**Solution 1a.** Say the Barracks are Barrack 1 and Barrack 2 and the latter is boosted. Let x_1, x_2 be the number of Archers trained from Barrack 1 and Barrack 2 respectively. Define x_3, x_4 for Giants and x_5, x_6 for Wizards similarly. We can formulate the linear program as follows.

Maximize | \begin{gathered}(100\times 22 - 400)(x_1+x_2) \\ +(100\times 43 - 940)(x_3+x_4) \\ +(100\times 170 - 3500)(x_5+x_6)\end{gathered} |
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Subject to | 44(x_1+x_2)+940(x_3+x_4)+156(x_5+x_6)\ge 18800 Hit points constraint |

(x_1+x_2)+5(x_3+x_4)+4(x_5+x_6)\le 220 Housing space constraint | |

0.4x_1+2x_3+8x_5 \le 45 Training time constraint for Barrack 1 | |

0.1x_2+0.5x_4+2x_6 \le 45 Training time constraint for Barrack 2 | |

x_1, \dots, x_6 \ge 0 |

**Solution 1b.** Insert a surplus variable p_1 and an artificial variable a_1 for the first constraint, and insert slack variables s_2, s_3, s_4 for the last three constraints respectively.

Maximize | 1800x_1+1800x_2+3360x_3+3360x_4+13500x_5+13500x_6 |
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Subject to | 44x_1+44x_2+940x_3+940x_4+156x_5+156x_6 - p_1 + a_1 = 18800 |

x_1+x_2+5x_3+5x_4+4x_5+4x_6 + s_2=220 | |

0.4x_1+2x_3+8x_5+s_3=45 | |

0.1x_2+0.5x_4+2x_6+s_4=45 | |

x_1, \dots, x_6, p_1, a_1, s_2, s_3, s_4 \ge 0 |

The artificial objective function is -a_1 = 44x_1+44x_2+940x_3+940x_4+156x_5+156x_6 - p_1 - 18800. Therefore we get the initial tableau.

x_1 | x_2 | x_3 | x_4 | x_5 | x_6 | p_1 | a_1 | s_2 | s_3 | s_4 | |
---|---|---|---|---|---|---|---|---|---|---|---|

44 | 44 | 940 | 940 | 156 | 156 | -1 | 1 | 18800 | |||

1 | 1 | 5 | 5 | 4 | 4 | 1 | 220 | ||||

0.4 | 2 | 8 | 1 | 45 | |||||||

0.1 | 0.5 | 2 | 1 | 45 | |||||||

-1800 | -1800 | -3360 | -3360 | -13500 | -13500 | 0 | |||||

-44 | -44 | -940 | -940 | -156 | -156 | 1 | -18800 |

We may choose any variable among x_1, \dots, x_6 as the entering variable as their coefficients in the artificial objective function are negative. We choose x_3 here. Now the replacement quantities are 18800/940 = 20, 220/5 = 44, 45/2 = 22.5. So the first entry in column 3 is the pivot. Couple of row operations yield:

x_1 | x_2 | x_3 | x_4 | x_5 | x_6 | p_1 | a_1 | s_2 | s_3 | s_4 | |
---|---|---|---|---|---|---|---|---|---|---|---|

11/235 | 11/235 | 1 | 1 | 39/235 | 39/235 | -1/940 | 1/940 | 20 | |||

36/47 | 36/47 | 0 | 0 | 149/47 | 149/47 | 1/188 | 1/188 | 1 | 120 | ||

72/235 | -22/235 | -2 | 1802/235 | -78/235 | 1/470 | -1/470 | 1 | 5 | |||

0.1 | 0.5 | 2 | 1 | 45 | |||||||

-77208/47 | -77208/47 | -608292/47 | -608292/47 | -168/47 | -168/47 | 67200 | |||||

1 | 0 |

**Solution 1c.** The dual linear program is as follows.

Minimize | 19200y_1+220y_2+45y_3+45y_4 |
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Subject to | 44y_1+y_2+0.4y_3 \ge 1800 |

44y_1 + y_2 + 0.1y_4 \ge 1800 | |

940y_1+5y_2+2y_3 \ge 3360 | |

940y_1 + 5y_2+0.5y_4 \ge 3360 | |

156y_1 + 4y_2+ 8y_3 \ge 13500 | |

156y_1 + 4y_2+2y_4 \ge 13500 | |

y_1 \le 0, y_2, y_3, y_4 \ge 0 |

**Solution 2. **Denote the cities by 1, 2, 3, 4. Let x_{ij} denote whether the skiers travel from city i to city j. In addition, let b_1, b_2 be the binary variables indicating whether they will take the non-stop/one-stop flight and the two-stop flight from Pittsburgh to Jackson respectively. Similarly define b_3, b_4 for the flights from Salt Lake City to Aspen.

Maximize | \begin{gathered}Mx_{11}+183x_{12}+(443b_1+335b_2)+328x_{14} \\ +183x_{21}+Mx_{22}+279x_{23}+(323b_3+317b_4) \\ +333x_{31}+279x_{32}+Mx_{33}+409x_{34}+\\ 423x_{41}+356x_{42}+403x_{43}+Mx_{44}\end{gathered} |
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Subject to | x_{i1}+x_{i2}+x_{i3}+x_{i4} = 1 for all i = 1, 2, 3, 4. |

x_{1j}+x_{2j}+x_{3j}+x_{4j} = 1 for all j = 1, 2, 3, 4. | |

\sum_{i \in D, j\notin D}x_{ij} \ge 1 for all proper subsets D. | |

b_1 + b_2 \le x_{13} | |

b_3 + b_4 \le x_{24} |