Problem 1:
For this problem, we consider the weight for each attribute as given in the question, namely taste, pricing and location of the proposed restaurant. Through the total expected utility, one can easily determine the best type of restaurant that Patricia should open.
Weight for taste aspect = 3
Weight for price aspect = 2
Weight for location aspect = 1
Total Expected Utility = (PA*UA) + (PB * UB)… (Pn * Un)
Total utility expected for the steak restaurant = ( 80*3) + (55*1) + (65*2) = 425
Total utility expected for the pizza restaurant = ( 70*3) + (80*1) + (50*2) = 390
Patricia is best suited to open the steak restaurant in the Los Angeles suburban area as the total expected utility for opening the steak restaurant is higher than that of the pizza restaurant.
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To determine the best applicable restaurant among the picks that Patricia has, it is important to determine the total expected utility for each and choose the one with the highest value.
Weight for taste = 1
Weight for price = 2
Weight for location = 3
Total Expected Utility = (PA*UA) + (PB * UB)… (Pn * Un)
Total utility expected for the steak restaurant = ( 80*1) + (55*3) + (65*2) = 375
Total utility expected for the pizza restaurant = ( 70*1) + (8*3) + (50*2) = 410
Patricia is best suited to open a pizza restaurant in the LA suburban area since the pizza restaurant has a higher total expected utility than the steak restaurant.
In this case, we consider preset probabilities for finding suitable restaurant locations to determine the best choice.
The availability of the venue for both locations is considered alongside the total utility for the steak restaurant as follows:
(0.7*425) + (0.3*375) = 410
As for the pizza restaurant, the venue availability is also considered:
(0.7*390) + (0.3*410) = 396
When the two values are considered, Patricia would best open the steak restaurant since the expected value for venue availability is higher compared to that of the pizza restaurant.
Finally, a consideration of the application of this theory to the real world is explained:
The application of this scenario in the real world is possible where the values remain constant. For the application of this scenario in the real world, one must choose the option with the highest expected value as this option offers the best operational scenario. Nonetheless, there are challenges with applying this theory entirely as it conflicts with the intuition of people. For example, where there is an assurance of $1 million dollars and a 50% chance of $3 million, the person is more likely to choose the assured value for $1 million dollars although total expected value will point to the latter option.
Problem 2:
Considering:
Qx = thousands of donuts;
Px = price per donut;
Py = average price per donut for other brands;
Ax = thousands of dollars spent on Newton’s donuts advertising.
Price elasticity is given by:
Price elasticity of demand is given by (dQx/dPx )* (Px/Qx). Price elasticity, substituting the values as Qx =37.9; .
P.E. = - 54 * 0.95/37.9 = - 1.35.
Considering the above value, demand of the product will decrease by 1.35% if the price increases by 1% if all factors remain constant.
To derive the inverse demand curve;
The inverse demand function is given by Px = f(Qx)
Qx = -14 - 54Px + 45Py + 0.62 Ax
The demand function is, therefore: Px = 1/54( -Qx - 14 + 45 Py + 0.62 Ax)
Substituting Py=0.64 and Ax=120,
Px is given by 1/54 ( -Qx +89.2).
Since price elasticity is -1.35 a decrease in price by 1%, demand will see a 1.35% increase, thereby increasing profit. Nevertheless, this can only occur until the elasticity goes back to 1.
If the company should inject $1000 into advertising, demand will increase by 620 donuts. Profits obtained from those sales are as follows:
620 x ( 0.95 - 0.15) = $ 496
If this is compared to the $1000 investment, it is even less than half the value. As a result, it is not worth increasing investment in advertising, as it will bring no returns.
References
Grant, R. M. (2016). Contemporary strategy analysis: Text and cases edition. New York: John Wiley & Sons.
