- b) Constraint E + F <= 290 is redundant as it has no part in forming the feasible region.
- c) From the graph we see that if the RHS of this constraint is gradually reduced the constraint will move towards the origin parallel to itself. If it moves sufficiently it will touch the feasible region and any further movement will start shaving the feasible region (it will become non-redundat). We can see that the first feasible point that the constrain will touch is the optimal point. This will happen when the RHS of the constraint is 118.4 + 152.6 = 271.2. Thus for the constraint to become active the RHS must be less than 271.2. When this happens point II is no longer feasible (eliminated by the modified constraint).
- d) This is the same question as above except we allow the coefficient of E in the constraint to change rather than the RHS. As the coefficient of E in the constraint increases the constraint line will move clockwise on the E axis and eventually make contact with the feasible region. Again verify that the optimal point is the first point that the constrain will touch as a result of this move and any further movement will shave off the feasible region ( or the constraint will become non-redundat). For this to happen we want the coefficient of E, say c, be such that c*E + 1*F = 290 passes through the optimal point. Thus c*(118.4) + 1*(152.6) = 290. Solving for c, we find c = 1.16. The constraint: 1.16E + F <=290 passes through the optimal point and any further increase in the coefficient of E in the constraint beyond 1.16 will rendere the constraint non-redundant.
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e) Notice from the graph that as the objective coeficient of E
is increases each contour becomes steeper. If the
increase is sufficient, the contour will become tangent to the
feasible region along the line segment II and III or the contour and the line
100E + 60F = 21,000 (on which points II and III lie) will have the same slope.
Equating the slopes, we have
- CE/1000 = - 100/60.
Solving for CE we find CE = 1667.
Another way to answer the question is to note that when alternative optima occurs the objective value will be the same at both points II (118.4, 152.6) and III (150,100). If the objective function is denoted by
CE*E + 1000*F, W e want CE*(150) + 1000*(100) = CE*(118.4) + 1000*(152.6)
solving this for CE, we find CE = 1667 as before.
- a) Both (E = 4.5 F= 7) and (E = 1.5 F = 9.0) will be optimal
- b) Objective function value is $15,000 at both points.
- a)Point x1=4; x2=0 is optimal (OV = 8)
- b) The question is referring to x2. As its profitability
increases, the contour becomes flatter. If it is as flat as 3x1+3x2=12,
point b (x1=3; x2=1) also becomes optimal. Allowing
the coefficient of X2 (say C2) change and using the
equality of the slopes of the objective function and 3x1+3x2=12 we have
-3/3 = -2/C2 . Solving for C2 we get C2 = 2. Therefore when the objective function becomes 2x1+ 2x2 point b becomes an alternative optimum where the value of x2>0.
Same conclusion could be reached from the equality of the OVs at these two points:
2*(4) + C2*(0) = 2*(3) + C2 * (1) Solving for C2 We find that the profitability of x2 has to become at least 2, before the point of optimality shifts to point b where x2 is no longer zero.
- c) When the Objective function becomes 2x1 + 2x2 both (4,0) and (3,1) are optimal with OV=8.
- d) Second constraint, x1 + 3x2 <=6, is inactive. Its RHS can increase indefinitely without affecting the optimal point. The RHS can be reduced until it passes through point (4,0). This happens for 1*(4) + 3*(0) = 4. Thus the constraint x1+3x2<=4 passes through point (4,0)and becomes active and any further reductions in its RHS will render point (4,0) infeasible and hence not optimal.
- e) A change in either direction will change the optimal solution.
- f) The constraint in part d) is inactive whereas the constraint in e) is active .
- g) Notice that this is an equality constraint therefore the feasible region consists only of the points on the darker line segment in the blue area. Thus the optimal point moves to point g where x1= 6/11 and x2 = 20/11.
- h)The optimal solution is now x1 =1 and x2 = 0.
. Increasing the RHS of a "<=" means that there will be more combinations of decision-variable values that satisfy the constraint. This means that one is relaxing (loosening) the constraint.
Increasing the RHS of a ">=" means that there will be fewer combinations of decision-variable values that satisfy the constraint. This means that one is tightening the constraint.
Tightening a constraint cannot enlarge the feasible set and may leave it unchanged or make it smaller
Loosening a constraint cannot diminish the constraint set and may leave it larger or unchanged.
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