Question

Part II (50 marks) The advertising alternatives for a company include television, radio, and newspaper ad- vertisements. The costs and estimates for audience coverage are given in table below Television Newspaper Radio 300 600 Cost per advert 2,000 Audience per advert 100, 000 40,000 18,000 l newspaper limits the number of weekly advertisements from a single company nce the advertising amon The loca to ten. Moreover, in order no more than half of the total number of advertisements should occur on the radio, and g the three types of media, to bala on. The weekly advertising budget is €18,200. The at least 10% should occur on televisi company want to maximise the total audience. Using 3 iterations of the simplex method to show that the optimal solution is 2. Basic 52 S3 S4 Sol'n 118,000 272 00 60,000 0 1,052,000 r2 3 Si z 100 0 0 0 1100O 272000 T2 0 1 0 0 T3 0 0 1 10 14 12 S4 0 0 0 where xi, 2 and t3 represent the number of advertisements in television, newspaper and radio, respectively (20 marks] 3. How do you know that the solution in Q2. is optimal? 2 marks 4. How many advertisements should be run in each of the three types of media to maximise the total audience? 3 marks)

Answer 1

Question 2:

Let the number of TV advertisements be x1

Let the number of Newspaper be x2

Let the number of Radio be x3

Maximize, Z = 100000*x1 + 40000*x2 + 18000*x3 (Audience advert)

Constraints:

x2 <= 10 (The number of newspaper adverts cannot be more than 10)

x3 <= 0.5(x1 + x2 + x3) [the total number of advertisements on radio should not be more than 50% of all advertisements]

x1 >= 0.1(x1 + x2 + x3) [the total number of TV advertisements must be atleast 10% of all of them]

2000x1 + 600x2 + 300x3 <= 18200 [budget for advertisement is equal to 18200]

Solving the equation using simplex tableau method

Tableau #1
x1 x2 x3 s1 s2 s3 s4 z   
0 1 0 1 0 0 0 0 10
-0.5 -0.5 0.5 0 1 0 0 0 0   
0.9 -0.1 -0.1 0 0 -1 0 0 0   
2000 600 300 0 0 0 1 0 18200   
-100000 -40000 -18000 0 0 0 0 1 0   

Tableau #2
x1 x2 x3 s1 s2 s3 s4 z   
0 1 0 1 0 0 0 0 10
-0.5 -0.5 0.5 0 1 0 0 0 0   
-0.9 0.1 0.1 0 0 1 0 0 0   
2000 600 300 0 0 0 1 0 18200   
-100000 -40000 -18000 0 0 0 0 1 0   

Tableau #3
x1 x2 x3 s1 s2 s3 s4 z   
0 1 0 1 0 0 0 0 10
0 -0.35 0.575 0 1 0 0.00025 0 4.55
0 0.37 0.235 0 0 1 0.00045 0 8.19
1 0.3 0.15 0 0 0 0.0005 0 9.1   
0 -10000 -3000 0 0 0 50 1 910000

Tableau #4
x1 x2 x3 s1 s2 s3 s4 z   
0 1 0 1 0 0 0 0 10
0 0 0.575 0.35 1 0 0.00025 0 8.05
0 0 0.235 -0.37 0 1 0.00045 0 4.49
1 0 0.15 -0.3 0 0 0.0005 0 6.1   
0 0 -3000 10000 0 0 50 1 1010000

Tableau #5
x1 x2 x3 s1 s2 s3 s4 z   
0 1 0 1 0 0 0 0 10
0 0 1 0.608696 1.73913 0 0.000434783 0 14
0 0 0 -0.513043 -0.408696 1 0.000347826 0 1.2   
1 0 0 -0.391304 -0.26087 0 0.000434783 0 4   
0 0 0 11826.1 5217.39 0 51.3043 1 1052000   

Question 3:

The solution to Q2 is optimal, since it is taking into consideration all the constraints and the maxima function toe evaluate

Question 4:

We need to run 4 TV advertisements, 10 newspaper and 12 radio advertisements to maximize the total audience

Total audience will be 1052000   

Note - Post any doubts/queries in comments section.