Problem 11-9 The Chamber of Commerce periodically sponsors public service semina
ID: 3294359 • Letter: P
Question
Problem 11-9 The Chamber of Commerce periodically sponsors public service seminars and programs. Currently, promotional plans are under way for this year's program. Advertising alternatives include television, radio, and online. Audience estimates, costs, and maximum media usage limitations are as shown: Television Radio Online Constraint Audience per advertisement Cost per advertisement Maximum media usage 100,000 16,000 35,000 1,500 $250 $650 14 10 24 To ensure a balanced use o advertising media, adio advertisements must not exceed 50% ofthe total number o advertisements authorized addition television should account rateas 10% of the total number of advertisements authorized (a) If the promotional budget is limited to $18,400, how many commercial messages should be run on each medium to maximize total audience contact? No of commercial Advertisement Alternatives Television Radio Online messages What is the allocation of the budget among the three media? Advertisement Alternatives Budget (s) Television RadioExplanation / Answer
Ans:
Let x1,x2,x3 are television,radio and online ads respectively.
Maximize z=100,000x1+16000x2+35000x3
s.t.
1500x1+250x2+650x3<=18400
x2<=0.5(x1+x2+x3)
-0.5x1+0.5x2-0.5x3<=0
x1>=0.1(x1+x2+x3)
-0.9x1+0.1x2+0.1x3<=0
x1<=10
x2<=24
x3<=14
subject to
The optimal solution value is Z = 1211000
X1 = 10
X2 = 11
X3 = 1
allocation of budget for
television=10*1500=15000
radio=11*250=2750
online=1*650=650
Total audience reached=1211000
b)now,if 100 dollar is added to the budget.
Budget constraint changes to
1500x1+250x2+650x3<=18500
The optimal solution value is Z = 1216666.67
X1 = 10
X2 = 11.11
X3 = 1.11
increase in audience=1216666-1211000=5666
MAXIMIZE: 100000 X1 + 16000 X2 + 35000 X3subject to
1500 X1 + 250 X2 + 650 X3 18400-0.5 X1 + 0.5 X2 -0.5 X3 0
-0.9 X1 + 0.1 X2 + 0.1 X3 0
1 X1 + 0 X2 + 0 X3 10
0 X1 + 1 X2 + 0 X3 24
0 X1 + 0 X2 + 1 X3 14 X1, X2, X3 0