Question

Use Huff’s law to compute the probability of consumers traveling from their homes to each of three shopping areas: Square footage of selling space: Location 1 = 15,000 Location 2 = 20,000 Location 3 = 25,000 Travel time: To location 1 = 12 minutes To location 2 = 18 minutes To location 3 = 25 minutes Effect of travel time on shopping trip is 2. For full credit you need to show your full equation AND explain how you used the equation to get your answers.

Answer 1

Answer:

the probability of the consumer’s shopping in location 1 is 50.6 %

the probability of the consumer’s shopping in location 2 is 29.9 %

the probability of the consumer’s shopping in location 3 is 19.4 %

Explanation:

Huff’s law is a mathematical model that takes consideration in the relation  between the patronage and distance from location of the shopping area.

The equation for this mathematical model can be expressed as :

$P_y = \dfrac{\dfrac{ S_j}{(T_y)^{\lambda} } }{ \sum \limits ^{n}_{f} \dfrac{S_j}{(T_y)^{\lambda }}}$

Where;

$P_y=$Probability of a consumer travelling from home (i) to shopping location (j)

$T_y$ = Travel time from consumer’s home (i)  to shopping location (j)

$\lambda$ = Dataset used to determine the effect of travel time in different kinds of shopping trips

n = Number of different shopping location.

NOW; from the given information.

for location 1 , the consumer shopping probability is :

$P_{i, 1} = \dfrac{ \dfrac{15000}{(12)^2} }{ \dfrac{15000}{(12)^2} + \dfrac{20000}{(18)^2} + \dfrac{25000}{(25)^2}}$

$P_{i, 1} = \dfrac{ \dfrac{15000}{144} }{ \dfrac{15000}{144} + \dfrac{20000}{324} + \dfrac{25000}{625} }$

$P_{i, 1} = \dfrac{104.17 }{ 104.17 + 61.73 +40.00 }$

$P_{i, 1} = \dfrac{104.17 }{ 205.9 }$

$\mathbf{P_{i,1} = 0.506 \ or \ 50.6 \% }$

Thus; the probability of the consumer’s shopping in location 1 is 50.6 %

for location 2 , the consumer shopping probability is :

$P_{i, 2} = \dfrac{ \dfrac{20000}{(18)^2} }{ \dfrac{15000}{(12)^2} + \dfrac{20000}{(18)^2} + \dfrac{25000}{(25)^2} }$

$P_{i, 2} = \dfrac{ \dfrac{20000}{324} }{ \dfrac{15000}{144} + \dfrac{20000}{324} + \dfrac{25000}{625} }$

$P_{i, 2} = \dfrac{61.73 }{ 104.17 + 61.73 +40.00 }$

$P_{i, 2} = \dfrac{61.73 }{ 205.9 }$

$\mathbf{P_{I,2} = 0.299 \ or \ 29.9 \%}$

Thus; the probability of the consumer’s shopping in location 2 is 29.9 %

for location 3 , the consumer shopping probability is :

$P_{i, 3} = \dfrac{ \dfrac{25000}{(25)^2} }{ \dfrac{15000}{(12)^2} + \dfrac{20000}{(18)^2} + \dfrac{25000}{(25)^2} }$

$P_{i, 3} = \dfrac{ \dfrac{25000}{625} }{ \dfrac{15000}{144} + \dfrac{20000}{324} + \dfrac{25000}{625} }$

$P_{i, 3} = \dfrac{40.00 }{ 104.17 + 61.73 +40.00 }$

$P_{i, 3} = \dfrac{40.00}{ 205.9 }$

$\mathbf{P_{i,3} = 0.194 \ or \ 19.4 \%}$

Thus; the probability of the consumer’s shopping in location 3 is 19.4 %