Question

In a marketing study, the researcher wants to minimize the cost of interviewing. These interviews will be done to households with children, without children, and to individuals during the day and during the night. The cost for each type of interview is presented in the following table:

Category	Cost per day	Cost per night
With children	$1500	$1900
Without children	$1200	$1550
Individual	$1100	$1250

The sample size is 800 people.

At least there must be 400 households with children.

At least there must be 200 homes without children.

There must be at least 100 individuals.

The night interviews are greater or equal to the interviews during the day.

In the group of households with children, 50% of those interviewed during the night are greater than 50% of the group of respondents during the day.

In the group of homes without children, 40% of those interviewed during the night is greater than 60% of the group interviewed during the day.

In the group of individuals, 30% of those interviewed during the night is greater than the 70% of the group interviewed during the day.

Using linear programming (Using RStudio)

1. Find the minimum expected cost.

2. Find the cost for each group day and night.

3. Find the number of people interviewed by each group.

4. Check if the restrictions are met.

[Note: In your answer you must submit the "script" of the RStudio used for your answer]

Answer 1

Given Table

1. Minimum Expected Cost

Let Total number of Households with children = Hc

Total number of Households without children = Hwc

Total number of individual households = Hi

At Night

People interviewed in night (Hc, Hwc, hi) denoted by (Inc, Inwc, Ini) respectively

number of Households with children People interviewed in night = Inc

number of Households without children People interviewed in night = Inwc

number of individual households People interviewed in night = Ini

At day

People interviewed during the day (Hc, Hwc, Hi) denoted by (idc, Idwc, Id) respectively.

number of Households with children People interviewed in day = Idc

number of Households without children People interviewed in day = Idwc

number of individual households People interviewed in day = Idi

We know, Hc + Hwc + Hi = 800;

Inc + Idc = Hc Inwc + Idwc = Hwc; Ini + Idl = Hi

Given :

Hc >= 400,

Hwc >= 200,

Hi >=100 and ;

Inc > Idc ; Inwc > 3/2 (Idwc); Ini > (7/3) ldi

Total Cost = Tc = (InC*1900 + Idc*1500) + (InwC*1550 + Idwc *1200) + (Ini*1250 + Idi*1100)

Using all equations above we reach the expression;

Therefore Minimum cost = 1082500

Inc = Idc = Hc/2 = 400/2 = 200;

Inwc = 3/5'w Hwc =120

Idwc = 2/3 'Inwc = 80;

Ini = 7/10 = 70;

Idi = 30

Therefore Minimum cost = 1082500

Now, Hc (min) = 400; Hwc (min) = 200;

2.Cost of each group

Total Cost in the Night (households with children)

= InC1900

= 200*1900

=380000

Total Cost in day (households with children)

= IdC1500

=200*1500

= 300000

Total Cost in the Night (households without children)

= InwC1550

=120*1550

=186000

Total cost in day (households without children)

= IdwC1300

=80*1200

= 96000

Total Cost in the Night (individual households)

=lni1250

=70*1250

= 87500

Total Cost in the day (individual households)

= Idi*1100

= 30*1100

=33000

3.Number of people interviewed by each group

Night = 380000 + 186000 + 87500 = 6535000

Day = 300000 +96000 + 33000      = 429000

4.Restriction are met

Yes, all restrictions have been met

Note : unable to send code if you hacve any doubts comment me