STAT3613 Tutorial 02

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The University of Hong Kong *

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3613

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Statistics

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Nov 24, 2024

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT2313/3613 Marketing Engineering Tutorial 2 Objectives: 1. Learn more about some response models (ADBUDG model and Logistic model) 2. Optimizing the profit of a company by SAS, based on the collected data 1 Review 1.1 ADBUDG model and Logistic model 1. ADBUDG Model Y = a + b X c X c + d a = minimum value of Y b = maximum value of Y - minimum value of Y Suppose Y = 1 when X = 1 (normalized data) and Y 0 = (slope at X = 1 ), then d = b 1 - a - 1 , c = y 0 ( d + 1) 2 b × d 2. Logistic Model Y = a 1 + exp( - ( b + cX )) + d a = maximum value of Y - minimum value of Y b = - ( the value of X at the point of inflection ) × c c = 4 × ( slope at the point of inflection ) /a d = minimum value of Y 1.2 Optimization of profit 1. Short-run profit profit = ( unit price - unit variable cost ) × sales volume - relevant costs = unit margin × quantity - relevant costs . 2. Long-run profit Present value of the profit stream PV = Z 0 + Z 1 r + Z 2 r 2 + Z 3 r 3 + . . . Z i is the profit at period i , r = 1 / (1 + d ) with d being the discount rate. 3. Multiple goals More than one objectives, there may be conflict. Therefore, optimize one (the most important) objective and to make all the others constraints. 1
2 Examples 2.1 Examples 1. Consider an ADBUDG model given as Y = a + b X c X c + d given that c > 0 and X 0 , show that (a) a = minimum value of Y (b) b = maximum value of Y - minimum value of Y (c) Suppose Y = 1 when X = 1 (normalized data) and Y 0 = ( slope at X = 1) , then d = b 1 - a - 1 , c = y 0 ( d + 1) 2 b × d Solutions Since 0 X c X c + d 1 a a + b X c X c + d a + b, therefore, min( Y ) = a, max( Y ) = a + b = b = max Y - min Y. Suppose Y = 1 when X = 1 , substitute Y = 1 and X = 1 , 1 = a + b (1) c (1) c + d d = b 1 - a - 1 . Furthermore, we can rewrite Y = a + b 1 1 + dX - c . Hence, Y 0 = b d (1 + dX - c ) - 1 dX = b d (1 + dX - c ) - 1 d (1 + dX - c ) d (1 + dX - c ) dX = - b (1 + dX - c ) - 2 · ( - cdX - c - 1 ) = bcd X c +1 (1 + dX - c ) 2 . Suppose Y 0 = y 0 when X = 1 , y 0 = bcd (1) c +1 (1 + d (1) - c ) 2 = c = y 0 ( d + 1) 2 bd . 2. Consider a logistic model given as Y = a 1 + exp( - ( b + cX )) + d given that c > 0 , show that (a) a = maximum value of Y - minimum value of Y 2
(b) b = - ( the value of X at the point of inflection ) × c (c) c = 4 × ( slope at the point of inflection ) /a (d) d = minimum value of Y Solutions Since 0 1 1 + e - ( b + cX ) 1 d a 1 + e - ( b + cX ) + d a + d, therefore, min( Y ) = d, max( Y ) = a + d = a = max Y - min Y, d = min( Y ) . At the point of infection, Y 00 = 0 , Y 0 = ace - ( b + cX ) (1 + e - ( b + cX ) ) - 2 Y 00 = ac 2 h 2 e - 2( b + cX ) (1 + e - ( b + cX ) ) - 3 - e - ( b + cX ) (1 + e - ( b + cX ) ) - 2 i = ac 2 e - ( b + cX ) (1 + e - ( b + cX ) ) - 2 h 2 e - ( b + cX ) (1 + e - ( b + cX ) ) - 1 - 1 i = ac 2 e - ( b + cX ) (1 + e - ( b + cX ) ) - 2 2 e b + cX + 1 - 1 At the point of infection, ( X = ˜ x , Y 0 = ˜ y 0 ), 0 = 2 e b + c ˜ x + 1 - 1 , = e b + c ˜ x = 1 . Thus, b + c ˜ x = 0 = b = - c ˜ x, and ˜ y 0 = ac (1) - 1 ( 1 + (1) - 1 ) - 2 = ac 4 = c = y 0 a Logistic Function: 3. For electronic company B, salesforce is to be allocated to 3 different products, the digital camera (DC), the mobile phone (MP) and the laptop (LP). The current sales for DC, MP and laptop are 5000, 10000 and 4000 respectively. The current salespeople allocated to DC are 25 and that for MP are 35, while that for laptop are 20. The average cost for each salesperson is $40. The margin of DC is $0.8 and that of MP and laptop are $0.7 and $0.9. Historical data of the sales (in %) for various numbers of salespeople (in %) are shown in the table. 3
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