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The Sea Wharf Restaurant would like to determine the best way to allocate a meat

ID: 392631 • Letter: T

Question





The Sea Wharf Restaurant would like to determine the best way to allocate a meathly advertising budget of $10,000 among newspaper, television, and radio advertising (You may assume that each month they spend their entire advertising budget) Management has decided that at least 25% of the budget must be spent on each type of media. Management also feels that the amount of money spent on local newspaper advertising must be at least twice the amount spent on radio advertising. A marketing consultant has developéd an index that measures audience exposure per ad on a scale from 0 to 100, with higher values implying greater audience exposure. The indices are: 50 points per newspaper ad X. . 80 points per radio ad X . 100 points per television ad The cost of placing each ad is $250 for a newspaper ad $400 for a radio ad $600 for a television ad Develop a linear programming model which would show the management of the Sea Wharf Restaurant the best way to advertise in order to maximize their audience exposure A. Define your variables in English.(points) B. Write out the mathematical formulation of your objective function. (7 points) Write out all the constraints for this LP model. (22 points)

Explanation / Answer

Let X1 be the number of ads in local newspaper, X2 in Radio, X3 for Television.

Total budget is $10,000.

Based on the research by marketing consultant, we got the points to reach the customer.

As the objective is to maximize the exposure, we need to Maximize,

50X1+80X2+100X3.

Of course, there are other constraints in terms of budget. 25% of budget should be utilized for each ad method.

250X1 >=2500 , i.e X1>=10

400X2 >=2500, i.e , X2>=6.25

600X3 >=2500 i,e, X3>=4.17

Newspaper ads should grab twice the budget to Radio.

Also, 250X1 >=2*400*X2.

It implies 250X1 >=800 X2

5X1-16X2>=0.

Hence, the Linear program is as below -

Maximise - 50X1+80X2+100X3.

Constraints are -

X1>=10

X2>=6.25

X3>=4.17

5X1-16X2>=0.

and, X1,X2,X3 should not be negative.

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