Question

41. Advertising. A department store has up to $20,000 to spend on television advertising for a sale. All ads will be placed with one television station. A 30-second ad costs $1,000 on daytime TV and is viewed by 14,000 potential customers, $2,000 on prime-time TV and is viewed by 24,000 potential customers, and $1,500 on late-night TV and is viewed by 18,000 potential customers. The television station will not accept a total of more than 15 ads in all three time periods. How many ads should be placed in each time period in order to maximize the number of potential customers who will see the ads? How many potential customers will see the ads? (Ignore repeated viewings of the ad by the same potential customer.) 42. Advertising. Repeat Problem 41 if the department store in- creases its budget to $24,000 and requires that at least half of the ads be placed during prime-time.
please answer the question 42

Answer 1

Let

x1 : the number of daytime ads

x2 : the number of prime time ads

x3 : the number of late night ads

Our objective is to

Max Z = 14000 x1 + 24000 x2 + 18000 x3

subject to

1000 x1 + 2000 x2 + 1500 x3 ≤ 24000 x1 + x2 + x3 ≤ 15 x2 ≥ 7

and x1,x2,x3≥0;

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

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 subtract surplus variable S3 and add artificial variable A1

After introducing slack,surplus,artificial variables

Max Z = 14000 x1 + 24000 x2 + 18000 x3 + 0 S1 + 0 S2 + 0 S3 - M A1

subject to

1000 x1 + 2000 x2 + 1500 x3 + S1 = 24000 x1 + x2 + x3 + S2 = 15 x2 - S3 + A1 = 7

and x1,x2,x3,S1,S2,S3,A1≥0

Iteration-1		Cj	14000	24000	18000	0	0	0	-M
B	CB	XB	x1	x2	x3	S1	S2	S3	A1	MinRatio XB/x2
S1	0	24000	1000	2000	1500	1	0	0	0	24000/2000=12
S2	0	15	1	1	1	0	1	0	0	15/1=15
A1	-M	7	0	(1)	0	0	0	-1	1	7/1=7→
Z=-7M		Zj	0	-M	0	0	0	M	-M
		Zj-Cj	-14000	-M-24000↑	-18000	0	0	M	0

Negative minimum Zj-Cj is -M-24000 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 7 and its row index is 3. So, the leaving basis variable is A1.

∴ The pivot element is 1.

Entering =x2, Departing =A1, Key Element =1

R3(new)=R3(old)

R1(new)=R1(old) - 2000R3(new)

R2(new)=R2(old) - R3(new)

Iteration-2		Cj	14000	24000	18000	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	MinRatio XB/S3
S1	0	10000	1000	0	1500	1	0	(2000)	10000/2000=5→
S2	0	8	1	0	1	0	1	1	8/1=8
x2	24000	7	0	1	0	0	0	-1	---
Z=168000		Zj	0	24000	0	0	0	-24000
		Zj-Cj	-14000	0	-18000	0	0	-24000↑

Negative minimum Zj-Cj is -24000 and its column index is 6. So, the entering variable is S3.

Minimum ratio is 5 and its row index is 1. So, the leaving basis variable is S1.

∴ The pivot element is 2000.

Entering =S3, Departing =S1, Key Element =2000

R1(new)=R1(old)÷2000

R2(new)=R2(old) - R1(new)

R3(new)=R3(old) + R1(new)

Iteration-3		Cj	14000	24000	18000	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	MinRatio XB/x1
S3	0	5	1/2	0	3/4	1/2000	0	1	5/1/2=10
S2	0	3	(1/2)	0	1/4	-1/2000	1	0	3/1/2=6→
x2	24000	12	1/2	1	3/4	1/2000	0	0	12/1/2=24
Z=288000		Zj	12000	24000	18000	12	0	0
		Zj-Cj	-2000↑	0	0	12	0	0

Negative minimum Zj-Cj is -2000 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 6 and its row index is 2. So, the leaving basis variable is S2.

∴ The pivot element is 1/2.

Entering =x1, Departing =S2, Key Element =1/2

R2(new)=R2(old) ×2

R1(new)=R1(old) - 1/2R2(new)

R3(new)=R3(old) - 1/2R2(new)

Iteration-4		Cj	14000	24000	18000	0	0	0
B	CB	XB	x1	x2	x3	S1	S2	S3	MinRatio
S3	0	2	0	0	12	4000/4000000	-1	1
x1	14000	6	1	0	12	-1/1000	2	0
x2	24000	9	0	1	12	4000/4000000	-1	0
Z=300000		Zj	14000	24000	19000	160000000000000/16000000000000	4000	0
		Zj-Cj	0	0	1000	160000000000000/16000000000000	4000	0

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=6,x2=9,x3=0

Max Z=300000