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Managerial Economics Assignment One
Economics (Unity University)
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Assignment One
Managerial Economics
2021
1. Consider a firm that is planning an advertising campaign for a new product. Goals set for
the campaign include exposure to at least 100,000 individuals, no fewer than 80,000 of
whom have an annual income of at least $50,000 and no fewer than 40,000 of whom are
single. For simplicity, assume that the firm has only radio and television media available
for this campaign. One television advertisement costs $10,000 and is expected to reach an
average audience of 20,000 persons. Ten thousand of these individuals will have an
income of $50,000 or more, and 4,000 will be single. A radio advertisement costs $6,000
and reaches a total audience of 10,000, all of whom have at least $50,000 in income.
Eight thousand of those exposed to a radio advertisement are single.
Advertising-Media-Relations
Primal Problem
The objective is to minimize the cost of the advertising campaign. Because total cost is merely
the sum of the amounts spent on radio and television advertisements, the objective function is:
Minimize Cost = $6,000R + $10,000TV
Where R and TV represent the number of radio and television ads, respectively, those are
employed in the advertising campaign.
This linear programming problem has three constraint equations, including the minimum
audience exposure requirement, the audience income requirement, and the marital status
requirement.
 The minimum audience exposure requirement states that the number of persons exposed
to radio ads plus the number exposed to television ads must be equal to or greater than
100,000 persons. Algebraically, 10,000 times the number of radio ads plus 20,000 times the
number of television advertisements must be equal to or greater than 100,000:
10,000R + 20,000TV ≥ 100,000
 The two remaining constraints can be constructed in a similar fashion from the data in Table.
The audience income constraint is written
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Assignment One
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2021
10,000R + 10,000TV ≥ 80,000 and
 The marital status constraint is given by
8,000R + 4,000TV ≥ 40,000
Combining the cost-minimization objective function with these three constraint conditions
written in equality form using slack variables gives the complete linear programming problem:
Minimize Cost = $6,000R + $10,000TV
Subject to
10,000R + 20,000TV – SA = 100,000
10,000R + 10,000TV – SI = 80,000
8,000R + 4,000TV – SS = 40,000
R, TV, SA, SI and SS>=0
SA, SI, and SS are slack variables indicating the extent to which minimums on total audience
exposure, exposure to individuals with incomes of at least $50,000, and exposure to single
individuals, respectively, have been exceeded. Note that each slack variable is subtracted from
the relevant constraint equation because greater-than-or-equal-to inequalities are involved.
Excess capacity or nonzero slack variables for any of the constraints mean that audience
exposure minimums have been exceeded.
The solution to this linear programming problem is easily obtained using a combination of
graphic and analytical methods.
TV Ads
10
9
8
10,000R + 20,000TV – SA = 100,000
7
6
5
4
3
2
1 – SI = 80,000
10,000R + 10,000TV
0
1
2
3
4
5
6
7
8
TV
0
5
0
8
0
10
R
10
0
8
0
5
0
Radio Ads
9
8,000R + 4,000TV – SS = 40,000
10
11
12
Figure illustrates this solution. The feasible space problem is bordered by the three constraint
equations and the nonnegative requirements. An isocost curve shows that costs are minimized at
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Assignment One
Managerial Economics
2021
point M, where the total audience exposure and income constraints are binding. With these
constraints binding, slack variables SA = SI = 0. Thus
10,000R + 20,000TV = 100,000
Minus 10,000R + 10,000TV = 80,000
10,000TV = 20,000
TV = 2
10,000R + 20,000(2)
= 100,000
10,000R = 100,000 – 40,000
10,000R = 60,000
R=6
The firm should employ six radio advertisements and two television advertisements to minimize
costs while still meeting audience exposure requirements. Just substituting the value R and TV in
the objective function we will get the total cost of Advertizing: the objective function was
Minimize Cost = $6,000R + $10,000TV
Cost = $6,000(6) + $10,000(2)
Cost = $56,000
Total cost for such a campaign is $56,000.
Dual Problem
The dual to the advertising-mix problem is a constrained-maximization problem, because the
primal is a minimization problem. The objective function of the dual is expressed in terms of
shadow prices or implicit values for the primal constraint conditions. The dual objective function
includes an implicit value, or shadow price, for the minimum audience exposure requirement, the
audience income requirement, and the marital status requirement. Because constraint limits in the
primal problem become the dual objective function coefficients, the dual objective function is
Maximize C* = 100,000VA + 80,000VI + 40,000VS
Where VA, VI, and VS are shadow prices for the minimum audience exposure, audience income,
and marital status requirements. Dual constraints are based on the two variables from the primal
objective function. Thus, there are two constraint conditions in the dual, the first associated with
radio advertisements and the second with television advertisements. Both constraints are of the
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Assignment One
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2021
less-than-or-equal-to type, because primal constraints are of the greater-than-or-equal-to type.
The radio advertising constraint limit is the $6,000 radio advertisements coefficient from the
primal objective function. Coefficients for each shadow price in this constraint equation are
given by the advertising effectiveness measures for a single radio advertisement. The coefficient
for the audience exposure shadow price, VA, is 10,000, the number of individuals reached by a
single radio advertisement. Similarly, the coefficient for VI is 10,000 and that for VS is 8,000.
Thus, the dual radio advertisements constraint is
10,000VA + 10,000VI + 8,000VS ≤ $6,000
The dual television advertising constraint is developed in the same fashion. Because each TV
advertisement reaches a total audience of 20,000, this is the coefficient for the VA variable in the
second dual constraint equation. Coefficients for VI and VS are 10,000 and 4,000, respectively,
because these are the numbers of high-income and single persons reached by one TV
advertisement. The $10,000 cost of a television advertisement is the limit to the second dual
constraint, which can be written
20,000VA + 10,000VI + 4,000VS ≤ $10,000
Solving the Dual
It is possible but difficult to solve this dual problem using a three- dimensional graph or the
simplex method. However, because the primal problem has been solved already, information
from this solution can be used to easily solve the dual. The solutions to the primal and dual of a
single linear programming problem are complementary, and the following must hold:
Primal Objective Variablei X Dual Slack Variablei
Primal Slack Variablei X
Because
both R and TV have
Dual Objective Variablei
nonzero
solutions
in
the
primal,
the
dual
slack
variables LR and LTV must equal zero at the optimal solution. Furthermore, because there is
excess audience exposure to the single marital status category in the primal solution, SS ≠ 0, and
the related dual shadow price variable VS must also equal zero in the optimal solution. This
leaves only VA and VI as two unknowns in the two-equation system of dual constraints:
10,000VA + 10,000VI = $6,000
20,000VA + 10,000VI = $10,000
To get the value for VA and VI :
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2021
- (10,000VA + 10,000VI = $6,000)
-10,000VA - 10,000VI = -$6,000
20,000VA + 10,000VI = $10,000
10,000VA = $4,000
VA= 0.40
10,000(0.40) + 10,000VI = $6,000
10,000VI = $6,000- 4,000
10,000VI = $2,000
VI = 0.20
Substituting the value $0.40 for VA in either constraint equation produces a value of $0.20
for VI. Finally, substituting the appropriate values for VA, VI, and VS into the dual objective
function gives a value of C* = $56,000 [= ($0.40 _ 100,000) + ($0.20 _ 80,000) + ($0 _ 40,000)].
This is the same figure as the $56,000 minimum cost solution to the primal.
Interpreting the Dual Solution
The primal solution tells management the minimum-cost advertising mix. The dual problem
results are equally valuable. Each dual shadow price indicates the change in cost that would
accompany a one-unit change in the various audience exposure requirements. These prices show
the marginal costs of increasing each audience exposure requirement by one unit.
For example, VA is the marginal cost of reaching the last individual in the overall audience. If
there were a one-person reduction in the total audience exposure requirement, a cost saving
of VA = $0.40 would be realized. The marginal cost of increasing total audience exposure from
100,000 to 100,001 individuals would also be 40¢. Shadow prices for the remaining constraint
conditions are interpreted in a similar manner. The shadow price for reaching individuals with
incomes of at least $50,000 is VI = $0.20, or 20¢. It would cost an extra 20¢ per person to reach
more high-income individuals. Azero value for VS, the marital status shadow price, means that
the proposed advertising campaign already reaches more than the 40,000 minimum required
number of single persons. Thus, a small change in the marital status constraint has no effect on
total costs. By comparing these marginal costs with the benefits derived from additional
exposure, management is able to judge the effectiveness of its media advertising campaign. If the
expected profit per exposure exceeds 40¢, it would prove profitable to design an advertising
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Assignment One
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2021
campaign for a larger audience. Likewise, if the expected return per exposure to high-income
individuals is greater than 20¢, promotion to this category of potential customers should be
increased.
Question 2: Minimization case
The Objective Function:
Minimize Z = $6x1 + 3x2
Model Constraints:
2x1 + 4x2  16 lb (nitrogen constraint)
4x1 + 3x2  24 lb (phosphate constraint)
X1, X2  0 (non-negativity constraint)
Where
X1 = bags of Super-gro fertilizer
X2 = bags of Crop-Quick fertilizer
Z= farmer total cost of purchasing fertilizer
A surplus variable is subtracted from a  constraint to convert it to an equation (=).
Subtracting slack variables in the farmer problem constraints:
Minimize Z = $6x1 + $3x2 + 0s1 + 0s2
Subject to:
2x1 + 4x2 – s1 = 16
4x1 + 3x2 – s2 = 24
x1 , x2 , s1 , s2  0
Crop-Quick fertilizer
A
X1 X2 S1 S2
10
9
0 8 16 0 24 8
B
7
X1 X2 S1 S2 Z 6
4.8 1.6 0 0 33.
5
6
4
C
3
X1 X2 S1 S2 Z
8
0 0 8 48 2
1
0
Z
4x1 + 3x2 – s2 = 24
1
2
3
4
5
6
7
8
9
2x1 + 4x2 – s1 = 16
10
11 12
Super-gro fertilizer
The optimal solution point of the minimization problem is
X1= 0 bags of Super-gro fertilizer
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Z = $6(0) + $3(8)
Z= 24, optimal solution
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Managerial Economics
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X2 = 8 bags of Crop-Quick fertilizer
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