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Multiobjective Value AnalysisPowerPoint Presentation

Multiobjective Value Analysis

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Multiobjective Value Analysis

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Multiobjective Value Analysis

- A procedure for ranking alternatives and selecting the most preferred
- Appropriate for multiple conflicting objectives and no uncertainty about the outcome of each alternative.

- Specify decision alternatives and objectives
- Evaluate objectives for each alternative

A prospective home buyer has visited four open houses in Medfield over the weekend. Some details on the four houses are presented in the following table.

- Determine a value function which combines the multiple objectives into a single measure of the overall value of each alternative.
- The simplest form of this function is a simple weighted sum of functions over each individual objective.

Estimating the single objective value functions

- Price - price ranges from roughly $300,000 to $600,000 dollars with lower amounts being preferred.
- Suppose that a decrease in price from $600,000 to $450,000 will increase value by the same amount as would a decrease in price from $450,000 to $300,000.

- This implies that over the range $300,000 to $600,00 the value function for price is linear and the value for each price alternative can be found by linear interpolation.
- First set v1(389,900)=1 and v1(599,000)=0.
- Then

- Number of bedrooms - the number of bedrooms for the four alternatives is 3, 4 or 5 with more bedrooms preferred to fewer.
- Thus v2(5)=1 and v2(3)=0.
- Suppose the increase in value in going from 3 to 4 bedrooms is twice the increase in value in going from 4 to 5 bedrooms.

- Then if the value increase in going from 4 to 5 bedrooms is x, the value increase in going from 3 bedrooms to 4 is 2x.
- And since the value increase in going from 3 bedrooms to 5 is 1, 2x+x=1.
- Thus x=1/3 and finally the v2(4)=0+2(1/3) =.67

- Number of bathrooms - The number of bathrooms for the four alternatives are 1.5, 2, 2.5, and 3 with more bathrooms being preferred to fewer bathrooms.
- Thus v3(3)=1 and v3(1.5)=0.
- Suppose that the increase in value in going from 1.5 to 2 bathrooms is small and about equal to the increase in value in going from 2.5 to 3 bathrooms. The increase in value in going from 2 to 2.5 bathrooms is more significant and is about twice this value.

- Then, the value increase in going from 1.5 to 2 bathrooms is x. The value increase in going from 2 to 2.5 bathrooms is 2x. And the value increase in going from 2.5 to 3 bathrooms is also x.
- The sum of the value increases x+2x+x=1 and x=1/4.
- So, v3(2)=0+x=0+1/4=.25, and v3(2.5)=0+x+2x=0+1/4+2/4=.75

- Style - there are three house styles available: Ranch, Colonial and Garrison Colonial.
- Suppose that Colonial, is most preferred, Ranch is least preferred and the value of Garrison Colonial is about mid-value.
- Then v4(Colonial)=1, v4(Garrison Colonial)=.5 and v4(Ranch)=0

Determine the weights

- Consider the value increase that would result from swinging each alternative (one at a time) from its worst value to its best value (e.g.. the value increase from swinging price from $599,000 to $389,900).
- Determine which swing results in the largest value increase, the next largest, etc..

- Suppose going from a Ranch to a Colonial results in the largest value increase, going from 3 to 5 bedrooms the second largest, going from 1.5 bathrooms to 3 bathrooms the next largest and swinging price from $599,000 to $389,900 results in the smallest value increase.

- Set the smallest value increase equal to w and set each other value increase as a multiple of w.
- Suppose the bathroom swing is twice as valuable as the price swing, the style swing is 3 times as valuable as the price swing and the bedroom swing falls about half way in between these two.

- Since the single objective value functions are scaled from 0 to 1 the weight for any objective is equal to its value increase for swinging from worst to best.
- And because we would like the multiobjective value function to be scaled from 0 to 1, the weights should sum to 1.

Determine the overall value of each alternative

Compute the weighted sum of the single objective values for each alternative.

- Rank the alternatives from high to low.

- The weighted sums provide a ranking of the alternatives. The most preferred alternative has the highest sum.
- The “ideal“ alternative would have a value of 1. The value for any alternative tell us how close it is to the theoretical ideal.