Linear Programming : Simplex Method (5 Problems)

1.    Consider the simplex tableau


x      y    u     v    w      M
              [ 1     0     3    0    0       0   |   10]
              [ 0     0     1    0    1        0  |    0]
              [ 0     1    -6     0   0        0  |    3]
              [ 0     0      8    1    0        0 |    7]    
              [ 0     0      5    0    0        1     4]

The tableau above is the final one in a problem to minimize -x + 2y.  The minimum value of -x + 2y is:

a. -10
b. -4
c. 0
d. 4
e. none of the above

2.  The inequality 2x - y + 5z (greater than or equal to symbol) -6 is equivalent to the inequality

a. 2x - y + 5z (less than or equal to symbol) -6
b. 2x - y + 5z (greater than or equal to  symbol) 6
c. -2x + y - 5z (less than or equal to symbol) -6
d. -2x + y -5z (greater than or equal to symbol) 6
e. none of the above

3. Consider the following linear programming problem:  A workshop of Peter's Potters makes vases and pitchers.  Profit on a vase is $3; profit on a pitcher is $4.  Each vase requires ½ hour of labor, each pitcher requires 1 hour of labor.  Each item requires 1 unit of time in the kiln.  Labor is limited to 4 hours per day and kiln time is limited to 6 units per day.  Initial and final tableaux  are shown in finding the production plan which will maximize profits: (x = number of vases and y = number of pitchers made per day).

                   x    y    u     v    M
                [ -1   1    1    0     0   | 4  ]
                [  2                          |     ]
                [ 1    1    0   1     0   |  6 ]  
               [-3   -4    0   0     1   |  0  ]
                         (initial)

                 x    y    u     v    M
                [ 0    1    2    -1     0    | 2   ]
                [ 1   0  -2     2      0   |  4  ]  
                [0    0   2      2      1   | 20 ]
                                         (final)
                          
If kiln time were decreased by one unit per day, determine the optimal performance schedule.

a. x = 3, y = 4
b. x = 4, y = 2
c. x = 6, y = 1
d. x = 2, y = 3
e. none of the above




4. In a linear programming problem in a standard from, the initial and final tableaux are given as below:

                 x    y    u     v    M
                [ 1   3     1    0     0    | 50  ]
                [ 1  5    0    1     0     |  70 ]  
                [-6  -24   0    0     1   |   0  ]
                         (initial)


                 x    y    u     v    M
                [ 0   1    2    -5    -3    | 20   ]
                                   2     2
                                  -1     1
                [ 1   0   2     2      0   |  10  ]  
               [0    0   3      3      1   | 360]
                                         (final)

Given that x (greater than or equal to sign) 0 and y (greater than or equal to sign) 0, if h units were added to the first resource, the maximum value of the objective function is

a. 360
b. 360 + 3h
c. 360 + h
d. 360 + h(to the 5th power)/2
e. none of the above


5. To solve the linear programming problem

Minimize 50x + 70y subject to:

x + y (greater than or equal to symbol) 6
3x + 5y (greater than or equal to symbol) 24 we can use the simplex method on its dual.
x (greater than or equal to symbol) 0 , y (greater than or equal to symbol) 0
The objective function of the dual is

a. u + 3v
b. u + 5v
c. 6u + 24v
d. 70u + 50v
e. none of the above

Attached file(s):

Linear inequalities.doc  View File