Question

d) Cathy carefully examined the four attributes and narrowed down the most important attributes for her as the life span and price. She is willing to pay up to $950 for an extra year of life span. Calculate $K_L$ and $K_P$ and recommend which car she would select from A, B, or C? Why? Now you need to create a Hypothetical object, Lets say \"G\" Now select a base object, Lets pick \"A\" SO Price of G = 8500 + 950 = 9450 and Life Span = 6 +1 = 7 $U_L(7) = (7 - 6) / (16 - 6) = 0.10$, $U_P(9450) = (9450 - 21000) / (8500 - 21000) = 0.924$ Cathy is now indifferent between \"A\" and \"G\". $EU(A) = EU(G)$ $K_P = K_L(0.1) + K_P(0.924)$ $K_L = K_P(0.76)$ $K_P = 0.568$ $K_L = 0.432$ $EU(A) = (0 \times 0.432) + (1 \times 0.568) = 0.568$ $EU(B) = (0.8 \times 0.432) + (0.4 \times 0.568) = 0.573$ $EU(C) = (1 \times 0.432) + (0 \times 0.568) = 0.432$ Car \"B\" would be chosen because EU(B) > EU(A) and EU(B) > EU(C)

          d) Cathy carefully examined the four attributes and narrowed down the most important
attributes for her as the life span and price. She is willing to pay up to $950 for an extra
year of life span. Calculate $K_L$ and $K_P$ and recommend which car she would select from
A, B, or C? Why?
Now you need to create a Hypothetical object, Lets say \"G\"
Now select a base object, Lets pick \"A\"
SO Price of G = 8500 + 950 = 9450 and Life Span = 6 +1 = 7
$U_L(7) = (7 - 6) / (16 - 6) = 0.10$,
$U_P(9450) = (9450 - 21000) / (8500 - 21000) = 0.924$
Cathy is now indifferent between \"A\" and \"G\".
$EU(A) = EU(G)$
$K_P = K_L(0.1) + K_P(0.924)$
$K_L = K_P(0.76)$
$K_P = 0.568$
$K_L = 0.432$
$EU(A) = (0 \times 0.432) + (1 \times 0.568) = 0.568$
$EU(B) = (0.8 \times 0.432) + (0.4 \times 0.568) = 0.573$
$EU(C) = (1 \times 0.432) + (0 \times 0.568) = 0.432$
Car \"B\" would be chosen because EU(B) > EU(A) and EU(B) > EU(C)

d) Cathy carefully examined the four attributes and narrowed down the most important
attributes for her as the life span and price. She is willing to pay up to 950 for an extra
year of life span. CalculateKLandKPand recommend which car she would select from
A, B, or C? Why?
Now you need to create a Hypothetical object, Lets say G̈N̈ow select a base object, Lets pick ÄS̈O Price of G = 8500 + 950 = 9450 and Life Span = 6 +1 = 7UL(7) = (7 - 6) / (16 - 6) = 0.10,UP(9450) = (9450 - 21000) / (8500 - 21000) = 0.924Cathy is now indifferent between Äänd G̈.̈EU(A) = EU(G)KP = KL(0.1) + KP(0.924)KL = KP(0.76)KP = 0.568KL = 0.432EU(A) = (0 ×0.432) + (1 ×0.568) = 0.568EU(B) = (0.8 ×0.432) + (0.4 ×0.568) = 0.573EU(C) = (1 ×0.432) + (0 ×0.568) = 0.432Car B̈ẅould be chosen because EU(B) > EU(A) and EU(B) > EU(C)

Added by Virginia C.

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