Solutions for HW#1 – MgtOp470 - Section 1 - Dr. Chen
Problem 1
a. What is the linear programming model for this problem? You need to define the decision
variables clearly.
Let X1 = number of regular pizzas produced
and X2 = number of deluxe pizzas produced
Max(profit)
X1 + 1.5X2
s.t.
1X1 + 1X2  150
.25X1 + .5X2  50
1X1  50
1X2  25
profit line(PL)
dough mix constraint (DM)
topping mix constraint (TM)
lower bound on # of regular pizzas (LR)
lower bound on # of deluxe pizzas (LD)
b. Find the optimal solution by the graphical method.
x2
150 D
ou
gh
Regular Pizzas
125
Optimal Solution
100
x1 = 100, x2 = 50
z = $175
75
50
Deluxe Pizzas
25
Top
pin
g
0
25
50
75
100
125
150
175
200
x1
Optimal solution is to produce 100 regular pizzas and 50 deluxe pizzas and achieve a $175
profit.
c. What are the values and interpretations of all slacks and surpluses at the optimal solution?
The slack for the DM constraint is 0. This means we used all of the dough mix.
The slack for the TM constraint is 0. This means we used all of the topping mix.
The surplus for the LR constraint is 50. This means we overproduced the minimum # of regular
pizzas by 50.
The surplus for the LD constraint is 25. This means we overproduced the minimum # of deluxe
pizzas by 25.
d. Which constraints are binding at the optimal solution?
The DM and TM constraints are binding (slack=surplus=0)
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Problem 2
Does the following linear program involve infeasibility, unboundedness, and/or alternative optimal
solutions? Explain.
x2
Max
4 x1  8x2
14
s.t.
2 x1  2 x 2  10
 1x1  1x 2  8
x1 , x2  0
12
Satisfies Constraint #2
10
8
Infeasibility
6
4
Satisfies Constraint #1
2
0
2
4
6
8
10
12
x1
It is infeasible because there is no common feasible region (you cannot satisfy all the constraints
at the same time).
It is NOT unbounded since, even though the upper area is unbounded, at any one of these
points, it is still infeasible because the first constraint is never satisfied in this area.
Problem 3
Does the following linear program involve infeasibility, unboundedness, and/or alternative optimal
solutions? Explain.
Max
s.t.
1x1  1x2
8 x1  6 x 2  24
4 x1  6 x 2  12
2 x2  4
x1 , x2  0
x2
4
Unbounded
3
2
1
The solution is unbounded since the objective is to
maximize and the feasible region is unbounded.
x1
0
1
2
3
2
Problem 4
Consider the linear program given below.
Max
2 x1
s.t.
 3x2
x1  x 2  10
2 x1  x 2  4
x1  3 x 2  24
2 x1  x 2  16
x1 , x2  0
a.
Solve this problem using the graphical solution procedure.
x2
Line B
x1 + x2 = 10
10
Objective Function
Optimal Solution
(3,7)
8
Line A
x1 + 3x2 = 24
6
Feasible
Region
4
2
x1
0
2
4
6
8
10
Optimal Value = 27
The optimal solution is (X1,X2) = (3,7) for an objective function value of 27.
b. Compute the ranges of c1 and c2 (one coefficient is fixed at a time) of the objective function
coefficients for x1 and x2, for the current optimal solution to remain optimal.
Solving the two intersecting constraint lines at the optimal solution (the equations x1 + x2 = 10, and
x1 + 3x2 = 24), the slopes of the lines are:
-1 and -1/3, so -1< -c1/c2 < -1/3
Keeping c2 = 3 (coefficient of X2 in the objective function), and solving for c1, then 1 < c1 < 3.
Keeping c1 = 2, (coefficient of X1 in the objective function), and solving for c2, then 2 < c2 < 6.
c. Suppose c1 is increased from 2 to 2.5 and c2 is increased from 3 to 3.5. Will the current optimal
solution change?
From b. above -1< -c1/c2 < -1/3, so with the changes -c1/c2 = -(2.5/3.5) = -5/7. Since this is still
between the limits of -1 and -1/3, the optimal solution will not change.
d. Suppose c1 is increased from 2 to 2.5. What is the new optimal solution?
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From b. above 2.5 is still in the range of c1, so the optimal solution will not change.
e. Suppose c2 is decreased from 3 to 1. What is the new optimal solution?
Since this is outside of the range given in b. above, and -c1/c2 now becomes more negative, the
profit line shifts clockwise to the next corner point which is at (X1,X2) = (6,4) and the objective
function value now becomes 16. However, we now have a condition of alternative optimal
solutions, since the point (X1,X2) = (0,8) also yields an optimal solution of 16. And in fact,
anywhere along the line from (6,4) to (0,8) is an optimal solution.
x2
10
Objective Function (2x1+x2)
8
5
4
6
4
Feasible
Region
6
3
Alternative Optimal solutions
2
1
2
x1
0
2
4
6
8
10
f. Compute the dual prices and the associated ranges for constraints 1 and 2 and interpret them.
Constraint 4
x2
Constraint 1
10
Point (c)
8
Point (a)
Point (b)
Constraint 3
6
Feasible
Region
4
Objective Function
Constraint 2 2
0
2
4
6
8
10
For the ranges of the RHS’s for constraints 1 and 2, we must determine how much we can adjust
the RHS’s while still maintaining the same binding constraints. In this case, constraints 1 and 3
are binding. For constraint 1 we can increase the RHS (parallel to the current constraint) until
another constraint also becomes binding. In this case it is constraint 4. At this point (a)
constraints 1, 3, and 4 are binding (if we move constraint any further than this, only constraints 1
and 4 will be binding). Since 3 and 4 are binding, we can find their intersection at Point (a):
4
2x1 + x2 = 16

x1 = 4.8
x1 + 3x2 = 24

x2 = 6.4
(X1, X2) = (4.8, 6.4). Plug this new optimal solution into the LHS of constraint 1 to determine the
new RHS. So:
Point (a)
X1 + X2 = 4.8 + 6.4 = 11.2
is the upper bound for the RHS of constraint 1. The lower bound will be achieved by moving
constraint 1 toward the origin, while maintaining constraints 1 and 3 binding. This can be done
(again the line moves parallel to the original constraint) until the intersection of constraints 1 and 3
also intersect the X2 axis. Then the two constraints are still binding, but so is the non-negativity
constraint for variable X2. This happens at Point (b):
Point (b)
x1 = 0

x1 = 0
x1 + 3x2 = 24

x2 = 8
(X1, X2) = (0, 8). Again plugging in this new optimal solution to constraint 1 we get:
X1 + X2 = 0 + 8 = 8
for the lower bound of the RHS of constraint 1.
Thus the range is (8, 11.2) for the RHS of constraint 1.
When considering constraint 2, it is not the binding constraint. (The two binding constraints are
still constraints 1 and 3.) If we move constraint 2 in the direction of the origin, those two
constraints remain binding. In fact, as long as we keep moving in the direction of the origin, we
can continue to - , and they will remain binding. So the lower bound for the RHS of constraint 2
is - . For the upper bound we move the constraint parallel to the original constraint, but away
from the origin. Constraints 1 and 3 remain binding until constraint 2 is binding as well at the
original optimal solution Point (c): (X1, X2) = (3,7). If we plug this solution into constraint 2 we get:
2X1 + X2 = 6 + 7 = 13
for the upper bound.
Thus the range is (- , 13) for the RHS of constraint 2.
The dual price for constraint 1 can be calculated using the RHS range for constraint 1 and the
associated objective function values. We have already determined that the RHS range has been
determined through solutions at (X1, X2) = (4.8, 6.4) and (0,8). The objective function value at
(4.8, 6.4) is:
2X1 + 3X2 = 2(4.8) + 3(6.4) = 28.8
The objective function value at (0, 8) is:
2X1 + 3X2 = 2(0) + 3(8) = 24
Since the RHS range is (8, 11.2), the dual price is determined by the change in objective function
value divided by the change in the RHS.
The dual price for constraint 1 is (28.8-24)/ (11.2-8) = 4.8/3.2 = 1.5.
For constraint 2, since this constraint is not binding over the RHS range, the optimal solution
doesn’t change with the shift in its RHS. Neither does the objective function.
Then the dual price for constraint 2 is 0.
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