Operations Management Group 2 - Module 2 HW

1. A commercial bakery has recorded sales (in dozen) for three products, shows as follows DAY BLUEBERRY MUFFINS CINNAMON BUNS CUPCAKES 1 30 18 45 2 34 17 26 3 32 19 23 4 34 19 23 5 35 22 22 6 30 23 48 7 34 23 29 8 36 25 20 9 29 24 14 10 31 26 18 11 35 27 47 12 31 28 26 13 37 29 27 14 24 31 24 15 33 33 22 A. Predict orders for the following day for each of the products using an appropriate naive method. Hint: Plot each data set DAY BLUEBERRY MUFFINS (STABLE) CINNAMON BUNS (TREND) CUPCAKES (SEASONAL) 16 33 35 47 Stable = Last value Trend = 33 + (33-31) = 35 Seasonal = Value of the last season forecast B. What should the use of sales data instead of demand imply? Sales data implies instead of demand because the sales would reflect the demand. Using the previous period, it estimates the next period's forecast. Assume that it will exhaust the production; demand equals sales if there is no shortage. 2. National Scan, Inc., sells radio frequency inventory tags. Monthly sales for a seventh-month period were as follows: 3. MONTH SALES (000 UNITS) FEBRUARY 19 MARCH 18 APRIL 15 MAY 20 JUNE 18 JULY 22 AUGUST 20 4. A. Plot the monthly data on a sheet of graph paper B. Forecast September sales volume using each of the following: a. The native approach MONTH SALES (000 UNITS) FEBRUARY 19 MARCH 18 19 APRIL 15 18 MAY 20 15 JUNE 18 20 JULY 22 18 AUGUST 20 22 SEPTEMBER b. A five month moving average NAIVE 20 FIVE MONTH MOVING AVE. MONTH SALES (000 UNITS) FEBRUARY 19 MARCH 18 APRIL 15 MAY 20 JUNE 18 JULY 22 18 AUGUST 20 18.6 SEPTEMBER 𝑀𝐴5 = 19 19+18+15+20+18 18 𝑀𝐴5 = 18 𝑀𝐴5 = 18=15+20+18+22 5 𝑀𝐴5 = 18. 6 𝑀𝐴5 = 15+20+18+22+20 5 𝑀𝐴5 = 19 c. A weighted average using .60 for August, .30 for July and .10 for june MONTH WEIGHTED AVE AUG 0.6 JUL 0.3 JUNE 0.1 = 0. 6(20) + 0. 3(22) + 0. 1(18) = 20. 4 d. Exponential smoothing with a smoothing constant equal to .20, assuming a March forecast of 19 (000) MONTH SALES (000 UNITS) FEBRUARY 19 EXPONENTIAL SMOOTHING MARCH 18 19 APRIL 15 18.8 MAY 20 18.04 JUNE 18 18.43 JULY 22 18.34 AUGUST 20 29.07 SEPTEMBER 19.26 e. A linear trend equation 2 MONTH SALES (000 UNITS) 𝑥𝑦 FEBRUARY 19 19 1 MARCH 18 36 4 APRIL 15 45 9 MAY 20 80 16 JUNE 18 90 25 JULY 22 132 36 AUGUST 20 140 49 𝑥 Slope 𝑚 = 𝑚 = 𝑛Σ𝑥𝑦−(Σ𝑥)(Σ𝑦) 2 2 𝑛Σ𝑥 −(Σ𝑥) 7(542)−28(132) 7(140)−28) 2 𝑚 = 0. 50 Intercept 𝑐 = 𝑐 = Σ𝑦−𝑚Σ𝑥 𝑛 132−(0.50)(28) 7 𝑐 = 16. 86 C. Which method seems least appropriate? Why? (Hint: Refer to your plot from part a.) After analyzing the plot, I've come to the conclusion that Naive Forecasting reflects the least appropriate forecast since it only uses the previous values to predict the next sales. This could rapidly decrease its accuracy considering the values used came from limited data, though it is the easiest way to predict a forecast. Althemore, the graph shows predicted increases and decreases in sales. D. What does the term sales rather than demand presume? When sales data is used instead of demand data, it is assumed that sales accurately reflect demand, because demand equals sales if there are no shortages, we are presuming that no stock-outs occurred. 5. An electrical contractor’s records during the last five weeks indicate the number of job requests: 6. . WEEK REQUESTS 1 2 3 4 5 20 22 18 21 22 7. 8. Predict the no. of requests for week 6 using each of the following methods: a. Naive WEEK JOB REQUEST FORECAST 1 20 2 22 20 3 18 22 4 21 18 5 21 6 22 b. Four-period moving average WEEK JOB REQUEST 1 20 2 22 3 18 4 21 5 22 4 MOVING AVE. FORECAST 20.25 6 20.75 c. Exponential Smoothing with α = 30; use 20 for week 2 forecast QUARTER JOB REQUESTS α = 30 FORECAST 1 20 2 20 3 18 20 4 21 19.4 5 22 19.88 20 6 20.52 9. . 10. 11. .. 12. . 13. Compose seasonal Relatives for this data using the SA method QUARTER 1 2 3 1 2 3 7 4 2 6 10 18 14 3 2 6 8 8 4 5 9 15 11 4 QUARTER QUARTER TOTAL QUARTER AVERAGE QUARTER RELATIVES 1 16 4 0.5 2 48 12 1.5 3 24 6 0.75 4 40 10 1.25 32 4 OVERALL AVERAGE = 32/4 = 8 14. A tourist center is open on weekends (Friday, Saturday, and Sunday). The owner-manager hopes to improve scheduling of part-time employees by determining seasonal relatives for each of these days. Data on recent traffic at the center have been tabulated and are shown in the following table: 15. WEEK 1 2 3 4 5 6 FRIDAY 149 154 152 150 159 163 SATURDAY 250 255 260 268 273 276 SUNDAY 166 162 171 173 176 183 a. a. Develop seasonal relatives for the shop using the centered moving average method. SEASON AVERAGE INDEX STANDARD INDEX FRIDAY 0.7891 0.7887 SATURDAY 1.3463 1.3456 SUNDAY 0.8662 0.8658 SEASONAL INDEX AVERAGE FRIDAY 0.79 SATURDAY 1.34 SUNDAY 0.87 b. Develop seasonal relatives for the shop using the SA method (see Example 8B) WEEK 1 2 3 4 5 FRIDAY 149 154 152 150 159 163 927 154.5 0.79 SATUR DAY 250 255 260 268 273 276 1582 236.67 1.34 SUNDAY 166 162 171 173 176 183 1031 171.83 0.87 TOTAL O.A 6 Q.T Q.A Q.R 590/3 196.67 c. Explain why the results of the two methods correlate the way they do. Both methods correlate with each other’s results because, in both methods they find the range of the mean of every season in order to get the seasonal relative. In using the centered moving method, the moving average is moved to the middle of the periods in order to find the range, while the simple average method finds the average of every season in order to get the range. 16. .Obtain estimates of daily relatives for the number of customers at a restaurant for the evening meal, given the following data. a. Use the centered moving average method. (Hint: Use a seven-day moving average.) b. b. Use the SA method. 21. A. Compute MAD for each set of forecasts. Given your results, which forecast appears to be more accurate? Explain. With the given data above, Forecast 2 Mean Absolute Deviation (MAD) is more accurate than the MAD for Forecast 1 as it is stated that the lower the value, the more accurate it is. The data above showed us that the MAD of forecast 1 is 3 and Forecast 2 with 4. 𝐹𝑂𝑅𝐸𝐶𝐴𝑆𝑇 1 = = 𝐹𝑂𝑅𝐸𝐶𝐴𝑆𝑇 2 = = 32/8 4 24/8 3 B. Compute the MSE for each set of forecasts. Given your results, which forecast appears to be more accurate? In terms of MSE, Forecast 2 is more accurate than Forecast 1 as it has a lower value than Forecast 1 with a MSE of 25.14 𝐹𝑂𝑅𝐸𝐶𝐴𝑆𝑇 1 = 176/7 = 25. 14 𝐹𝑂𝑅𝐸𝐶𝐴𝑆𝑇 2 = 106/7 = 15. 14 C. In practice, either MAD or MSE would be employed to compute forecast errors. What factors might lead a manager to choose one rather than the other? Forecast errors can be calculated by utilizing both the Mean Absolute Deviation (MAD) and the Mean Standard Deviation (MSE). The mean absolute deviation (MAD) can operate in calculating the average forecast error since a value of 0 indicates that the forecast is accurate to the highest level. Moreover, the Mean Standard Deviation (MSE) is also helpful as it categorizes the forecasting errors. The fact that one of the methodologies has already been used to calculate forecast errors is one of the variables that can drive a manager to prefer one over the other. Then there's the possibility that one of the users of both ways is already familiar with one of them. A manager's decision may also be influenced by the fact that they compare the two. D. Compute MAPE for each data set. Which forecast appears to be more accurate? With the help of the data above, The 0.26% MAPE of Forecast 2 is more accurate than the Forecast 1 MAPE since it has a lower value. 𝑀𝐴𝑃𝐸 𝐹𝑂𝑅𝐸𝐶𝐴𝑆𝑇 1 = 3. 53 % 𝑀𝐴𝑃𝐸 𝐹𝑂𝑅𝐸𝐶𝐴𝑆𝑇 2 = 0. 26 % 22. The manager of a seafood restaurant was asked to establish a pricing policy on lobster dinners. Experimenting with prices produced the following data: a. Plot the data and a regression line on the same graph Regression Line (ŷ = 19.51x + 315.98) b. Determine the correlation coefficient and interpret it Correlation coefficient = -0.9849 Interpretation: The price of the lobster dinners has a very high negative correlation with the average number sold per day. This means that as the price increases, the more the average number sold per day decreases in turn. Formula: 28. The manager of a travel agency has been using a seasonally adjusted forecast to predict demand for packaged tours. The actual and predicted values are as follows: a. Compute MAD for the fifth period, then update it period by period using exponential smoothing with α =. 3 FORECAST ABSOLUTE ERROR ERROR MAD SMOOTHED MAD PERIOD DEMAND PREDICTED 1 129 124 5 5 5.00 2 194 200 -6 6 5.50 3 156 150 6 6 5.67 4 91 94 -3 3 5.00 5 85 80 5 5 5.00 5.00 6 132 140 -8 8 5.50 5.90 7 126 128 -2 2 5.00 4.73 8 126 124 2 2 4.63 3.91 9 95 100 -5 5 4.67 4.23 10 149 150 -1 1 4.30 3.27 11 98 94 4 4 4.27 3.49 12 85 80 5 5 4.33 3.94 13 137 140 -3 3 4.23 3.66 14 135 128 7 7 4.36 4.36 b. Compute a tracking signal for periods 5 through 14 using the initial and updated MADs. If limits of ± 4 are used, what can you conclude? PERIOD FORECA ABSOLU TE ERROR DEMAN PREDICT ST D ED ERROR MAD SMOOTH ED MAD CUMULA TRACKING TIVE SIGNAL ERROR 1 129 124 5 5 5.00 5 2 194 200 -6 6 5.50 -1 3 156 150 6 6 5.67 5 4 91 94 -3 3 5.00 5 85 80 5 5 5.00 5.00 7 1.40 6 132 140 -8 8 5.50 5.90 -1 -0.17 7 126 128 -2 2 5.00 4.73 -3 -0.63 8 126 124 2 2 4.63 3.91 -1 -0.26 9 95 100 -5 5 4.67 4.23 -6 -1.42 10 149 150 -1 1 4.30 3.27 -7 -2.14 11 98 94 4 4 4.27 3.49 -3 -8.86 12 85 80 5 5 4.33 3.94 2 0.51 13 137 140 -3 3 4.23 3.66 -1 -2.27 14 135 128 7 7 4.36 4.36 5 1.15 2 The data presented above shows that the tracking signal range of the forecast lies from -2.14 to 1.40. With this, it can be concluded that bias did not occur and cannot be found in the data due to the tracking data staging within the acceptable limit of ± 4