Data for this months advertising problem for a certain company is described in the table below: Media
Question:
Data for this month’s advertising problem for a certain company is described in the table below:
Media Type | Number of Customers Reached per time unit of an Advertisement | Cost ($) per time unit of an Advertisement | Maximum Time Available per Month | Exposure Quality Units per time unit of an Advertisement |
Daytime TV | 1000 | 1500 | 15 | 65 |
Evening TV | 2000 | 3000 | 10 | 90 |
Daily newspaper | 1500 | 400 | 25 | 40 |
Sunday newspaper magazine | 2500 | 1000 | 4 | 60 |
Radio | 300 | 100 | 30 | 20 |
At most $30,000 can be spent on advertisements this month. There must be at least 10 time units of TV ads. At most $18,000 may be spent on TV ads. At least 50,000 customers must be reached. The objective is to maximize total exposure quality.
The SOLVER formulation and output is shown below.
DTV | ETV | DN | SN | R | ||||
10 | 0 | 25 | 2 | 30 | ||||
65 | 90 | 40 | 60 | 20 | 2370 | |||
Available DTV | 1 | 10 | <= | 15 | ||||
Available ETV | 1 | 0 | <= | 10 | ||||
Available DN | 1 | 25 | <= | 25 | ||||
Available SN | 1 | 2 | <= | 4 | ||||
Available R | 1 | 30 | <= | 30 | ||||
Budget | 1500 | 3000 | 400 | 1000 | 100 | 30000 | <= | 30000 |
TV restriction 1 | 1 | 1 | 10 | >= | 10 | |||
TV restriction 2 | 1500 | 3000 | 15000 | <= | 18000 | |||
Customers reached | 1000 | 2000 | 1500 | 2500 | 300 | 61500 | >= | 50000 |
The optimal value of the objective function is 2370.
Microsoft Excel Sensitivity Report | |||||||
Variable Cells | |||||||
Final | Reduced | Objective | Allowable | Allowable | |||
Cell | Name | Value | Cost | Coefficient | Increase | Decrease | |
$C$8 | DTV | 10 | 0 | 65 | 25 | 65 | |
$D$8 | ETV | 0 | RC | 90 | 65 | 1E+30 | |
$E$8 | DN | 25 | 0 | 40 | 1E+30 | 16 | |
$F$8 | SN | 2 | 0 | 60 | 40 | 16.66666667 | |
$G$8 | R | 30 | 0 | 20 | 1E+30 | 14 | |
Constraints | |||||||
Final | Shadow | Constraint | Allowable | Allowable | |||
Cell | Name | Value | Price | R.H. Side | Increase | Decrease | |
$H$10 | Available DTV | 10 | SP | 15 | 1E+30 | AD | |
$H$11 | Available ETV | 0 | 0 | 10 | 1E+30 | 10 | |
$H$12 | Available DN | 25 | 16 | 25 | 5 | 5 | |
$H$13 | Available SN | 2 | 0 | 4 | 1E+30 | 2 | |
$H$14 | Available R | 30 | 14 | 30 | 20 | 20 | |
$H$15 | Budget | 30000 | 0.06 | 30000 | 2000 | 2000 | |
$H$16 | TV restriction 1 | 10 | -25 | 10 | 1.333333333 | 1.333333333 | |
$H$17 | TV restriction 2 | 15000 | 0 | 18000 | 1E+30 | 3000 | |
$H$18 | Customers reached | FA | 0 | 50000 | 11500 | 1E+30 |
Note that each question below is to be considered independently of all others.
The Company would like to spend at least 20% of the advertisement cost on the radio. Provide the algebraic formulation of this constraint.
Some values in the Sensitivity report were deleted by the Professor. Complete the table below: Write a brief description of each item deleted using LP terminology and record its corresponding value that belongs in the Sensitivity report.
Number | Item | Description | Value |
1 | RC | ||
2 | SP | ||
3 | AD | ||
4 | FA |
What is the new optimal value of the objective function, and what are the optimal values of the decision variables (describe verbally the results via a managerial statement), if the objective function coefficient for DN were to decrease from 40 to 30? Justify.
If the right-hand side of the second constraint (Available ETV) were to increase from 10 to 13, what would be the new optimal value of the objective function? Justify.
If the right-hand side of the seventh constraint (TV restriction 1) were decreased from 10 to 9.5, what would be the new optimal value of the objective function? Justify.
Business Forecasting with ForecastX
ISBN: 978-0073373645
6th edition
Authors: Holton wilson, barry keating, john solutions inc