1-3. Solve the linear programs, where all

x_{1} |
x_{2} |
-x_{3} |
-x_{4} |
x_{5} |
-1 | |

10^{11} |
2 | 0 | 0 | 6 | -10^{-12} |
=- x_{6} |

0 | -1 | -2 | 0 | 3 | 10^{13} |
= -x_{7} |

-1 | 0 | 0 |
4 | -4 | 2^{300} |
= -x_{8} |

2 | 0 | 3^{-300} |
0 | 1 | 2 | = f -> min |

Solution.

x_{1} |
x_{2} |
x_{3} |
x_{4} |
x_{5} |
-1 | |

10^{11} |
2 | 0 | 0 | 6 | -10^{-12} |
=- x_{6} |

0 | -1 | 2 | 0 | 3 | 10^{13} |
= -x_{7} |

-1 | 0 | 0 |
-4 | -4 | 2^{300} |
= -x_{8} |

-2 | 0 | 3^{-300} |
0 | 1 | 2 | = -f -> max |

The first row is bad, so the program is infeasible.

2.

x_{1} |
x_{2} |
x_{3} |
x_{4} |
x_{5} |
-1 | |

1 | 2 | -3 | -5 | 6 | 2 | = -x_{6} |

0 | -10^{-100} |
-2 | 0 | -3 | -10^{-300} |
= x_{7} |

-1 | 0 | 2 | -4 | -4 | -2 | = x_{8} |

2 | 0 | 3 | -1 | 1 | 0 |
= x_{9} |

-1 |
-2 |
-10^{-100} |
1 |
0 |
0 |
= f -> max |

x_{1} |
x_{2} |
x_{3} |
x_{4} |
x_{5} |
-1 | |

1 | 2 | -3 | -5 | 6 | 2 | = -x_{6} |

0 | 10^{-100} |
2 | 0 | 3 | 10^{-300} |
= -x_{7} |

1 | 0 | -2 | 4 | 4 | 2 | =- x_{8} |

-2 | 0 | -3 | 1 | -1 | 0 | = -x_{9} |

-1 |
-2 |
-10^{-100} |
1 |
0 |
0 |
= f -> max |

This tableau is feasible. From the first row, we see that 0 <=

Setting these 4 variables to be 0 (which implies that

Here is the optimal solution found by computer:

max = (4 + 17 e^3 - 9 e^4)/18

at x2 = x5 = x7 = x8 = x9 = 0,

x1 = (2 - 5 e^3)/9, x3 = e^3/2, x4 = (8 + 7 e^3)/18, x6 = 4 + 4 e^3,

where e = 10

**3.**

x_{1}/30000001 |
x_{1}/3000002 |
x_{3}/3000003 |
x_{4}/3000004 |
x_{5}/3000005 |
-1 | |

1/300001 | -1/3000002 | 1/300003 | 1/300004 | 0 | 2 | = -x_{6} |

-1/3000002 |
-1/3000002 | -2/3003 | 0 | 0 |
1 | = -x_{7} |

1/3001 | 0 | 2/3003 | -1/3000002 |
-4/3005 | 2 | = -x_{8} |

2/301 | 0 | 1/3000002 | 0 | 1/305 | 2 | = f -> min |

*Solution. *Combine the first two columns
and multiply the last row by -1 to obtain a standard tableau. This standard tableau is optimal, so

max=2 (hence min(f) = -2) at *x*_{1}= *x*_{3}=*x*_{4}=*x*_{5}= 0, *x*_{6}=2, *x*_{7}= 1, *x*_{8} = 2.

4-5. Solve matrix games:

1 | 3 | 5-10^{-100} |
7 | 1 | 3 | 2 |

8 | 8+10^{-100} |
9 | 8+10^{-100} |
9 | 8 | 9 |

1 | -5 | 4 | 3 | 4 | 8 | 10^{-100} |

3 | 5 | 6 | -3 | -10^{-100} |
1 | 3 |

8+10^{-100} |
9 |
9 |
8 | 9 |
8 | 9 -10^{-100} |

*Solution. *

There is a saddle point at row 2, column 6 with the payoff = 8. Another equilibrium is at row 5, column 6

**5.**

0 | 0 | 0 | -2 | -2 | -1 | -2 |

0 | 2 | 0 | -1 | -2 | -1 | -1 |

0 | 3 | 3 | 0 | -2 | 0 | -1 |

3-10^{-100} |
0 | 3 | 0 | 2 | 3 | 0 |

0 | 3 | 1 | 4 | 3 | 0 | 4 |

*Solution. *Rows 1 and 2 are domimated by Row 5, so we
cross them out. Column 3 is dominated by Column 6, so
Column 3 goes.

Now the first remaining row is dominated by the last row, so it goes.
By additional dominations,

we are reduced to

c1 | c2 | c5 | |

r4 | 3-e | 0 | 2 |

r5 | 0 | 3 | 3 |

where e = 10^{-100}. By graphical method, optimal strategies are [3, 3-e]^{T}/(6-e)
and [3, 3- e, 0]/(6-e)
and the value of game is 3(3-e)/(6-e).

For the original game, the value is the same, and optimal
strategies
are

[0,0,0, 3,3-e]^{T}/(6-e)
and [3, 3-e,0, 0, 0, 0, 0]/(6-e) . The c5 column
above is dominated by c2 column, so it can be crossed out -- T.B.