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BernardToh

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- Homework Statement
- An interview is to be conducted and there are three family types that have been identified for the interview; 1-Child Family, 2-Child Family, and 3-Child Family.

At least 2200 families have to be interviewed for the entire interview exercise.

There are 3 interview timings; Weekdays Morning, Weekdays Evening, and Saturday Morning

The costs of the interviews are as follows:

Family Type Weekdays Morning Weekdays Evening Saturdays

One-Child $20 $22 $20

Two-child $23 $25 $23

Three-child $26 $28 $26

The total number of respondents during weekdays evening interviews must be at least equals to the total number of respondents during weekdays morning interviews.

The minimum number of respondents for each family types for weekdays interview are:

One child - 650

Two-child - 600

Three-child - 700

The combined total of two and three child households being interviewed on Saturday has to be at least 110.

Below is the allocation ratio for weekdays evening interviews to weekdays morning interviews per family type are:

One child - 55:45

Two child - 60:40

Three child - 60:40

Question - Do up a Linear Programming model based on the information given above to minimize the cost of the interview exercise.

- Relevant Equations
- Linear Programming

I've tried formulating the LP model for the question above and would like to check if I'm doing anything incorrectly.

Below is my LP model.

Let X1 be the number of one-child family interviewed on weekdays morning

Let X2 be the number of one-child family interviewed on weekdays evening

Let X3 be the number of one-child family interviewed on saturday

Let X4 be the number of two-child family interviewed on weekdays morning

Let X5 be the number of two-child family interviewed on weekdays evening

Let X6 be the number of two-child family interviewed on saturday

Let X7 be the number of three-child family interviewed on weekdays morning

Let X8 be the number of three-child family interviewed on weekdays evening

Let X9 be the number of three-child family interviewed on saturday

Let Z be the total cost of the interviews

Z = 20X1 + 22X2 + 20X3 + 23X4 + 25X5 + 23X6 + 26X7 + 28X8 + 26X9

Constraints

X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 =>2200

X1 + X2 => 650

X4 + X5 => 600

X7 + X8 => 700

X6 + X9 => 110

X2 + X5 + X8 => X1 + X4 + X7

X1 = 0.45(X1 + X2)

X2 = 0.55(X1 + X2)

X4 = 0.4(X4 + X5)

X5 = 0.6(X4 + X5)

X7 = 0.4(X7 + X8)

X8 = 0.6(X7 + X8)

Below is my LP model.

Let X1 be the number of one-child family interviewed on weekdays morning

Let X2 be the number of one-child family interviewed on weekdays evening

Let X3 be the number of one-child family interviewed on saturday

Let X4 be the number of two-child family interviewed on weekdays morning

Let X5 be the number of two-child family interviewed on weekdays evening

Let X6 be the number of two-child family interviewed on saturday

Let X7 be the number of three-child family interviewed on weekdays morning

Let X8 be the number of three-child family interviewed on weekdays evening

Let X9 be the number of three-child family interviewed on saturday

Let Z be the total cost of the interviews

Z = 20X1 + 22X2 + 20X3 + 23X4 + 25X5 + 23X6 + 26X7 + 28X8 + 26X9

Constraints

X1 + X2 + X3 + X4 + X5 + X6 + X7 + X8 + X9 =>2200

X1 + X2 => 650

X4 + X5 => 600

X7 + X8 => 700

X6 + X9 => 110

X2 + X5 + X8 => X1 + X4 + X7

X1 = 0.45(X1 + X2)

X2 = 0.55(X1 + X2)

X4 = 0.4(X4 + X5)

X5 = 0.6(X4 + X5)

X7 = 0.4(X7 + X8)

X8 = 0.6(X7 + X8)

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