3 x4 x5 S1 S2 S3 S4 S5 Basis Cj 1 1 1 1 1 0 0 0 0 0 Amount Trade ratio x5 1 0.14 0.12 0.002 0.05 1 0.026 0 0 0 0 0.026 0.026 0.002=13 S2 0 -3.64 6.94 4.01 1.37 0 -0.96 1 0 0 0 0.045 0.045 3.94=0.011 S3 0 -0.99 3.92 4.32 2.16 0 -0.94 0 1 0 0 0.061 0.061 4.25=0.014 S4 0 1.26 1.79 2.62 6.08 0 -0.89 0 0 1 0 0.115 0.115 2.55=0.045 S5 0 -3.52 3.40 5.19 3.66 0 -0.95 0 0 0 1 0.045 0.045 4.24 =0.011 𝒕𝒂𝒌𝒆 𝒐𝒖𝒕 Zj 0.14 0.12 0.002 0.05 1 0.026 0 0 0 0 0.026 Cj – Zj 0.36 0.88 0.998 0.95 0 -0.026 0 0 0 0 Bring in r Computing of values for S2 Row x1 = 1.51 – 0.14  36.75 = - 3.64 x2 = 6.94 – 0.12  36.75 = 2.53 x3 = 4.01 – 0.002  36.75 = 3.94 x4 = 1.37 – 0.05  36.75 = 0.47 x5 = 6.94 – 1  36.75 = 0 S1 = 0 – 0.026  36.75 = - 0.96 S2 = 1 – 0  36.75 = 1 S3 = 0 – 0  36.75 = 0 S4 = 0 – 0  36.75 = 0 S5 = 0 – 0  36.75 = 0 Amount = S2 = 1 – 0.026  36.75 = 0.045 Computing of values for S3 Row x1 = 4.07 – 0.14  36.11 = - 3.64 x2 = 3.92 – 0.12  36.11 = 2.53 x3 = 4.32 – 0.002  36.11 = 3.94 x4 = 2.16 – 0.05  36.11 = 0.47 x5 = 36.11 – 1  36.11 = 0 S1 = 0 – 0.026  36.11 = - 0.94 S2 = 0 – 0  36.11 = 0 S3 = 1 – 0  36.11 = 1 S4 = 0 – 0  36.11 = 0 S5 = 0 – 0  36.11 = 0 Amount = S3 = 1 – 0.026  36.11 = 0.061 Computing of values for S4 Row x1 = 6.03 – 0.14  34.05 = 1.26 x2 = 1.97 – 0.12  34.05 = -2.12 x3 = 2.62 – 0.002  34.05 = 2.55 x4 = 6.08 – 0.05  34.05 = 4.38 x5 = 34.05 – 1  34.05 = 0 S1 = 0 – 0.026  34.05 = -0.89 S2 = 0 – 0  34.05 = 0 S3 = 0 – 0  34.05 = 0 S4 = 1 – 0  34.05 = 1 S5 = 0 – 0  34.05 = 0 Amount = S4 = 1 – 0.026  34.05 = 0.115 Computing of values for S5 Row x1 = 1.62 – 0.14  36.72 = -3.52 x2 = 3.40 – 0.12  36.72 = -1.01 x3 = 5.19 – 0.002  36.72 = 4.24 x4 = 3.66 – 0.05  36.72 = 1.82 x5 = 36.72 – 1  36.72 = 0 S1 = 0 – 0.026  36.72 = -0.95 S2 = 0 – 0  36.72 = 0 S3 = 0 – 0  36.72 = 0 S4 = 0 – 0  36.72 = 0 S5 = 1 – 0  36.72 = 1 Amount = S5 = 1 – 0.026  36.72 = 0.045 World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 660 Table 9 Third (3rd) Simplex table Variables x1 x2 x3 x4 x5 S1 S2 S3 S4 S5 Basis Cj 1 1 1 1 1 0 0 0 0 0 Amount Trade ratio x3 1 -0.83 -0.23 1 0.43 0 -0.22 0 0 0 0.24 0.011 0.011 −0.83=0.013 x5 1 0.14 0.12 0 0.05 1 -0.026 0 0 0 -0.005 0.026 0.026 0.14=0.186 S2 0 -0.37 3.44 0 -1.22 0 -0.09 1 0 0 -0.95 0.002 0.002 0.37=0.0054 S3 0 2.54 0.57 0 -1.48 0 -0.005 0 1 0 -1.02 0.014 0.014 2.54 =0.0055 S4 0 3.38 -1.53 0 3.28 0 -0.33 0 0 1 -0.61 0.087 0.087 3.38=0.0257 Zj -0.69 -0.11 1 0.48 1 -0.19 0 0 0 0.24 0.037 Cj – Zj 1.69 1.11 0 0.52 0 0.19 0 0 0 -0.24 Bring in Computing of values for X5 Row x1 = 0.14 – (-0.83  0.002) = 1.14 x2 = 0.12 – (-0.23  0.002) = 0.12 x3 = 0.002 – 1  0.002 = 0 x4 = 0.05 – 0.43  0.002 = 0.05 x5 = 1 – 0  0.002 = 1 S1 = – 0.026 – (-0.22  0.002) = - 0.026 S2 = 0 – 0  0.002 = 0 S3 = 0 – 0  0.002 = 0 S4 = 0 – 0  0.002 = 0 S5 = 0 – 0.24  0.002 = -0.0005 Amount = 0.026 – 0.011 0.002 = 0.026 Computing of values for S2 Row x1 = -3.64 – (-0.83  3.94) = -0.37 x2 = 0.12 – (-0.23  3.94) = 3.44 x3 = 3.94 – 1  3.94 = 0 x4 = 0.47 – 0.43  3.94 = -1.22 x5 = 0 – 0  3.94 = 0 S1 = – 0.96 – (-0.22  3.94) = - 0.09 S2 = 1 – 0  3.94 = 1 S3 = 0 – 0  3.94 = 0 S4 = 0 – 0  3.94 = 0 S5 = 0 – 0.24  3.94 = -0.95 Amount = 0.045 – 0.011 3.94 = 0.002 Computing of values for S3 Row x1 = -0.99 – (-0.83  4.25) = 2.54 x2 = 0.41 – (-0.23  4.25) = 0.57 x3 = 4.25 – 1  4.25 = 0 x4 = 0.35 – 0.43  4.25 = -1.48 x5 = 0 – 0  4.25 = 0 S1 = – 0.94 – (-0.22  4.25) = - 0.05 S2 = 0 – 0  4.25 = 0 S3 = 1 – 0  4.25 = 1 S4 = 0 – 0  4.25 = 0 S5 = 0 – 0.24  4.25 = -1.02 Amount = 0.061 – 0.011 4.25 = 0.014 Computing of values for S4 Row x1 = 1.26 – (-0.83  2.55) = 3.38 x2 = -2.12 – (-0.23  2.55) = -1.53 x3 = 2.55 – 1  2.55 = 0 x4 = 4.38 – 0.43  2.55 = 3.28 x5 = 0 – 0  2.55 = 0 S1 = – 0.89 – (-0.22  2.55) = - 0.33 S2 = 0 – 0  2.55 = 0 S3 = 0 – 0  2.55 = 0 S4 = 1 – 0  2.55 = 1 S5 = 0 – 0.24  2.55 = -0.61 Amount = 0.061 – 0.011 2.55 = 0.087 World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 661 Table 10 Fourth (4th) Simplex table Variables x1 x2 x3 x4 x5 S1 S2 S3 S4 S5 Basis Cj 1 1 1 1 1 0 0 0 0 0 Amount Trade ratio x1 1 1 0.22 0 -0.58 0 -0.02 0 0.39 0 -0.40 0.0055 0.0055 −0.58=0.009 x3 1 0 -0.05 1 0.05 0 -0.22 0 0.32 0 -0.09 0.016 0.016 0.05=0.32 x5 1 0 0.09 0 0.13 1 0.026 1 -0.05 0 0.06 0.025 0.025 0.13=0.192 S3 0 0 3.52 0 -1.43 0 -0.09 0 0.14 0 -1.10 0.004 0.004 −1.43=0.0028 S4 0 0 -2.27 0 5.24 0 -0.32 0 -1.32 1 0.74 0.068 0.068 5.24 =0.013 take out Zj 1 0.26 1 -0.4 1 -0.20 0 0.66 0 -0.53 0.047 Cj – Zj 0 0.74 0 1.4 0 0.20 0 -0.66 0 0.53 Bring in Computing of values for X3 Row x1 = -0.83 – (1  – 0.83) = 0 x2 = -0.23 – 0.22  –0.83 = -0.047 x3 = 1–0  –0.83 = 1 x4 = –0.43– (–0.58 –0.83) = 0.051 x5 = 0 – 0  –0.83 = 0 S1 = – 0.22 – (-0.002  –0.83) = - 0.22 S2 = 1 – 0  –0.83 = 0 S3 = 0 – 0.39  –0.83 = 0.32 S4 = 0 – 0  –0.83 = 0 S5 = 0 – 0.24 –(–0.40 –0.83) = -0.09 Amount = 0.011 – 0.0055 –0.83 = 0.016. Computing of values for S2 Row x1 = -0.37 – (1  – 0.37) = 0 x2 = 3.44 – 0.22  –0.37 = 3.52 x3 = 0–0  –0.37 = 0 x4 = –1.22 – (–0.58 –0.37) = – 1.43 x5 = 0 – 0  –0.37 = 0 S1 = – 0.09 – (– 0.002  –0.37) = – 0.091 S2 = 1 – 0  –0.37 = 1 S3 = 0 – 0.39  –0.37 = 0.14 S4 = 0 – 0  –0.37 = 0 S5 = 0 – 0.95 –(–0.40 –0.37) = -1.10 Amount = 0.002 – 0.0055 –0.37 = 0.004. Computing of values for X5 Row x1 = 0.14 – 1  – 0.14 = 0 x2 = 0.12 – 0.22  0.14 = 0.09 x3 = 0 – 0  0.14 = 0 x4 = 0.05 – (–0.58 0.14) = 0.13 x5 = 1 – 0  0.14 = 1 S1 = 0.026 – (– 0.002  0.14) = – 0.026 S2 = 0 – 0  0.14 = 0 S3 = 0 – 0.39  0.14 = 0.05 S4 = 0 – 0  0.14 = 0 S5 = – 0.0005 –(–0.40  0.14) = 0.06 Amount = 0.026 – 0.0055 0.14 = 0.025. Computing of values for S4 Row x1 = 3.38 – 1  – 3.38 = 0 x2 = –1.53 – 0.22  3.38 = –2.27 x3 = 0 – 0  3.38 = 0 x4 = 3.28 – (–0.58 3.38) = 5.24 x5 = 0 – 0  3.38 = 0 S1 = – 0.33 – (– 0.002  3.38) = – 0.32 S2 = 0 – 0  3.38 = 0 S3 = 0 – 0.39  3.38 = –1.32 S4 = 1 – 0  3.38 = 1 S5 = – 0.61 –(–0.40  3.38) = 0.74 Amount = 0.087 – 0.0055 3.38 = 0.068. World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 662 Table 11 Fifth (5th) Simplex table Variables x1 x2 x3 x4 x5 S1 S2 S3 S4 S5 Basis Cj 1 1 1 1 1 0 0 0 0 0 Amount Trade ratio x4 1 1 -0.43 0 1 0 -0.06 0 -0.25 0.19 0.14 0.013 0.013 −0.43=−0.009 x1 1 0 -0.03 0 0 0 0.03 0 0.25 0.11 -0.32 0.013 0.013 −0.03=−0.455 x3 1 0 -0.03 1 0 0 -0.22 0 0.33 -0.01 -0.1 0.015 0.015 −0.03=−0.5 x5 1 0 0.15 0 0 1 0.03 0 0.02 -0.03 0.04 0.023 0.023 0.15=0.153 S2 0 0 2.91 0 0 0 -0.18 1 -0.22 0.27 -0.90 0.023 0.023 2.91 =0.0079 take out Zj 1 -0.34 1 1 1 -0.22 0 0.35 0.26 -0.24 0.064 Cj – Zj 0 1.34 0 0 0 0.22 0 -0.35 -0.26 0.24 Bring in Computing of values for X1 Row x1 = 1 – 0  – 0.58 = 1 x2 = -0.22 – (–0.43  –0.58) = 0.03 x3 = 0 – 0  –0.58 = 0 x4 = 0.58 – 1  –0.58 = 0 x5 = 0 – 0  –0.58 = 0 S1 = – 0.002 – 0.06  –0.58 = 0.03 S2 = 0 – 0  –0.58 = 0 S3 = 0.39 – (– 0.25  –0.58) = 0.25 S4 = 0 – 0.19  –0.58 = 0.11 S5 = – 0.40 – 0.14 –0.58 = -0.32 Amount = 0.0055 – 0.013  –0.58 = 0.013. Computing of values for X5 Row x1 = 0 – 0  0.13 = 0 x2 = 0.09 – (–0.43  0.13) = 0.15 x3 = 0 – 0  0.13 = 0 x4 = 0.13 – 1  0.13 = 0 x5 = 1 – 0  0.13 = 1 S1 = 0.026 –(– 0.06  0.13) = 0.03 S2 = 0 – 0  0.13 = 0 S3 = –0.05 – (– 0.25  0.13) = 0.02 S4 = 0 – 0.19  0.13 = – 0.025 S5 = – 0.06 – 0.14  0.13 = 0.042 Amount = 0.025 – 0.013  0.13 = 0.023. Computing of values for X3 Row x1 = 0 – 0  0.05 = 0 x2 = –0.05 – (–0.43  0.05) = –0.03 x3 = 1 – 0  0.05 = 1 x4 = 0.05 – 1  0.05 = 0 x5 = 0 – 0  0.05 = 0 S1 = –0.22 –(– 0.06  0.05) = –0.22 S2 = 0 – 0  0.05 = 0 S3 = 0.32 – (– 0.25  0.05) = 0.33 S4 = 0 – 0.19  0.05 = – 0.010 S5 = – 0.00 – 0.14  0.05 = –0.10 Amount = 0.016 – 0.013  0.05 = 0.015. Computing of values for S2 Row x1 = 0 – 0  – 1.43 = 0 x2 = 3.52 – (–0.43 – 1.43) = 2.91 x3 = 0 – 0  – 1.43 = 0 x4 = 1.43 – 1  – 1.43 = 0 x5 = 0 – 0  – 1.43 = 0 S1 = – 0.09 –(– 0.06  – 1.43) = 0.18 S2 = 1 – 0  – 1.43 = 1 S3 = 0.14 – (– 0.25  – 1.43) = – 0.22 S4 = 0 – 0.19  – 1.43 = – 0.25 S5 = – 1.10 – 0.14  –1.43 = 0.042 Amount = 0.004 – 0.013  – 1.43 = 0.023. World Journal of Advanced Research and Reviews, 2025, 26(03), 639-667 663 Table 12 Sixth (6th) Simplex Table (Optimal Solution) Variables x1 x2 x3 x4 x5 S1 S2 S3 S4 S5 Basis Cj 1 1 1 1 1 0 0 0 0 0 Amount x2 1 0 1 0 0 0 -0.06 0.34 -0.08 0.09 -0.31 0.008 x4 1 1 0 0 1 0 -0.03 0.15 0.22 0.23 0.07 0.0164 x1 1 0 0 0 0 0 0.03 0.01 0.25 0.11 -0.33 0.0132 x3 1 0 0 1 0 0 -0.22 0.01 0.33 0.01 0.33 0.0152 x5 1 0 0 0 0 1 0.04 -0.05 0.03 -0.04 -0.09 0.0218 Zj 1 1 1 1 1 -0.24 0.46 0.75 0.4 -0.92 0.0746 Cj – Zj 0 0 0 0 0 0.24 -0.46 -0.75 -0.4 0.92 Computing of values for X4 Row x1 = 1 – 0  – 0.43 = 1 x2 = -0.43 – 1  –0.43 = 0 x3 = 0 – 0  –0.43 = 0 x4 = 1 – 0  –0.43 = 1 x5 = 0 – 0  –0.43 = 0 S1 = – 0.006 – 0.06  –0.43 = 0.03 S2 = 0 – 0.34  –0.43 = 0.15 S3 = – 0.25 – (– 0.25  –0.43) = 0.22 S4 = 0.19 – 0.09  –0.43 = 0.23 S5 = – 0.14 –(–0.31 –0.43 = 0.007 Amount= 0.013 – 0.008  –0.43 = 0.0164. Computing of values for X1 Row x1 = 0 – 0  – 0.03 = 0 x2 = -0.03 – 1  –0.03 = 0 x3 = 0 – 0  –0.03 = 0 x4 = 0 – 0  –0.03 = 0 x5 = 0 – 0  –0.03 = 0 S1 = – 0.03 – (–0.06  –0.03) = 0.028 S2 = 0 – 0.34  –0.03 = 0.01 S3 = 0.25 – (– 0.08  –0.03) = 0.25 S4 = 0.11 – 0.09  –0.03 = 0.11 S5 = – 0.32 – (–0.31 –0.03) = -0.33 Amount= 0.013 – 0.003  –0.03 = 0.0132. Computing of values for X3 Row x1 = 0 – 0  – 0.03 = 0 x2 = –0.03  1  – 0.03 = 0 x3 = 1 – 0  –0.03 = 1 x4 = 0 – 0  –0.03 = 0 x5 = 0 – 0  –0.03 = 0 S1 = –0.22 –(– 0.06  0.13) = –0.22 S2 = 0 – 0.34  –0.03 = 0.01 S3 = –0.33 – (– 0.08  –0.03) = 0.33 S4 = –0.01 – 0.09  –0.03 = – 0.007 S5 = – 0.1 –(– 0.31  –0.03) = –0.38 Amount = 0.015–0.008  –0.03 = 0.0152. Computing of values for X5 Row x1 = 0 – 0  – 0.15 = 0 x2 = 0.15 – 1  0.15 = 0 x3 = 0 – 0  0.15 = 0 x4 = 0 – 0  0.15 = 0 x5 = 1 – 0  0.15 = 1 S1 = – 0.03 – (–0.06  0.15) = 0.04 S2 = 0 – 0.34  0.15 = – 0.05 S3 = 0.02 – (– 0.08  0.15) = 0.03 S4 = –0.03 – 0.09  0.15 = 0.04 S5 = 0.04 –(–0.31  0.15) = 0.09 Amount= 0.023 – 0.008  0.15 = 0.0218. The optimal solution from the game model simplex method of linear programming in Table 12 shows that 𝑥1 = 0.0132, 𝑥2 = 0.008, 𝑥3 = 0.0152, 𝑥4 = 0.0164 and 𝑥5 = 0.0218. The expected value of the game obtained from the relation Zq = 1𝑉= 0.0746. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒,𝑉= 1 0.0746.=13.404826 ≈13.405