STAT3613 Tutorial 02

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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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% of current sales % of current salesforce Digital camera Mobile phone Laptop 0.0 0.50 0.18 0.15 0.5 0.62 0.44 0.40 1.0 1.00 1.00 1.00 1.5 1.25 1.22 1.18 2.0 1.35 1.28 1.22 2.5 1.40 1.29 1.24 3.0 1.48 1.36 1.25 (a) Calibrate ADBUDG models to predict the sales for DC, MP and laptop respectively. (b) Set up functions to calculate the total net profit. (c) Find the optimal salespeople allocation to maximize the total net profit. (d) If the total number of salespeople is restricted by 110, find the optimal allocation. Solutions (a) The parameters of three models are estimated by SAS " proc model ", DC MP LP a 0.4949 0.1789 0.1498 b 1.0117 0.1694 1.1033 c 2.6557 3.0003 3.5111 d 1.0334 0.4300 0.2986 R2 0.9982 0.9987 1.0000 (b) For each product, Net profit = Sales volume × Unit Margin - Cost of salesperson. Net profit (DC) = 5000 × % of current sales (DC) × 0 . 8 - 25 × % current salesforce (DC) × 40 , Net profit (MP) = 10000 × % of current sales (MP) × 0 . 7 - 35 × % current salesforce (MP) × 40 , Net profit (LP) = 4000 × % of current sales (LP) × 0 . 9 - 20 × % current salesforce (LP) × 40 , The total net profit is Net profit (DC) + Net profit (MP) + Net profit (LP). (c) The optimization is calculated by SAS " proc nlp ", and the result is given below: DC MP LP % of salesforce 1.7115 1.5725 1.3845 Salesforce 42.7870 55.0391 27.6907 % of sales 1.3056 1.2319 1.1572 Sales 6528.1994 12319.4779 4628.6540 Gross profit 5222.5595 8623.6345 4165.7886 Net profit 3511.0812 6422.0691 3058.1587 Total net profit 12991.3089 For maximum total net profit, the number of salespeople is 43 for digital camera, 55 for mobile phone and 28 for laptop. (d) The optimization result under constraint is: 4
DC MP LP % of salesforce 1.4506 1.3942 1.2468 Salesforce 36.2657 48.7977 24.9365 % of sales 1.2256 1.1882 1.1197 Sales 6127.8538 11882.4754 4478.6151 Gross profit 4902.2831 8317.7328 4030.7536 Net profit 3451.6538 6365.8240 3033.2917 Total net profit 12850.7694 For maximum total net profit, the number of salespeople is 36 for digital camera, 49 for mobile phone and 25 for laptop. 5