Dorian Auto manufactures luxury cars and trucks. The company believes that its most likely customers are high-income women and men. To reach these groups, Dorian Auto has embarked on an ambitious TV advertising campaign and has decided to purchase 1-minute commercial spots on two types of programs: comedy shows and football games.
Each comedy commercial is seen by 7 million income women and 2 million high-income men. Each football commercial is seen by 2 million high-high-income women and 12 million high-income men. A 1-minute comedy ad costs $50,000, and a 1-minute football ad costs $100,000. Dorian would like the commercials to be seen by at least 28 million high-income women and 24 million high-income men. Use linear programming to deter-mine how Dorian Auto can meet its advertising requirements at minimum cost.
Solution Dorian must decide how many comedy and football ads should be purchased, so the de-cision variables are
x1 number of 1-minute comedy ads purchased x2 number of 1-minute football ads purchased
Then Dorian wants to minimize total advertising cost (in thousands of dollars).
Total advertising cost cost of comedy ads cost of football ads
com ce o d
s y t
ad
50x1 100x2
total football ads
cost
total comedy ads Dorian Auto
E X A M P L E 2
Thus, Dorian’s objective function is
min z 50x1 100x2 (9)
Dorian faces the following constraints:
Constraint 1 Commercials must reach at least 28 million high-income women.
Constraint 2 Commercials must reach at least 24 million high-income men.
To express Constraints 1 and 2 in terms of x1 and x2, let HIW stand for high-income women viewers and HIM stand for high-income men viewers (in millions).
HIW
com He I d W
y ad
foo H
tb I a
W ll ad
7x1 2x2
HIM
com He I d
M
y ad
foo H
tb I a
M ll ad
2x1 12x2
Constraint 1 may now be expressed as
7x1 2x2 28 (10)
and Constraint 2 may be expressed as
2x1 12x2 24 (11)
The sign restrictions x1 0 and x2 0 are necessary, so the Dorian LP is given by:
min z 50x1 100x2
s.t. 7x1 2x2 28 (HIW) s.t. 2x1 12x2 24 (HIM)
x1, x2 0
This problem is typical of a wide range of LP applications in which a decision maker wants to minimize the cost of meeting a certain set of requirements. To solve this LP graphically, we begin by graphing the feasible region (Figure 4). Note that (10) is satis-fied by points on or above the line AB (AB is part of the line 7x1 2x2 28) and that
total football ads total
comedy ads
total football ads total
comedy ads
z = 600 z = 320
(10)
(4, 4) B
E A C D x2
x1 4
2
(11) 6
8 10 12 14
2 4 6 8 10 12 14
F I G U R E 4 Graphical Solution of Dorian Problem
(11) is satisfied by the points on or above the line CD (CD is part of the line 2x1 12x2 24). From Figure 4, we see that the only first-quadrant points satisfying both (10) and (11) are the points in the shaded region bounded by the x1axis, CEB, and the x2axis.
Like the Giapetto problem, the Dorian problem has a convex feasible region, but the feasible region for Dorian, unlike Giapetto’s, contains points for which the value of at least one variable can assume arbitrarily large values. Such a feasible region is called an un-bounded feasible region.
Because Dorian wants to minimize total advertising cost, the optimal solution to the problem is the point in the feasible region with the smallest z-value. To find the optimal solution, we need to draw an isocost line that intersects the feasible region. An isocost line is any line on which all points have the same z-value (or same cost). We arbitrarily choose the isocost line passing through the point (x1 4, x2 4). For this point, z 50(4) 100(4) 600, and we graph the isocost line z 50x1 100x2 600.
We consider lines parallel to the isocost line 50x1 100x2 600 in the direction of decreasing z (southwest). The last point in the feasible region that intersects an isocost line will be the point in the feasible region having the smallest z-value. From Figure 4, we see that point E has the smallest z-value of any point in the feasible region; this is the optimal solution to the Dorian problem. Note that point E is where the lines 7x1 2x2 28 and 2x1 12x2 24 intersect. Simultaneously solving these equations yields the op-timal solution (x1 3.6, x2 1.4). The optimal z-value can then be found by substitut-ing these values of x1and x2into the objective function. Thus, the optimal z-value is z 50(3.6) 100(1.4) 320 $320,000. Because at point E both the HIW and HIM con-straints are satisfied with equality, both concon-straints are binding.
Does the Dorian model meet the four assumptions of linear programming outlined in Section 3.1?
For the Proportionality Assumption to be valid, each extra comedy commercial must add exactly 7 million HIW and 2 million HIM. This contradicts empirical evidence, which indicates that after a certain point advertising yields diminishing returns. After, say, 500 auto commercials have been aired, most people have probably seen one, so it does little good to air more commercials. Thus, the Proportionality Assumption is violated.
We used the Additivity Assumption to justify writing (total HIW viewers) (HIW viewers from comedy ads) (HIW viewers from football ads). In reality, many of the same people will see a Dorian comedy commercial and a Dorian football commercial. We are double-counting such people, and this creates an inaccurate picture of the total num-ber of people seeing Dorian commercials. The fact that the same person may see more than one type of commercial means that the effectiveness of, say, a comedy commercial depends on the number of football commercials. This violates the Additivity Assumption.
If only 1-minute commercials are available, then it is unreasonable to say that Dorian should buy 3.6 comedy commercials and 1.4 football commercials, so the Divisibility As-sumption is violated, and the Dorian problem should be considered an integer program-ming problem. In Section 9.3, we show that if the Dorian problem is solved as an integer programming problem, then the minimum cost is attained by choosing (x1 6, x2 1) or (x1 4, x2 2). For either solution, the minimum cost is $400,000. This is 25% higher than the cost obtained from the optimal LP solution.
Because there is no way to know with certainty how many viewers are added by each type of commercial, the Certainty Assumption is also violated. Thus, all the assumptions of lin-ear programming seem to be violated by the Dorian Auto problem. Despite these drawbacks, analysts have used similar models to help companies determine their optimal media mix.†
†Lilien and Kotler (1983).
P R O B L E M S
Group A
1 Graphically solve Problem 1 of Section 3.1.
2 Graphically solve Problem 4 of Section 3.1.
3 Leary Chemical manufactures three chemicals: A, B, and C. These chemicals are produced via two production processes: 1 and 2. Running process 1 for an hour costs $4 and yields 3 units of A, 1 of B, and 1 of C. Running process 2 for an hour costs $1 and produces 1 unit of A and 1 of B.
To meet customer demands, at least 10 units of A, 5 of B, and 3 of C must be produced daily. Graphically determine a daily production plan that minimizes the cost of meeting Leary Chemical’s daily demands.
4 For each of the following, determine the direction in which the objective function increases:
a z 4x1 x2
b z x1 2x2
c z x1 3x2
5 Furnco manufactures desks and chairs. Each desk uses 4 units of wood, and each chair uses 3. A desk contributes
$40 to profit, and a chair contributes $25. Marketing restrictions require that the number of chairs produced be at least twice the number of desks produced. If 20 units of wood are available, formulate an LP to maximize Furnco’s profit. Then graphically solve the LP.
6 Farmer Jane owns 45 acres of land. She is going to plant each with wheat or corn. Each acre planted with wheat yields $200 profit; each with corn yields $300 profit. The labor and fertilizer used for each acre are given in Table 1.
One hundred workers and 120 tons of fertilizer are available.
Use linear programming to determine how Jane can maximize profits from her land.
3.3 Special Cases
The Giapetto and Dorian problems each had a unique optimal solution. In this section, we encounter three types of LPs that do not have unique optimal solutions.
1 Some LPs have an infinite number of optimal solutions (alternative or multiple opti-mal solutions).
2 Some LPs have no feasible solutions (infeasible LPs).
3 Some LPs are unbounded: There are points in the feasible region with arbitrarily large (in a max problem) z-values.