- a)
optimal x1=1 1/11; x2 = 3 7/11 OV= 12 8/11
b)Constraints 2 and 3 are active 1 and 4 are inactive
- c) Slack and surpluses are calculated by plugging in the
optimal x1 and x2 values in the inequalities
- constraint 1: (3x1+6x2>=18) surplus = 7
1/11 ( how much the left hand side axceeds the RHS)
- constraint 2: (5x1+4x2>=20) surplus = 0
(active constraint)
- constraint 3: (8x1+2x2>=16) surplus = 0
(active constraint)
- constraint 4: (7x1+6x2<=42) slack = 12 6/11
(how much the left hand side is smaller than the right hand side)
- d) there are four extreme points
- e)
Optimal points are
- (x =1 1/11; x2=3 7/11) and
- (x1=2 2/3; x2 =1 2/3) as well as all points in between with
OV = 60
Notice that in this objective function x1 and x2 has coefficients that
have the same ratio to one another as in the second constraint, causing
the cost contours and the second constraint to have the same slope.
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-
4-17
- a)
- b)The redundant constraint is E+F<=290
- c)As the right side of this constraint is reduced, it moves
towards the feasible region, and touches the extrem point: (E = 118.4 F=
152.63) first. We can calculate the value of the RHS for which the
constraint will pass through this point by simply plugging in the E and
F values in the constraint, giving a value of 271. therefore the minimum
reduction in the RHS of E+F<=290 is 290 - 271 = 19.
- d)As the coefficient of E in E+F =290 increases it becomes
steeper (pegged at the F axis) when it becomes steep enogh, it will pass
through the optimal point E=118.4, F=152.6. The value of the
coefficient for which this will happen can be calculated by solving aE +
F =290 at point E=118.4 and F=152.6 for the value of a.
Solving 118.4a + 152.6 = 290 yields a =1.16047). Thus the minimum
increase in the coefficient of e to make the constraint active is 0.16047
- e) As the coefficient of E in the objective function
increases the contour becomes steeper. When it becomes as steep as
constraint: 100E+ 60F=21000, there will be alternative optima. When the
objective function contour will have the same slope as 100E+60F=21000 is
determined from the equality:-Ce/1000 = - 100/60.
Solving for Ce we find Ce = 1667. Note that when indeed the coefficient
of E becomes 1667, both the obj function and the equation 100E + 60F =
21,000 have the same slopes, namely -1667/1000 and -100/60 respectively.
Also notice that if the coefficient of E becomes greater than 1667, the
contour will become more steep than the constraint, making point III the
unique optimal.
[go back] [any
Questions?]
- 4-29 More, loosening
- 4-30 Fewer, Tightening
- 4-31 Enlarge, smaller, unchanged
- 4-32 Diminish, laregr,unchanged
-
- 4-36
a) 2
and 5 
