# STATS & OPERATIONS TASK Harvard Case Solution & Analysis

## STATS & OPERATIONS TASK Case Solution

Question 1

a).

The 3 months moving average forecast has been generated in the excel spreadsheet and the forecast for month 49 is 49,633.

b).

The MAD, MSE and MAPE for this model are as follows:

 MAD 1333.703704 MSE 81863642.68 MAPE 0.00086%

c).

The linear trend model has been generated in the excel spreadsheet and the forecast for month 49 is 39.338.

d).

The MAD, MSE and MAPE for this model are as follows:

 MAD -1161.77565 MSE 66165128.11 MAPE -0.00073%

e).

The exponential smoothing model has been generated in the excel spreadsheet based upon the optimal smoothing constant. This has been also calculated which is 0.54. At this smoothing constant, the MAD value is minimized. This means that at this smoothing constant, the distance between the mean data value and each value of the data set is minimized and thus this reduces the deviations in the data.

f).

Based on the exponential smoothing model the month 49 forecast value is 23,923.

g).

The MAD, MSE and MAPE for this model are as follows:

 MAD 763.180 MSE 28552154.69 MAPE 0.00048%

h).

Among all the three forecasting models, the most recommended forecast for DataNet is based upon the moving average is the exponential smoothing model due to lower MAD and MAPE.

Question 2

a).

The time to produce a single truck is 665 seconds or 11.08 minutes. This is the sum of all the times at each station.

b).

Bottleneck is insert chip with 90 seconds, which makes it the highest time among all the stations.

c).

The steady state capacity of the assembly line is:

60/11.08 = 5.41 toy trucks per hour.

d).

The utilization of the worker at station 2 is:

85/665 = 12.78%.

e).

The total time required to produce 100 trucks would be:

100x665sec = 66500 sec or 1103.33 minutes or 18.47 hours.

f).

Step 1: First of all, the time reduction would be done for the bottleneck station which is station 3. We would reduce 15 seconds of the time of this station for a total of \$1500. The remaining investment would be now 4500-1500 = 3000. The total trucks per hour produced at this step would be 3600/650= 5.53 trucks per hour.

Step 2: After this the next bottleneck would be at station 2, therefore we would reduce its time by 10 seconds with a total investment of \$ 600. The remaining funds would be now 3000-600=2400. The total trucks per hour produced at this step would be 3600/640= 5.625 trucks per hour.

Step 3: At this step we would now reduce the processing time for station 7 and 9 by 5 seconds which have the highest processing time. The total investment would be 400+300=700. The remaining funds would be now 2400-700= 1700. The total trucks per hour produced at this step would be 3600/630= 5.71 trucks per hour.

Step 4: In this last step we would now be reducing 5 seconds for the maximum stations possible with the highest processing time. This has been done for station 1, 2, 3 and 7. The total investment would be 400+300+500+400= 1600. The remaining funds would be 100. The total trucks per hour produced at this step would be 3600/610= 5.90 trucks per hour.

The total investment would be \$ 4400. ..................

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