Question

The Westchester 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:

Constraint	Television	Radio	Online
Audience per advertisement	110,000	18,000	30,000
Cost per advertisement	$1,500	$250	$500
Maximum media usage	11	21	13

To ensure a balanced use of advertising media, radio advertisements must not exceed 50% of the total number of advertisements authorized. In addition, television should account for at least 10% of the total number of advertisements authorized.

(a)	If the promotional budget is limited to $23,000, how many commercial messages should be run on each medium to maximize total audience contact? If your answer is zero enter “0”.
	Advertisement Alternatives No of commercial messages Television Radio Online
	What is the allocation of the budget among the three media?
	Advertisement Alternatives Budget ($) Television $ Radio $ Online $
	What is the total audience reached?
(b)	By how much would audience contact increase if an extra $100 were allocated to the promotional budget? Round your answer to the nearest whole number.
	Increase in audience coverage of approximately

Answer 1

Solution: First we will formulate the given problem as a Linear Programming Model:

Decision Variables:

Let decision variables:

T = number of television spot advertisements,

R = number of radio advertisements,

O = number of Online advertisements

Objective Function:

Here, an objective is to maximize the total audience contact, hence the objective function=

Max Z = 110000 T + 18000 R + 30000 O

Subject to Constraints:

C1 = 1500 T + 250 R + 500 O ≤ 23000 (Promotional Budget)

C2 = T ≤ 11 (Max. no. of television adv)

C3 = R ≤ 21 (Max no. of radio adv)

C4 = O ≤ 13 (Max no. of online adv)

C5 = R ≤ 0.50 ( T + R + O) (Radio adv should not increase 50% of total adv)

= R ≤ 0.50 T + 0.50 R + 0.50 O

= R - 0.50 T - 0.50 R - 0.50 O ≤ 0

= -0.50 T + 0.50 R - 0.50 O ≤ 0

C6 = T ≥ 0.10 (T+ R + O) (TV adv should be at least 10% of the total adv)

= T ≥ 0.10 T + 0.10 R + 0.10 O

= T - 0.10 T - 0.10 R - 0.10 O ≥ 0

= 0.90 T - 0.10 R - 0.10 O ≥ 0

Non-Negativity Condition =

T, R, O ≥ 0

Answer a) Solve the Obtained LP Model:

As no specific information is given in the question, we will solve the obtained LP model by using the Simplex Method as mentioned in the below steps:

Step 1: The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate
1. As the constraint-1 is of type '≤' we should add slack variable S1
2. As the constraint-2 is of type '≤' we should add slack variable S2
3. As the constraint-3 is of type '≤' we should add slack variable S3
4. As the constraint-4 is of type '≤' we should add slack variable S4
5. As the constraint-5 is of type '≤' we should add slack variable S5
6. As the constraint-6 is of type '≥' we should subtract surplus variable S6 and add artificial variable A1

After introducing slack, surplus, artificial variables:

Step 2: Prepare first iteration table:

Negative minimum Zj-Cj is -0.9M-110000 and its column index is 1. So, the entering variable is T.
Minimum ratio is 0 and its row index is 6. So, the leaving basis variable is A1.
∴ The pivot element is 0.9.
Entering =T, Departing =A1, Key Element =0.9

Step 3: Prepare the second iteration table:

Negative minimum Zj-Cj is -122222.2222 and its column index is 9. So, the entering variable is S6.
The minimum ratio is 9.9 and its row index is 2. So, the leaving basis variable is S2.
∴ The pivot element is 1.1111.
Entering =S6, Departing =S2, Key Element =1.1111

Step 4: Prepare the third iteration table:

Negative minimum Zj-Cj is -30000 and its column index is 3. So, the entering variable is O.
Minimum ratio is 13 and its row index is 1. So, the leaving basis variable is S1.
∴ The pivot element is 500.
Entering =O, Departing =S1, Key Element =500

Step 5: Prepare the fourth iteration table:

Negative minimum Zj-Cj is -3000 and its column index is 2. So, the entering variable is R.
Minimum ratio is 16 and its row index is 5. So, the leaving basis variable is S5.
∴ The pivot element is 0.75.
Entering =R, Departing =S5, Key Element =0.75

Step 6: Prepare fifth iteration table:

Since all Zj-Cj ≥ 0

Hence, the optimal solution arrives with the value of variables as :

Max Z = 110000 (11) + 18000 (16) + 30000 (5) = 1648000

Answer b) Here, the promotional budget is increased by 100 $. Hence, the new LP model =

Max Z = 110000 T + 18000 R + 30000 O

subject to,

C1 = 1500 T + 250 R + 500 O ≤ 23100 (Promotional Budget)

C2 = T ≤ 11 (Max. no. of television adv)

C3 = R ≤ 21 (Max no. of radio adv)

C4 = O ≤ 13 (Max no. of online adv)

C5 = -0.50 T + 0.50 R - 0.50 O ≤ 0 (Radio adv should not increase 50% of total adv)

C6 = 0.90 T - 0.10 R - 0.10 O ≥ 0 (TV adv should be at least 10% of the total adv)

T, R, O ≥ 0

Now, we will solve this model by following the exact steps as we did in the previous answer:

Step 1: The problem is converted to canonical form by adding slack, surplus and artificial variables as appropriate
1. As the constraint-1 is of type '≤' we should add slack variable S1
2. As the constraint-2 is of type '≤' we should add slack variable S2
3. As the constraint-3 is of type '≤' we should add slack variable S3
4. As the constraint-4 is of type '≤' we should add slack variable S4
5. As the constraint-5 is of type '≤' we should add slack variable S5
6. As the constraint-6 is of type '≥' we should subtract surplus variable S6 and add artificial variable A1

After introducing slack, surplus, artificial variables

Step 2: Prepare first iteration table:

Negative minimum Zj-Cj is -0.9M-110000 and its column index is 1. So, the entering variable is T.
Minimum ratio is 0 and its row index is 6. So, the leaving basis variable is A1.
∴ The pivot element is 0.9.
Entering =T, Departing =A1, Key Element =0.9

Step 3: Prepare second iteration table:

Negative minimum Zj-Cj is -122222.2222 and its column index is 9. So, the entering variable is S6.
Minimum ratio is 9.9 and its row index is 2. So, the leaving basis variable is S2.
∴ The pivot element is 1.1111.
Entering =S6, Departing =S2, Key Element =1.1111

Step 4: Prepare third iteration table:

Negative minimum Zj-Cj is -30000 and its column index is 3. So, the entering variable is O.
Minimum ratio is 13 and its row index is 4. So, the leaving basis variable is S4.
∴ The pivot element is 1.
Entering =O, Departing =S4, Key Element =1

Step 5: Prepare fourth iteraion table:

Negative minimum Zj-Cj is -18000 and its column index is 2. So, the entering variable is R.
Minimum ratio is 0.4 and its row index is 1. So, the leaving basis variable is S1.
∴ The pivot element is 250.
Entering =R, Departing =S1, Key Element =250

Step 6: Prepare fifth iteration table:

Negative minimum Zj-Cj is -6000 and its column index is 7. So, the entering variable is S4.
Minimum ratio is 7.8667 and its row index is 5. So, the leaving basis variable is S5.
∴ The pivot element is 1.5.
Entering =S4, Departing =S5, Key Element =1.5

Step 7: Prepare 6th iteration table:

Since all Zj-Cj ≥ 0

Hence, the optimal solution arrives with the value of variables as :

Max Z = 110000 (11) + 18000 (16.1333) + 30000 (5.1333) = 1654400 (Approx)

Hence, an increase in audience coverage = 1654400 - 1648000 = 6400