Operations Research  Linear Programming
Economics Application  Cost Minimization
by
Elmer G. Wiens
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Graphical Solution 
Simplex Algorithm 
Primal Simplex Tableaux Generator 
Dual Simplex Algorithm 
Dual Simplex Tableaux Generator
Economics Application  Profit Maximization 
Economics Application  Least Cost Formula
firm's least cost decision  sensitivity analysis  dual simplex method  dual simplex tableaux least cost solution  postoptimality analysis  dual solution interpretation  references
The Egwald's Coffee Import Decision.
Egwald's Coffee, a purveyor of speciality coffee, imports Yemen Mocha and Ethopian Harrar coffee beans from the Middle East, and Java and Sumatra coffee beans from Indonesia. Egwald has a standing order with coffee merchants for at least one hundred bags (60 kilos per bag) of "green coffee beans" every three months for each singleorigin coffee, plus additional coffee beans as required, all at the prevailing prices.
After a secret process of aging, roasting, and blending, Egwald markets this coffee, generically known as MochaJava, under the trade name of MoJo Coffee Egwald. Egwald blends the roasted beans according to four formulae. Formula 1 combines 1 part Mocha with 2 parts Java; Formula 2 combines 1 part Mocha with 2 parts Sumatra; Formula 3 combines 1 part Harrar with 2 parts Java; Formula 4 combines 1 part Harrar with 2 parts Sumatra. At the beginning of each quarter of the year, Egwald estimates the demand for MoJo Coffee Egwald, and orders quantities of green coffee beans to replenish inventories based on the prevailing coffee bean prices, taking into consideration the differentials in direct costs associated with processing beans according to each formula.
Egwald's decision involves minimizing the cost of producing MoJo  a least cost formulation problem  subject to expected quarterly demand from coffee lovers for MoJo, the price of green coffee beans, and processing costs for each blend of MoJo coffee.
Define the variables of the problem according to the following table:
Formula  Mocha  Harrar  Java  Sumatra  MoJo 
Blend 1  x_{1}   x_{5}   x_{9} 
Blend 2  x_{2}    x_{7}  x_{10} 
Blend 3   x_{3}  x_{6}   x_{11} 
Blend 4   x_{4}   x_{8}  x_{12} 
The first line in the table reads that the formula for blend 1 of MoJo combines Mocha and Java roasted coffee beans; the second line reads that the formula for blend 2 combines Mocha and Sumatra roasted coffee beans; etc.
The second table illustrates Egwald Coffee's quarterly standing default order from each singleorigin coffee supplier:
Dual Simplex Solution.
The primal objective of the Egwald's Coffee is to minimize costs, by ordering appropriate amounts of coffee beans. The firm's primal production decision is captured in the firm's Primal Linear Programming Problem, where I multiplied the coefficients of the objective function, in $ per sack, by 1 to turn the minimization problem into a maximization problem.
Primal Linear Program
Maximize the Objective Function (P)

P =  165.75 * x_{1} + 165.75 * x_{2} + 152.49 * x_{3} + 152.49 * x_{4} + 92.82 * x_{5} + 92.82 * x_{6} + 99.45 * x_{7} + 99.45 * x_{8} + 14.59 * x_{9} + 10.61 * x_{10} + 11.93 * x_{11} + 9.28 * x_{12} 

subject to the constraints
M:  1 * x_{1} +  1 * x_{2} +  0 * x_{3} +  0 * x_{4} +  0 * x_{5} +  0 * x_{6} +  0 * x_{7} +  0 * x_{8} +  0 * x_{9} +  0 * x_{10} +  0 * x_{11} +  0 * x_{12}  >=  100 
H:  0 * x_{1} +  0 * x_{2} +  1 * x_{3} +  1 * x_{4} +  0 * x_{5} +  0 * x_{6} +  0 * x_{7} +  0 * x_{8} +  0 * x_{9} +  0 * x_{10} +  0 * x_{11} +  0 * x_{12}  >=  100 
J:  0 * x_{1} +  0 * x_{2} +  0 * x_{3} +  0 * x_{4} +  1 * x_{5} +  1 * x_{6} +  0 * x_{7} +  0 * x_{8} +  0 * x_{9} +  0 * x_{10} +  0 * x_{11} +  0 * x_{12}  >=  100 
S:  0 * x_{1} +  0 * x_{2} +  0 * x_{3} +  0 * x_{4} +  0 * x_{5} +  0 * x_{6} +  1 * x_{7} +  1 * x_{8} +  0 * x_{9} +  0 * x_{10} +  0 * x_{11} +  0 * x_{12}  >=  100 
B_{1}:  1 * x_{1} +  0 * x_{2} +  0 * x_{3} +  0 * x_{4} +  1 * x_{5} +  0 * x_{6} +  0 * x_{7} +  0 * x_{8} +  1 * x_{9} +  0 * x_{10} +  0 * x_{11} +  0 * x_{12}  >=  0 
B_{2}:  0 * x_{1} +  1 * x_{2} +  0 * x_{3} +  0 * x_{4} +  0 * x_{5} +  0 * x_{6} +  1 * x_{7} +  0 * x_{8} +  0 * x_{9} +  1 * x_{10} +  0 * x_{11} +  0 * x_{12}  >=  0 
B_{3}:  0 * x_{1} +  0 * x_{2} +  1 * x_{3} +  0 * x_{4} +  0 * x_{5} +  1 * x_{6} +  0 * x_{7} +  0 * x_{8} +  0 * x_{9} +  0 * x_{10} +  1 * x_{11} +  0 * x_{12}  >=  0 
B_{4}:  0 * x_{1} +  0 * x_{2} +  0 * x_{3} +  1 * x_{4} +  0 * x_{5} +  0 * x_{6} +  0 * x_{7} +  1 * x_{8} +  0 * x_{9} +  0 * x_{10} +  0 * x_{11} +  1 * x_{12}  >=  0 
MJ:  0 * x_{1} +  0 * x_{2} +  0 * x_{3} +  0 * x_{4} +  0 * x_{5} +  0 * x_{6} +  0 * x_{7} +  0 * x_{8} +  1 * x_{9} +  1 * x_{10} +  1 * x_{11} +  1 * x_{12}  >=  1000 
F_{1}:  2 * x_{1} +  0 * x_{2} +  0 * x_{3} +  0 * x_{4} +  1 * x_{5} +  0 * x_{6} +  0 * x_{7} +  0 * x_{8} +  0 * x_{9} +  0 * x_{10} +  0 * x_{11} +  0 * x_{12}  =  0 
F_{2}:  0 * x_{1} +  2 * x_{2} +  0 * x_{3} +  0 * x_{4} +  0 * x_{5} +  0 * x_{6} +  1 * x_{7} +  0 * x_{8} +  0 * x_{9} +  0 * x_{10} +  0 * x_{11} +  0 * x_{12}  =  0 
F_{3}:  0 * x_{1} +  0 * x_{2} +  2 * x_{3} +  0 * x_{4} +  0 * x_{5} +  1 * x_{6} +  0 * x_{7} +  0 * x_{8} +  0 * x_{9} +  0 * x_{10} +  0 * x_{11} +  0 * x_{12}  =  0 
F_{4}:  0 * x_{1} +  0 * x_{2} +  0 * x_{3} +  2 * x_{4} +  0 * x_{5} +  0 * x_{6} +  0 * x_{7} +  1 * x_{8} +  0 * x_{9} +  0 * x_{10} +  0 * x_{11} +  0 * x_{12}  =  0 

x_{1} >= 0; x_{2} >= 0; x_{3} >= 0; x_{4} >= 0; x_{5} >= 0; x_{6} >= 0; x_{7} >= 0; x_{8} >= 0; x_{9} >= 0; x_{10} >= 0; x_{11} >= 0; x_{12} >= 0; 

Using matrix notation this is: 
Primal Linear Program
Maximize the Objective Function (P)
P = c^{T} x subject to
A x >= b, x >= 0

The standard form for the initial dual simplex tableau is:
Initial Tableau 
b  A  I_{3} 
0  c^{T}  O_{3}^{T} 
In the dual simplex algorithm, a sequence of tableaux are calculated. A given Tableau^{k} has the form:
Tableau^{k} 
ß  U  B 
µ  t^{T}  y^{T} 

= 

after consolidating the center 2 by 2 block of matrices.
Phase 0  drive all artificial variables (associated with = constraints) to zero, i.e. eliminate them from the basis;
Phase I  find a tableau with Ø >= 0, i.e. a feasible dual program;
Phase II  generate tableaux that decrease the value of µ turning ß >= 0, without dropping back into Phase 0 or I, i.e. find a feasible basic dual program that minimizes the objective function D.

The Tableaux of the Dual Simplex Method.
Phase 0: Drive the artificial variables from the basis.
Tableau^{0} 

b^{0} 
x^{0}_{1} 
x^{0}_{2} 
x^{0}_{3} 
x^{0}_{4} 
x^{0}_{5} 
x^{0}_{6} 
x^{0}_{7} 
x^{0}_{8} 
x^{0}_{9} 
x^{0}_{10} 
x^{0}_{11} 
x^{0}_{12} 
s^{0}_{1} 
s^{0}_{2} 
s^{0}_{3} 
s^{0}_{4} 
s^{0}_{5} 
s^{0}_{6} 
s^{0}_{7} 
s^{0}_{8} 
s^{0}_{9} 
s*^{0}_{10} 
s*^{0}_{11} 
s*^{0}_{12} 
s*^{0}_{13} 
row sum 
M^{0}  100  1  1  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  101 
H^{0}  100  0  0  1  1  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  101 
J^{0}  100  0  0  0  0  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  101 
S^{0}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{0}  0  1  0  0  0  1  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0 
B_{2}^{0}  0  0  1  0  0  0  0  1  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0 
B_{3}^{0}  0  0  0  1  0  0  1  0  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0 
B_{4}^{0}  0  0  0  0  1  0  0  0  1  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0 
MJ^{0}  1000  0  0  0  0  0  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  1003 
F_{1}^{0}  0  2  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0 
F_{2}^{0}  0  0  2  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0 
F_{3}^{0}  0  0  0  2  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0 
F_{4}^{0}  0  0  0  0  2  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0 
P^{0}  0  165.8  165.8  152.5  152.5  92.8  92.8  99.5  99.5  14.6  10.6  11.9  9.3  0  0  0  0  0  0  0  0  0  0  0  0  0  1067.4 
 0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{0}: [s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}, s_{7}, s_{8}, s_{9}, s_{10}, s_{11}, s_{12}, s_{13}, ]. Value of Objective Function = 0.
Proceed to the next tableau as follows:
Phase 0: Drive the artificial variables from the basis.
A. In Tableau^{0}:
1. Select an artificial variable in the basis: s^{*}_{13}.
2. Select a nonzero element in row F_{4} as pivot: x_{13,4} = 2. The pivot column = 4.
B. To create Tableau^{1}:
3. Compute row F_{4}^{1} = F_{4}^{0} / (2).
4. Subtract multiples of row F_{4}^{1} from all other rows of Tableau^{0} so that column x^{1}_{4} = e_{13} in Tableau^{1}.
Tableau^{1} 

b^{1} 
x^{1}_{1} 
x^{1}_{2} 
x^{1}_{3} 
x^{1}_{4} 
x^{1}_{5} 
x^{1}_{6} 
x^{1}_{7} 
x^{1}_{8} 
x^{1}_{9} 
x^{1}_{10} 
x^{1}_{11} 
x^{1}_{12} 
s^{1}_{1} 
s^{1}_{2} 
s^{1}_{3} 
s^{1}_{4} 
s^{1}_{5} 
s^{1}_{6} 
s^{1}_{7} 
s^{1}_{8} 
s^{1}_{9} 
s*^{1}_{10} 
s*^{1}_{11} 
s*^{1}_{12} 
s*^{1}_{13} 
row sum 
M^{1}  100  1  1  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  101 
H^{1}  100  0  0  1  0  0  0  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0.5  101 
J^{1}  100  0  0  0  0  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  101 
S^{1}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{1}  0  1  0  0  0  1  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0 
B_{2}^{1}  0  0  1  0  0  0  0  1  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0 
B_{3}^{1}  0  0  0  1  0  0  1  0  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0 
B_{4}^{1}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
MJ^{1}  1000  0  0  0  0  0  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  1003 
F_{1}^{1}  0  2  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0 
F_{2}^{1}  0  0  2  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0 
F_{3}^{1}  0  0  0  2  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0 
F_{4}^{1}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{1}  0  165.8  165.8  152.5  0  92.8  92.8  99.5  175.7  14.6  10.6  11.9  9.3  0  0  0  0  0  0  0  0  0  0  0  0  76.2  1067.4 
P^{1} / F_{3}^{1}  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{1}: [s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}, s_{7}, s_{8}, s_{9}, s_{10}, s_{11}, s_{12}, x_{4}, ]. Value of Objective Function = 0.
Proceed to the next tableau as follows:
Phase 0: Drive the artificial variables from the basis.
A. In Tableau^{1}:
1. Select an artificial variable in the basis: s^{*}_{12}.
2. Select a nonzero element in row F_{3} as pivot: x_{12,3} = 2. The pivot column = 3.
B. To create Tableau^{2}:
3. Compute row F_{3}^{2} = F_{3}^{1} / (2).
4. Subtract multiples of row F_{3}^{2} from all other rows of Tableau^{1} so that column x^{2}_{3} = e_{12} in Tableau^{2}.
Tableau^{2} 

b^{2} 
x^{2}_{1} 
x^{2}_{2} 
x^{2}_{3} 
x^{2}_{4} 
x^{2}_{5} 
x^{2}_{6} 
x^{2}_{7} 
x^{2}_{8} 
x^{2}_{9} 
x^{2}_{10} 
x^{2}_{11} 
x^{2}_{12} 
s^{2}_{1} 
s^{2}_{2} 
s^{2}_{3} 
s^{2}_{4} 
s^{2}_{5} 
s^{2}_{6} 
s^{2}_{7} 
s^{2}_{8} 
s^{2}_{9} 
s*^{2}_{10} 
s*^{2}_{11} 
s*^{2}_{12} 
s*^{2}_{13} 
row sum 
M^{2}  100  1  1  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  101 
H^{2}  100  0  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0.5  0.5  101 
J^{2}  100  0  0  0  0  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  101 
S^{2}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{2}  0  1  0  0  0  1  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0 
B_{2}^{2}  0  0  1  0  0  0  0  1  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0 
B_{3}^{2}  0  0  0  0  0  0  1.5  0  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0 
B_{4}^{2}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
MJ^{2}  1000  0  0  0  0  0  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  1003 
F_{1}^{2}  0  2  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0 
F_{2}^{2}  0  0  2  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0 
F_{3}^{2}  0  0  0  1  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 
F_{4}^{2}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{2}  0  165.8  165.8  0  0  92.8  169.1  99.5  175.7  14.6  10.6  11.9  9.3  0  0  0  0  0  0  0  0  0  0  0  76.2  76.2  1067.4 
P^{2} / F_{2}^{2}  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{2}: [s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}, s_{7}, s_{8}, s_{9}, s_{10}, s_{11}, x_{3}, x_{4}, ]. Value of Objective Function = 0.
Proceed to the next tableau as follows:
Phase 0: Drive the artificial variables from the basis.
A. In Tableau^{2}:
1. Select an artificial variable in the basis: s^{*}_{11}.
2. Select a nonzero element in row F_{2} as pivot: x_{11,2} = 2. The pivot column = 2.
B. To create Tableau^{3}:
3. Compute row F_{2}^{3} = F_{2}^{2} / (2).
4. Subtract multiples of row F_{2}^{3} from all other rows of Tableau^{2} so that column x^{3}_{2} = e_{11} in Tableau^{3}.
Tableau^{3} 

b^{3} 
x^{3}_{1} 
x^{3}_{2} 
x^{3}_{3} 
x^{3}_{4} 
x^{3}_{5} 
x^{3}_{6} 
x^{3}_{7} 
x^{3}_{8} 
x^{3}_{9} 
x^{3}_{10} 
x^{3}_{11} 
x^{3}_{12} 
s^{3}_{1} 
s^{3}_{2} 
s^{3}_{3} 
s^{3}_{4} 
s^{3}_{5} 
s^{3}_{6} 
s^{3}_{7} 
s^{3}_{8} 
s^{3}_{9} 
s*^{3}_{10} 
s*^{3}_{11} 
s*^{3}_{12} 
s*^{3}_{13} 
row sum 
M^{3}  100  1  0  0  0  0  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0.5  0  0  101 
H^{3}  100  0  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0.5  0.5  101 
J^{3}  100  0  0  0  0  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  101 
S^{3}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{3}  0  1  0  0  0  1  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0 
B_{2}^{3}  0  0  0  0  0  0  0  1.5  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0 
B_{3}^{3}  0  0  0  0  0  0  1.5  0  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0 
B_{4}^{3}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
MJ^{3}  1000  0  0  0  0  0  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  1003 
F_{1}^{3}  0  2  0  0  0  1  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  0 
F_{2}^{3}  0  0  1  0  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0 
F_{3}^{3}  0  0  0  1  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 
F_{4}^{3}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{3}  0  165.8  0  0  0  92.8  169.1  182.3  175.7  14.6  10.6  11.9  9.3  0  0  0  0  0  0  0  0  0  0  82.9  76.2  76.2  1067.4 
P^{3} / F_{1}^{3}  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{3}: [s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}, s_{7}, s_{8}, s_{9}, s_{10}, x_{2}, x_{3}, x_{4}, ]. Value of Objective Function = 0.
Proceed to the next tableau as follows:
Phase 0: Drive the artificial variables from the basis.
A. In Tableau^{3}:
1. Select an artificial variable in the basis: s^{*}_{10}.
2. Select a nonzero element in row F_{1} as pivot: x_{10,1} = 2. The pivot column = 1.
B. To create Tableau^{4}:
3. Compute row F_{1}^{4} = F_{1}^{3} / (2).
4. Subtract multiples of row F_{1}^{4} from all other rows of Tableau^{3} so that column x^{4}_{1} = e_{10} in Tableau^{4}.
Phase II: Goal: get ß >= 0.
Tableau^{4} 

b^{4} 
x^{4}_{1} 
x^{4}_{2} 
x^{4}_{3} 
x^{4}_{4} 
x^{4}_{5} 
x^{4}_{6} 
x^{4}_{7} 
x^{4}_{8} 
x^{4}_{9} 
x^{4}_{10} 
x^{4}_{11} 
x^{4}_{12} 
s^{4}_{1} 
s^{4}_{2} 
s^{4}_{3} 
s^{4}_{4} 
s^{4}_{5} 
s^{4}_{6} 
s^{4}_{7} 
s^{4}_{8} 
s^{4}_{9} 
s*^{4}_{10} 
s*^{4}_{11} 
s*^{4}_{12} 
s*^{4}_{13} 
row sum 
M^{4}  100  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0.5  0.5  0  0  101 
H^{4}  100  0  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0.5  0.5  101 
J^{4}  100  0  0  0  0  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  101 
S^{4}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{4}  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0  0 
B_{2}^{4}  0  0  0  0  0  0  0  1.5  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0 
B_{3}^{4}  0  0  0  0  0  0  1.5  0  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0 
B_{4}^{4}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
MJ^{4}  1000  0  0  0  0  0  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  1003 
F_{1}^{4}  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0  0 
F_{2}^{4}  0  0  1  0  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0 
F_{3}^{4}  0  0  0  1  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 
F_{4}^{4}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{4}  0  0  0  0  0  175.7  169.1  182.3  175.7  14.6  10.6  11.9  9.3  0  0  0  0  0  0  0  0  0  82.9  82.9  76.2  76.2  1067.4 
P^{4} / MJ^{4}  0  0  0  0  0  0  0  0  0  14.6  10.6  11.9  9.3  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{4}: [s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}, s_{7}, s_{8}, s_{9}, x_{1}, x_{2}, x_{3}, x_{4}, ]. Value of Objective Function = 0.
Proceed to the next tableau as follows:
Phase 0: Complete.
Phase I: Complete.
Phase II: Goal: get ß >= 0.
A. In Tableau^{4}:
1. Select a pivot row, row, with b^{4}_{row} < 0: row = 9 associated with b^{4}_{9} = 1000.
2. Compute the ratios Ø / MJ as per the last row. Discard ratios which are not positive and ratios associated with artificial variables.
Select the column with the least positive ratio as the pivot column: col = 12 associated with 9.3. Thus x_{9,12} = 1 is the pivot; variable s_{9} will leave the basis; variable x_{12} will enter the basis.
B. To create Tableau^{5}:
3. Compute row MJ^{5} = MJ^{4} / (1).
3. Subtract multiples of row MJ^{5} from all other rows of Tableau^{4} so that column x_{12} = e_{9} in Tableau^{5}.
Tableau^{5} 

b^{5} 
x^{5}_{1} 
x^{5}_{2} 
x^{5}_{3} 
x^{5}_{4} 
x^{5}_{5} 
x^{5}_{6} 
x^{5}_{7} 
x^{5}_{8} 
x^{5}_{9} 
x^{5}_{10} 
x^{5}_{11} 
x^{5}_{12} 
s^{5}_{1} 
s^{5}_{2} 
s^{5}_{3} 
s^{5}_{4} 
s^{5}_{5} 
s^{5}_{6} 
s^{5}_{7} 
s^{5}_{8} 
s^{5}_{9} 
s*^{5}_{10} 
s*^{5}_{11} 
s*^{5}_{12} 
s*^{5}_{13} 
row sum 
M^{5}  100  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0.5  0.5  0  0  101 
H^{5}  100  0  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0.5  0.5  101 
J^{5}  100  0  0  0  0  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  101 
S^{5}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{5}  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0  0 
B_{2}^{5}  0  0  0  0  0  0  0  1.5  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0 
B_{3}^{5}  0  0  0  0  0  0  1.5  0  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0 
B_{4}^{5}  1000  0  0  0  0  0  0  0  1.5  1  1  1  0  0  0  0  0  0  0  0  1  1  0  0  0  0.5  1003 
MJ^{5}  1000  0  0  0  0  0  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  1003 
F_{1}^{5}  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0  0 
F_{2}^{5}  0  0  1  0  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0 
F_{3}^{5}  0  0  0  1  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 
F_{4}^{5}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{5}  9282  0  0  0  0  175.7  169.1  182.3  175.7  5.3  1.3  2.7  0  0  0  0  0  0  0  0  0  9.3  82.9  82.9  76.2  76.2  8242.4 
P^{5} / B_{4}^{5}  0  0  0  0  0  0  0  0  117.1  5.3  1.3  2.7  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{5}: [s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}, s_{7}, s_{8}, x_{12}, x_{1}, x_{2}, x_{3}, x_{4}, ]. Value of Objective Function = 9282.
Proceed to the next tableau as follows:
Phase 0: Complete.
Phase I: Complete.
Phase II: Goal: get ß >= 0.
A. In Tableau^{5}:
1. Select a pivot row, row, with b^{5}_{row} < 0: row = 8 associated with b^{5}_{8} = 1000.
2. Compute the ratios Ø / B_{4} as per the last row. Discard ratios which are not positive and ratios associated with artificial variables.
Select the column with the least positive ratio as the pivot column: col = 10 associated with 1.3. Thus x_{8,10} = 1 is the pivot; variable s_{8} will leave the basis; variable x_{10} will enter the basis.
B. To create Tableau^{6}:
3. Compute row B_{4}^{6} = B_{4}^{5} / (1).
3. Subtract multiples of row B_{4}^{6} from all other rows of Tableau^{5} so that column x_{10} = e_{8} in Tableau^{6}.
Tableau^{6} 

b^{6} 
x^{6}_{1} 
x^{6}_{2} 
x^{6}_{3} 
x^{6}_{4} 
x^{6}_{5} 
x^{6}_{6} 
x^{6}_{7} 
x^{6}_{8} 
x^{6}_{9} 
x^{6}_{10} 
x^{6}_{11} 
x^{6}_{12} 
s^{6}_{1} 
s^{6}_{2} 
s^{6}_{3} 
s^{6}_{4} 
s^{6}_{5} 
s^{6}_{6} 
s^{6}_{7} 
s^{6}_{8} 
s^{6}_{9} 
s*^{6}_{10} 
s*^{6}_{11} 
s*^{6}_{12} 
s*^{6}_{13} 
row sum 
M^{6}  100  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0.5  0.5  0  0  101 
H^{6}  100  0  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0.5  0.5  101 
J^{6}  100  0  0  0  0  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  101 
S^{6}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{6}  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0  0 
B_{2}^{6}  1000  0  0  0  0  0  0  1.5  1.5  1  0  1  0  0  0  0  0  0  1  0  1  1  0  0.5  0  0.5  1003 
B_{3}^{6}  0  0  0  0  0  0  1.5  0  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0 
B_{4}^{6}  1000  0  0  0  0  0  0  0  1.5  1  1  1  0  0  0  0  0  0  0  0  1  1  0  0  0  0.5  1003 
MJ^{6}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
F_{1}^{6}  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0  0 
F_{2}^{6}  0  0  1  0  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0 
F_{3}^{6}  0  0  0  1  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 
F_{4}^{6}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{6}  10608  0  0  0  0  175.7  169.1  182.3  173.7  4  0  1.3  0  0  0  0  0  0  0  0  1.3  10.6  82.9  82.9  76.2  75.6  9572.4 
P^{6} / B_{2}^{6}  0  0  0  0  0  0  0  121.6  115.8  4  0  1.3  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{6}: [s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}, s_{7}, x_{10}, x_{12}, x_{1}, x_{2}, x_{3}, x_{4}, ]. Value of Objective Function = 10608.
Proceed to the next tableau as follows:
Phase 0: Complete.
Phase I: Complete.
Phase II: Goal: get ß >= 0.
A. In Tableau^{6}:
1. Select a pivot row, row, with b^{6}_{row} < 0: row = 6 associated with b^{6}_{6} = 1000.
2. Compute the ratios Ø / B_{2} as per the last row. Discard ratios which are not positive and ratios associated with artificial variables.
Select the column with the least positive ratio as the pivot column: col = 11 associated with 1.3. Thus x_{6,11} = 1 is the pivot; variable s_{6} will leave the basis; variable x_{11} will enter the basis.
B. To create Tableau^{7}:
3. Compute row B_{2}^{7} = B_{2}^{6} / (1).
3. Subtract multiples of row B_{2}^{7} from all other rows of Tableau^{6} so that column x_{11} = e_{6} in Tableau^{7}.
Tableau^{7} 

b^{7} 
x^{7}_{1} 
x^{7}_{2} 
x^{7}_{3} 
x^{7}_{4} 
x^{7}_{5} 
x^{7}_{6} 
x^{7}_{7} 
x^{7}_{8} 
x^{7}_{9} 
x^{7}_{10} 
x^{7}_{11} 
x^{7}_{12} 
s^{7}_{1} 
s^{7}_{2} 
s^{7}_{3} 
s^{7}_{4} 
s^{7}_{5} 
s^{7}_{6} 
s^{7}_{7} 
s^{7}_{8} 
s^{7}_{9} 
s*^{7}_{10} 
s*^{7}_{11} 
s*^{7}_{12} 
s*^{7}_{13} 
row sum 
M^{7}  100  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0.5  0.5  0  0  101 
H^{7}  100  0  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0.5  0.5  101 
J^{7}  100  0  0  0  0  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  101 
S^{7}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{7}  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0  0 
B_{2}^{7}  1000  0  0  0  0  0  0  1.5  1.5  1  0  1  0  0  0  0  0  0  1  0  1  1  0  0.5  0  0.5  1003 
B_{3}^{7}  1000  0  0  0  0  0  1.5  1.5  1.5  1  0  0  0  0  0  0  0  0  1  1  1  1  0  0.5  0.5  0.5  1003 
B_{4}^{7}  0  0  0  0  0  0  0  1.5  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0 
MJ^{7}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
F_{1}^{7}  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0  0 
F_{2}^{7}  0  0  1  0  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0 
F_{3}^{7}  0  0  0  1  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 
F_{4}^{7}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{7}  11934  0  0  0  0  175.7  169.1  180.3  171.7  2.7  0  0  0  0  0  0  0  0  1.3  0  2.7  11.9  82.9  82.2  76.2  74.9  10902.4 
P^{7} / B_{3}^{7}  0  0  0  0  0  0  112.7  120.2  114.5  2.7  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{7}: [s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, x_{11}, s_{7}, x_{10}, x_{12}, x_{1}, x_{2}, x_{3}, x_{4}, ]. Value of Objective Function = 11934.
Proceed to the next tableau as follows:
Phase 0: Complete.
Phase I: Complete.
Phase II: Goal: get ß >= 0.
A. In Tableau^{7}:
1. Select a pivot row, row, with b^{7}_{row} < 0: row = 7 associated with b^{7}_{7} = 1000.
2. Compute the ratios Ø / B_{3} as per the last row. Discard ratios which are not positive and ratios associated with artificial variables.
Select the column with the least positive ratio as the pivot column: col = 9 associated with 2.7. Thus x_{7,9} = 1 is the pivot; variable s_{7} will leave the basis; variable x_{9} will enter the basis.
B. To create Tableau^{8}:
3. Compute row B_{3}^{8} = B_{3}^{7} / (1).
3. Subtract multiples of row B_{3}^{8} from all other rows of Tableau^{7} so that column x_{9} = e_{7} in Tableau^{8}.
Tableau^{8} 

b^{8} 
x^{8}_{1} 
x^{8}_{2} 
x^{8}_{3} 
x^{8}_{4} 
x^{8}_{5} 
x^{8}_{6} 
x^{8}_{7} 
x^{8}_{8} 
x^{8}_{9} 
x^{8}_{10} 
x^{8}_{11} 
x^{8}_{12} 
s^{8}_{1} 
s^{8}_{2} 
s^{8}_{3} 
s^{8}_{4} 
s^{8}_{5} 
s^{8}_{6} 
s^{8}_{7} 
s^{8}_{8} 
s^{8}_{9} 
s*^{8}_{10} 
s*^{8}_{11} 
s*^{8}_{12} 
s*^{8}_{13} 
row sum 
M^{8}  100  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0.5  0.5  0  0  101 
H^{8}  100  0  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0.5  0.5  101 
J^{8}  100  0  0  0  0  1  1  0  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0  101 
S^{8}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{8}  1000  0  0  0  0  1.5  1.5  1.5  1.5  0  0  0  0  0  0  0  0  1  1  1  1  1  0.5  0.5  0.5  0.5  1003 
B_{2}^{8}  0  0  0  0  0  0  1.5  0  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0 
B_{3}^{8}  1000  0  0  0  0  0  1.5  1.5  1.5  1  0  0  0  0  0  0  0  0  1  1  1  1  0  0.5  0.5  0.5  1003 
B_{4}^{8}  0  0  0  0  0  0  0  1.5  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0 
MJ^{8}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
F_{1}^{8}  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0  0 
F_{2}^{8}  0  0  1  0  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0 
F_{3}^{8}  0  0  0  1  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 
F_{4}^{8}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{8}  14586  0  0  0  0  175.7  165.1  176.4  167.7  0  0  0  0  0  0  0  0  0  4  2.7  5.3  14.6  82.9  80.9  74.9  73.6  13562.3 
P^{8} / B_{1}^{8}  0  0  0  0  0  117.1  110.1  117.6  111.8  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{8}: [s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, x_{11}, x_{9}, x_{10}, x_{12}, x_{1}, x_{2}, x_{3}, x_{4}, ]. Value of Objective Function = 14586.
Proceed to the next tableau as follows:
Phase 0: Complete.
Phase I: Complete.
Phase II: Goal: get ß >= 0.
A. In Tableau^{8}:
1. Select a pivot row, row, with b^{8}_{row} < 0: row = 5 associated with b^{8}_{5} = 1000.
2. Compute the ratios Ø / B_{1} as per the last row. Discard ratios which are not positive and ratios associated with artificial variables.
Select the column with the least positive ratio as the pivot column: col = 6 associated with 110.1. Thus x_{5,6} = 1.5 is the pivot; variable s_{5} will leave the basis; variable x_{6} will enter the basis.
B. To create Tableau^{9}:
3. Compute row B_{1}^{9} = B_{1}^{8} / (1.5).
3. Subtract multiples of row B_{1}^{9} from all other rows of Tableau^{8} so that column x_{6} = e_{5} in Tableau^{9}.
Tableau^{9} 

b^{9} 
x^{9}_{1} 
x^{9}_{2} 
x^{9}_{3} 
x^{9}_{4} 
x^{9}_{5} 
x^{9}_{6} 
x^{9}_{7} 
x^{9}_{8} 
x^{9}_{9} 
x^{9}_{10} 
x^{9}_{11} 
x^{9}_{12} 
s^{9}_{1} 
s^{9}_{2} 
s^{9}_{3} 
s^{9}_{4} 
s^{9}_{5} 
s^{9}_{6} 
s^{9}_{7} 
s^{9}_{8} 
s^{9}_{9} 
s*^{9}_{10} 
s*^{9}_{11} 
s*^{9}_{12} 
s*^{9}_{13} 
row sum 
M^{9}  100  0  0  0  0  0.5  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0.5  0.5  0  0  101 
H^{9}  233.3  0  0  0  0  0.5  0  0.5  0  0  0  0  0  0  1  0  0  0.3  0.3  0.3  0.3  0.3  0.2  0.2  0.3  0.3  233.3 
J^{9}  566.7  0  0  0  0  0  0  1  1  0  0  0  0  0  0  1  0  0.7  0.7  0.7  0.7  0.7  0.3  0.3  0.3  0.3  567.7 
S^{9}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{9}  666.7  0  0  0  0  1  1  1  1  0  0  0  0  0  0  0  0  0.7  0.7  0.7  0.7  0.7  0.3  0.3  0.3  0.3  668.7 
B_{2}^{9}  1000  0  0  0  0  1.5  0  1.5  1.5  0  0  1  0  0  0  0  0  1  1  0  1  1  0.5  0.5  0  0.5  1003 
B_{3}^{9}  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0  0 
B_{4}^{9}  0  0  0  0  0  0  0  1.5  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0 
MJ^{9}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
F_{1}^{9}  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0  0 
F_{2}^{9}  0  0  1  0  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0 
F_{3}^{9}  333.3  0  0  1  0  0.5  0  0.5  0.5  0  0  0  0  0  0  0  0  0.3  0.3  0.3  0.3  0.3  0.2  0.2  0.3  0.2  334.3 
F_{4}^{9}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{9}  124644  0  0  0  0  10.6  0  11.3  2.7  0  0  0  0  0  0  0  0  110.1  114  112.7  115.4  124.6  27.8  25.9  19.9  18.6  123950.5 
P^{9} / M^{9}  0  0  0  0  0  21.2  0  22.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{9}: [s_{1}, s_{2}, s_{3}, s_{4}, x_{6}, x_{11}, x_{9}, x_{10}, x_{12}, x_{1}, x_{2}, x_{3}, x_{4}, ]. Value of Objective Function = 124644.
Proceed to the next tableau as follows:
Phase 0: Complete.
Phase I: Complete.
Phase II: Goal: get ß >= 0.
A. In Tableau^{9}:
1. Select a pivot row, row, with b^{9}_{row} < 0: row = 1 associated with b^{9}_{1} = 100.
2. Compute the ratios Ø / M as per the last row. Discard ratios which are not positive and ratios associated with artificial variables.
Select the column with the least positive ratio as the pivot column: col = 5 associated with 21.2. Thus x_{1,5} = 0.5 is the pivot; variable s_{1} will leave the basis; variable x_{5} will enter the basis.
B. To create Tableau^{10}:
3. Compute row M^{10} = M^{9} / (0.5).
3. Subtract multiples of row M^{10} from all other rows of Tableau^{9} so that column x_{5} = e_{1} in Tableau^{10}.
Tableau^{10} 

b^{10} 
x^{10}_{1} 
x^{10}_{2} 
x^{10}_{3} 
x^{10}_{4} 
x^{10}_{5} 
x^{10}_{6} 
x^{10}_{7} 
x^{10}_{8} 
x^{10}_{9} 
x^{10}_{10} 
x^{10}_{11} 
x^{10}_{12} 
s^{10}_{1} 
s^{10}_{2} 
s^{10}_{3} 
s^{10}_{4} 
s^{10}_{5} 
s^{10}_{6} 
s^{10}_{7} 
s^{10}_{8} 
s^{10}_{9} 
s*^{10}_{10} 
s*^{10}_{11} 
s*^{10}_{12} 
s*^{10}_{13} 
row sum 
M^{10}  200  0  0  0  0  1  0  1  0  0  0  0  0  2  0  0  0  0  0  0  0  0  1  1  0  0  202 
H^{10}  133.3  0  0  0  0  0  0  0  0  0  0  0  0  1  1  0  0  0.3  0.3  0.3  0.3  0.3  0.3  0.3  0.3  0.3  132.3 
J^{10}  566.7  0  0  0  0  0  0  1  1  0  0  0  0  0  0  1  0  0.7  0.7  0.7  0.7  0.7  0.3  0.3  0.3  0.3  567.7 
S^{10}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{10}  466.7  0  0  0  0  0  1  0  1  0  0  0  0  2  0  0  0  0.7  0.7  0.7  0.7  0.7  0.7  0.7  0.3  0.3  466.7 
B_{2}^{10}  700  0  0  0  0  0  0  0  1.5  0  0  1  0  3  0  0  0  1  1  0  1  1  1  1  0  0.5  700 
B_{3}^{10}  300  0  0  0  0  0  0  1.5  0  1  0  0  0  3  0  0  0  1  0  0  0  0  1  1.5  0  0  303 
B_{4}^{10}  0  0  0  0  0  0  0  1.5  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0  0  0 
MJ^{10}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
F_{1}^{10}  100  1  0  0  0  0  0  0.5  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  0.5  0  0  101 
F_{2}^{10}  0  0  1  0  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0  0  0 
F_{3}^{10}  233.3  0  0  1  0  0  0  0  0.5  0  0  0  0  1  0  0  0  0.3  0.3  0.3  0.3  0.3  0.3  0.3  0.3  0.2  233.3 
F_{4}^{10}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{10}  126765.6  0  0  0  0  0  0  0.7  2.7  0  0  0  0  21.2  0  0  0  110.1  114  112.7  115.4  124.6  17.2  15.2  19.9  18.6  126093.3 
P^{10} / S^{10}  0  0  0  0  0  0  0  0.7  2.7  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{10}: [x_{5}, s_{2}, s_{3}, s_{4}, x_{6}, x_{11}, x_{9}, x_{10}, x_{12}, x_{1}, x_{2}, x_{3}, x_{4}, ]. Value of Objective Function = 126765.6.
Proceed to the next tableau as follows:
Phase 0: Complete.
Phase I: Complete.
Phase II: Goal: get ß >= 0.
A. In Tableau^{10}:
1. Select a pivot row, row, with b^{10}_{row} < 0: row = 4 associated with b^{10}_{4} = 100.
2. Compute the ratios Ø / S as per the last row. Discard ratios which are not positive and ratios associated with artificial variables.
Select the column with the least positive ratio as the pivot column: col = 7 associated with 0.7. Thus x_{4,7} = 1 is the pivot; variable s_{4} will leave the basis; variable x_{7} will enter the basis.
B. To create Tableau^{11}:
3. Compute row S^{11} = S^{10} / (1).
3. Subtract multiples of row S^{11} from all other rows of Tableau^{10} so that column x_{7} = e_{4} in Tableau^{11}.
Tableau^{11} 

b^{11} 
x^{11}_{1} 
x^{11}_{2} 
x^{11}_{3} 
x^{11}_{4} 
x^{11}_{5} 
x^{11}_{6} 
x^{11}_{7} 
x^{11}_{8} 
x^{11}_{9} 
x^{11}_{10} 
x^{11}_{11} 
x^{11}_{12} 
s^{11}_{1} 
s^{11}_{2} 
s^{11}_{3} 
s^{11}_{4} 
s^{11}_{5} 
s^{11}_{6} 
s^{11}_{7} 
s^{11}_{8} 
s^{11}_{9} 
s*^{11}_{10} 
s*^{11}_{11} 
s*^{11}_{12} 
s*^{11}_{13} 
row sum 
M^{11}  100  0  0  0  0  1  0  0  1  0  0  0  0  2  0  0  1  0  0  0  0  0  1  1  0  0  101 
H^{11}  133.3  0  0  0  0  0  0  0  0  0  0  0  0  1  1  0  0  0.3  0.3  0.3  0.3  0.3  0.3  0.3  0.3  0.3  132.3 
J^{11}  466.7  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  1  0.7  0.7  0.7  0.7  0.7  0.3  0.3  0.3  0.3  466.7 
S^{11}  100  0  0  0  0  0  0  1  1  0  0  0  0  0  0  0  1  0  0  0  0  0  0  0  0  0  101 
B_{1}^{11}  466.7  0  0  0  0  0  1  0  1  0  0  0  0  2  0  0  0  0.7  0.7  0.7  0.7  0.7  0.7  0.7  0.3  0.3  466.7 
B_{2}^{11}  700  0  0  0  0  0  0  0  1.5  0  0  1  0  3  0  0  0  1  1  0  1  1  1  1  0  0.5  700 
B_{3}^{11}  150  0  0  0  0  0  0  0  1.5  1  0  0  0  3  0  0  1.5  1  0  0  0  0  1  1.5  0  0  151.5 
B_{4}^{11}  150  0  0  0  0  0  0  0  1.5  0  1  0  0  0  0  0  1.5  0  1  0  0  0  0  0.5  0  0  151.5 
MJ^{11}  0  0  0  0  0  0  0  0  1.5  0  0  0  1  0  0  0  0  0  0  0  1  0  0  0  0  0.5  0 
F_{1}^{11}  50  1  0  0  0  0  0  0  0.5  0  0  0  0  1  0  0  0.5  0  0  0  0  0  0  0.5  0  0  50.5 
F_{2}^{11}  50  0  1  0  0  0  0  0  0.5  0  0  0  0  0  0  0  0.5  0  0  0  0  0  0  0.5  0  0  50.5 
F_{3}^{11}  233.3  0  0  1  0  0  0  0  0.5  0  0  0  0  1  0  0  0  0.3  0.3  0.3  0.3  0.3  0.3  0.3  0.3  0.2  233.3 
F_{4}^{11}  0  0  0  0  1  0  0  0  0.5  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0.5  0 
P^{11}  126831.9  0  0  0  0  0  0  0  2  0  0  0  0  21.2  0  0  0.7  110.1  114  112.7  115.4  124.6  17.2  15.2  19.9  18.6  126160.3 
P^{11} / ^{11}  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 
Basis for Tableau^{11}: [x_{5}, s_{2}, s_{3}, x_{7}, x_{6}, x_{11}, x_{9}, x_{10}, x_{12}, x_{1}, x_{2}, x_{3}, x_{4}]. Value of Objective Function = 126831.9.
Phase 0: Complete.
Phase I: Complete.
Phase II: Complete.
Primal Solution: [x_{5}, s_{2}, s_{3}, x_{7}, x_{6}, x_{11}, x_{9}, x_{10}, x_{12}, x_{1}, x_{2}, x_{3}, x_{4}] = [100, 133.3, 466.7, 100, 466.7, 700, 150, 150, 0, 50, 50, 233.3, 0]; P = 126831.9.
(Primal x variables not in the basis have a value of 0). Dual Solution: [y_{M}, y_{H}, y_{J}, y_{S}, y_{B1}, y_{B2}, y_{B3}, y_{B4}, y_{MJ}, y_{F1}, y_{F2}, y_{F3}, y_{F4}] = [21.2, 0, 0, 0.7, 110.1, 114, 112.7, 115.4, 124.6, 17.2, 15.2, 19.9, 18.6]; D = 126831.9.
Egwald Coffee's Least Cost Import Solution.
Sacks of coffee imported and blends produced.
Formula  Mocha  Harrar  Java  Sumatra  MoJo 
Blend 1  50   100   150 
Blend 2  50    100  150 
Blend 3   233.33  466.67   700 
Blend 4   0   0  0 
Total 
100 
233.33 
566.67 
100 
1000 
Purchasing and processing costs
Formula  Mocha  Harrar  Java  Sumatra  MoJo  Total 
Blend 1  8287.5   9282   2187.9 
19757.4 
Blend 2  8287.5    9945  1591.2 
19823.7 
Blend 3   35581  43316   8353.8 
87250.8 
Blend 4   0   0  0 
0 
Total 
16575 
35581 
52598 
9945 
12132.9 
126831.9 
Post Optimality Analysis.
Over what ranges of the values of the coefficients of the model, such as costs (
c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}, c_{7}, c_{8}, c_{9}, c_{10}, c_{11}, c_{12}), or the constraints on inputs and outputs (
b_{M}, b_{H}, b_{J}, b_{S}, b_{B1}, b_{B2}, b_{B3}, b_{B4}, b_{MJ}, b_{F1}, b_{F2}, b_{F3}, b_{F4}), will the above basic feasible solution remain optimal? By the phrase "remain optimal," I mean that the variables in the basis of the current optimal basic solution remain in the basis of the new optimal basic solution, after the values of the coefficients are altered. As long as the list of variables in each basis is the same, even though the values of the variables may change with the changes in the coefficients of the model, the current basic solution is said to remain optimal.
The primal variables,
x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}, x_{8}, x_{9}, x_{10}, x_{11}, x_{12}, take on the values
50, 50, 233.33, 0, 100, 466.67, 100, 0, 150, 150, 700, 0, with per bag costs of
165.75, 165.75, 152.49, 152.49, 92.82, 92.82, 99.45, 99.45, 14.59, 10.61, 11.93, 9.28, and per pound costs of
1.25, 1.25, 1.15, 1.15, 0.7, 0.7, 0.75, 0.75, 0.11, 0.08, 0.09, 0.07. The basis variables associated with rows (1 to 13) of Tableau^{11} are the variables (
x_{5}, s_{2}, s_{3}, x_{7}, x_{6}, x_{11}, x_{9}, x_{10}, x_{12}, x_{1}, x_{2}, x_{3}, x_{4}
) whose associated objective function coefficients are (
92.82, 0, 0, 99.45, 92.82, 11.93, 14.59, 10.61, 9.28, 165.75, 165.75, 152.49, 152.49).
Objective Function Coefficients  NonBasic Variables
By how much can one decrease the cost (increase the profit) of
x_{8} before this variable will enter the basis of the new basic optimal solution. To determine the entry cost for x_{8}, one computes the opportunity cost of producing one unit of x_{8} = the increase in costs from producing one unit of x_{8} = z_{8}, using standard L.P. notation. The value of z_{8} is calculated by weighting each entry of the column associated with x_{8} in Tableau^{11}, x_{i, 8}, by the objective function coefficient associated with the row of that entry:
z_{8} =
92.82 * 1 + 0 * 0 + 0 * 0 + 99.45 * 1 + 92.82 * 1 + 11.93 * 1.5 + 14.59 * 1.5 + 10.61 * 1.5 + 9.28 * 1.5 + 165.75 * 0.5 + 165.75 * 0.5 + 152.49 * 0.5 + 152.49 * 0.5 = 97.46.
Per bag costs for x_{8} can decrease from 99.45 to 97.46, and the cost per pound of x_{8} can decrease from 0.75 to 0.735, before x_{8} will enter the basis in the new optimal primal solution.
Objective Function Coefficients  Basic Variables
By how much can one increase the cost of variable x_{5} before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list.
To determine the basis upset price for x_{5}, one needs to know the increase in costs that result from increasing the value of each nonbasic variable by one unit, i.e. the opportunity cost, z_{j} of producing one unit of each nonbasic variable variable, x_{j}.
One also needs to know by how much the value of each nonbasis variable will increase from a one unit decrease in the basic variable x_{5}, i.e. the tradeoff rate x_{1, j}, from row 1 of Tableau^{11} associated with basis variable x_{5}, between the variable x_{5} and each nonbasis variable, x_{j}.
Call zc_{j} the difference between z_{j} and the objective function coefficient c_{j} for variable x_{j}, divided by x_{1, j} from Tableau^{11}. (For slack variables, the objective function coefficient is 0.) If zc_{j} > 0, then the nonbasic variable x_{j} is a candidate to replace x_{5}, when the cost of x_{5} (or equivalently, the objective function coefficient, c_{5}) is increased by zc_{j}. Taking the least positive zc_{j} over all candidate nonbasic variables yields the possible increase in the cost of x_{5} before this variable leaves the basis of the existing optimal solution, i.e. a different primal or slack variable will be associated with row 1.
For the basic variable x_{5} associated with row 1 of Tableau^{11}, the candidate nonbasic variables — primal variables x_{j}, or slack variables s_{j} — and their zc_{j} values are given by:
x_{j}: zc_{j} = (z_{j}  c_{j}) / x_{1, j}, or
s_{j}: zc_{j} = (z_{j}  0) / x_{1, j}.
The zc_{j} values for x_{5} associated with row 1 of Tablueau^{11} are:
s_{4}: zc_{16} = (0.66  0) /
1 = 0.66
s_{10}: zc_{22} = (17.24  0) /
1 = 17.24
s_{11}: zc_{23} = (15.25  0) /
1 = 15.25
The least positive zc_{j} is zc_{16} = 0.66. Consequently, costs per bag for x_{5} can increase from 92.82 to
93.48, and the costs per pound of x_{5} can increase from 0.7 to 0.705, before the existing basis is upset. (x_{5} might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution).
Similarily, by how much can one increase the cost of variable x_{7} before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list.
The zc_{j} values for x_{7} associated with row 4 of Tablueau^{11} are:
x_{8}: zc_{8} = (97.46  99.45) /
1 = 1.99
The least positive zc_{j} is zc_{8} = 1.99. Consequently, costs per bag for x_{7} can increase from 99.45 to
101.44, and the costs per pound of x_{7} can increase from 0.75 to 0.765, before the existing basis is upset. (x_{7} might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution).
Similarily, by how much can one increase the cost of variable x_{6} before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list.
The zc_{j} values for x_{6} associated with row 5 of Tablueau^{11} are:
x_{8}: zc_{8} = (97.46  99.45) /
1 = 1.99
s_{1}: zc_{13} = (21.22  0) /
2 = 10.61
s_{12}: zc_{24} = (19.89  0) /
0.33 = 59.67
s_{13}: zc_{25} = (18.56  0) /
0.33 = 55.69
The least positive zc_{j} is zc_{8} = 1.99. Consequently, costs per bag for x_{6} can increase from 92.82 to
94.81, and the costs per pound of x_{6} can increase from 0.7 to 0.715, before the existing basis is upset. (x_{6} might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution).
Similarily, by how much can one increase the cost of variable x_{11} before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list.
The zc_{j} values for x_{11} associated with row 6 of Tablueau^{11} are:
x_{8}: zc_{8} = (97.46  99.45) /
1.5 = 1.33
s_{1}: zc_{13} = (21.22  0) /
3 = 7.07
s_{13}: zc_{25} = (18.56  0) /
0.5 = 37.13
The least positive zc_{j} is zc_{8} = 1.33. Consequently, costs per bag for x_{11} can increase from 11.93 to
13.26, and the costs per pound of x_{11} can increase from 0.09 to 0.1, before the existing basis is upset. (x_{11} might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution).
Similarily, by how much can one increase the cost of variable x_{9} before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list.
The zc_{j} values for x_{9} associated with row 7 of Tablueau^{11} are:
s_{4}: zc_{16} = (0.66  0) /
1.5 = 0.44
s_{5}: zc_{17} = (110.06  0) /
1 = 110.06
s_{10}: zc_{22} = (17.24  0) /
1 = 17.24
s_{11}: zc_{23} = (15.25  0) /
1.5 = 10.17
The least positive zc_{j} is zc_{16} = 0.44. Consequently, costs per bag for x_{9} can increase from 14.59 to
15.03, and the costs per pound of x_{9} can increase from 0.11 to 0.113, before the existing basis is upset. (x_{9} might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution).
Similarily, by how much can one increase the cost of variable x_{10} before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list.
The zc_{j} values for x_{10} associated with row 8 of Tablueau^{11} are:
x_{8}: zc_{8} = (97.46  99.45) /
1.5 = 1.33
s_{6}: zc_{18} = (114.04  0) /
1 = 114.04
The least positive zc_{j} is zc_{8} = 1.33. Consequently, costs per bag for x_{10} can increase from 10.61 to
11.93, and the costs per pound of x_{10} can increase from 0.08 to 0.09, before the existing basis is upset. (x_{10} might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution).
The basic variable x_{12} is degenerate. No analysis is available here. Similarily, by how much can one increase the cost of variable x_{1} before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list.
The zc_{j} values for x_{1} associated with row 10 of Tablueau^{11} are:
s_{4}: zc_{16} = (0.66  0) /
0.5 = 1.33
s_{11}: zc_{23} = (15.25  0) /
0.5 = 30.5
The least positive zc_{j} is zc_{16} = 1.33. Consequently, costs per bag for x_{1} can increase from 165.75 to
167.08, and the costs per pound of x_{1} can increase from 1.25 to 1.26, before the existing basis is upset. (x_{1} might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution).
Similarily, by how much can one increase the cost of variable x_{2} before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list.
The zc_{j} values for x_{2} associated with row 11 of Tablueau^{11} are:
x_{8}: zc_{8} = (97.46  99.45) /
0.5 = 3.98
The least positive zc_{j} is zc_{8} = 3.98. Consequently, costs per bag for x_{2} can increase from 165.75 to
169.73, and the costs per pound of x_{2} can increase from 1.25 to 1.28, before the existing basis is upset. (x_{2} might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution).
Similarily, by how much can one increase the cost of variable x_{3} before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list.
The zc_{j} values for x_{3} associated with row 12 of Tablueau^{11} are:
x_{8}: zc_{8} = (97.46  99.45) /
0.5 = 3.98
s_{1}: zc_{13} = (21.22  0) /
1 = 21.22
s_{13}: zc_{25} = (18.56  0) /
0.17 = 111.38
The least positive zc_{j} is zc_{8} = 3.98. Consequently, costs per bag for x_{3} can increase from 152.49 to
156.47, and the costs per pound of x_{3} can increase from 1.15 to 1.18, before the existing basis is upset. (x_{3} might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution).
The basic variable x_{4} is degenerate. No analysis is available here.
If you change the data on the form in the sensitivity analysis section above and click "Submit L.P.", this web page will display the tableaux for the new optimal solution. Continue this process until you convince yourself of the validity of the postoptimality analysis.
Economic Interpretation of the Dual Solution.
The firm's dual problem is to determine the costs, denoted by y_{M}, y_{H}, y_{J}, y_{S}, and y_{MJ} per minimum number of sacks of Mocha, Harrar, Java, Sumatra, and MoJo required, maximizing the total value of coffee beans priced at these shadow costs, yet remain within the costs per sack of for each variable, x_{1} to x_{12}, valued at each shadow price. The firm's dual production decision is captured in the firm's Dual Linear Programming Problem.

Dual Linear Program
Maximize the Objective Function (D)

D =  100  * y_{M} +  100  * y_{H} +  100  * y_{J} +  100  * y_{S} +  0  * y_{B1} +  0  * y_{B2} +  0  * y_{B3} +  0  * y_{B4} +  1000  * y_{MJ} +  0  * y_{F1} +  0  * y_{F2} +  0  * y_{F3} +  0  * y_{F4} 
subject to:

x_{1}:  1  * y_{M} +  0  * y_{H} +  0  * y_{J} +  0  * y_{S} +  1  * y_{B1} +  0  * y_{B2} +  0  * y_{B3} +  0  * y_{B4} +  0  * y_{MJ} +  2  * y_{F1} +  0  * y_{F2} +  0  * y_{F3} +  0  * y_{F4}  <=  165.75 
x_{2}:  1  * y_{M} +  0  * y_{H} +  0  * y_{J} +  0  * y_{S} +  0  * y_{B1} +  1  * y_{B2} +  0  * y_{B3} +  0  * y_{B4} +  0  * y_{MJ} +  0  * y_{F1} +  2  * y_{F2} +  0  * y_{F3} +  0  * y_{F4}  <=  165.75 
x_{3}:  0  * y_{M} +  1  * y_{H} +  0  * y_{J} +  0  * y_{S} +  0  * y_{B1} +  0  * y_{B2} +  1  * y_{B3} +  0  * y_{B4} +  0  * y_{MJ} +  0  * y_{F1} +  0  * y_{F2} +  2  * y_{F3} +  0  * y_{F4}  <=  152.49 
x_{4}:  0  * y_{M} +  1  * y_{H} +  0  * y_{J} +  0  * y_{S} +  0  * y_{B1} +  0  * y_{B2} +  0  * y_{B3} +  1  * y_{B4} +  0  * y_{MJ} +  0  * y_{F1} +  0  * y_{F2} +  0  * y_{F3} +  2  * y_{F4}  <=  152.49 
x_{5}:  0  * y_{M} +  0  * y_{H} +  1  * y_{J} +  0  * y_{S} +  1  * y_{B1} +  0  * y_{B2} +  0  * y_{B3} +  0  * y_{B4} +  0  * y_{MJ} +  1  * y_{F1} +  0  * y_{F2} +  0  * y_{F3} +  0  * y_{F4}  <=  92.82 
x_{6}:  0  * y_{M} +  0  * y_{H} +  1  * y_{J} +  0  * y_{S} +  0  * y_{B1} +  0  * y_{B2} +  1  * y_{B3} +  0  * y_{B4} +  0  * y_{MJ} +  0  * y_{F1} +  0  * y_{F2} +  1  * y_{F3} +  0  * y_{F4}  <=  92.82 
x_{7}:  0  * y_{M} +  0  * y_{H} +  0  * y_{J} +  1  * y_{S} +  0  * y_{B1} +  1  * y_{B2} +  0  * y_{B3} +  0  * y_{B4} +  0  * y_{MJ} +  0  * y_{F1} +  1  * y_{F2} +  0  * y_{F3} +  0  * y_{F4}  <=  99.45 
x_{8}:  0  * y_{M} +  0  * y_{H} +  0  * y_{J} +  1  * y_{S} +  0  * y_{B1} +  0  * y_{B2} +  0  * y_{B3} +  1  * y_{B4} +  0  * y_{MJ} +  0  * y_{F1} +  0  * y_{F2} +  0  * y_{F3} +  1  * y_{F4}  <=  99.45 
x_{9}:  0  * y_{M} +  0  * y_{H} +  0  * y_{J} +  0  * y_{S} +  1  * y_{B1} +  0  * y_{B2} +  0  * y_{B3} +  0  * y_{B4} +  1  * y_{MJ} +  0  * y_{F1} +  0  * y_{F2} +  0  * y_{F3} +  0  * y_{F4}  <=  14.59 
x_{10}:  0  * y_{M} +  0  * y_{H} +  0  * y_{J} +  0  * y_{S} +  0  * y_{B1} +  1  * y_{B2} +  0  * y_{B3} +  0  * y_{B4} +  1  * y_{MJ} +  0  * y_{F1} +  0  * y_{F2} +  0  * y_{F3} +  0  * y_{F4}  <=  10.61 
x_{11}:  0  * y_{M} +  0  * y_{H} +  0  * y_{J} +  0  * y_{S} +  0  * y_{B1} +  0  * y_{B2} +  1  * y_{B3} +  0  * y_{B4} +  1  * y_{MJ} +  0  * y_{F1} +  0  * y_{F2} +  0  * y_{F3} +  0  * y_{F4}  <=  11.93 
x_{12}:  0  * y_{M} +  0  * y_{H} +  0  * y_{J} +  0  * y_{S} +  0  * y_{B1} +  0  * y_{B2} +  0  * y_{B3} +  1  * y_{B4} +  1  * y_{MJ} +  0  * y_{F1} +  0  * y_{F2} +  0  * y_{F3} +  0  * y_{F4}  <=  9.28 

y_{L} >= 0; y_{K} >= 0; y_{M} >= 0


Using matrix notation this is: 
Dual Linear Program
Maximize the Objective Function (D)
D = b^{T} y subject to
y^{T} A <= c^{T}, y >= 0

The solution to this minimization problem is:
Dual Solution: [y_{M}, y_{H}, y_{J}, y_{S}, y_{B1}, y_{B2}, y_{B3}, y_{B4}, y_{MJ}, y_{F1}, y_{F2}, y_{F3}, y_{F4}] = [21.2, 0, 0, 0.7, 110.1, 114, 112.7, 115.4, 124.6, 17.2, 15.2, 19.9, 18.6]; D = 126831.9.
End of the Linear Programming Dual Simplex Method
Linear Programming References.

Sposito, Vincent. Linear and Nonlinear Programming. Ames, Iowa: Iowa UP, 1975.

Restrepo, Rodrigo A. Theory of Games and Programming: Mathematics Lecture Notes. Vancouver: University of B.C., 1967.

Restrepo, Rodrigo A. Linear Programming and Complementarity. Vancouver: University of B.C., 1994.

Taha, Hamdy A. Operations Research: An Introduction. 2nd ed. New York: MacMillan, 1976.
 Wu, Nesa, and Richard Coppins. Linear Programming and Extensions. New York: McGrawHill, 1981.
