$10 1 Sanotronics Simulation with 100 Trials 2 3 Data 4. Selling Price per Unit $249 5 Adm/Adv Cost $1 000 000 6 7 Direct Labor Cost Parts Cost (Normal Distribution) 8 Mean $90 Lower End Upper End of 9 Probability of Interval Interval Cost per Unit Standrard Deviation 10 0,1 0,0 0,1 $43 11 0,21 0,1 0,3 $44 12 0,4 0,3 0.7 $45 Demand (Uniform Distribution) 13 0,2 0,7 0.9 $46 Lower Bound 7500 14 0,1 0.9 1,0 $47 Upper Bound 30000 15 Expected Direct Labor Cost = $ 45,00 16 17 Simulation Trials 18 19 (1) (2) (3) (4) (5) (6) (7) Radom Direct Labor Radom First-Year 20 Number Cost per Unit Radom Number Parts Cost per Unit Number Demand 21 1 0,385 0,909 0,5201 22 2 0,528 0,195 0,049 23 3 0,037 0,574 0,963 24 4 0,587 0,028 0,554 25 5 0,087 0,848 0,782 115 95 0.572 0,756 0,066 116 96 0,098 0,753 0,545 117 97 0,024 0,822 0,562 118 98 0,392 0,703 0,335 119 99 0,670 0,560 0,507 120 100 0,381 0,137 0,831 121 Average This is the Sanotronics simulation with 100 trials that was covered in the lecture. Here, we want to illustrate how the results can change when different distributions are used for some of the random variables. Suppose the following changes occur in the assumptions. - Parts cost follows a normal distribution with mean of $90 and standard deviation of $10. - Demand now follows a uniform distribution between 7,500 and 30,000 units. The appropriate changes have been made in the Excel sheet. Do not change the given random numbers. Round all answers up to 2 decimal places. 1) What is the direct labor cost per unit for trial 1? 2) What is the parts cost per unit for trial 1? 3) What is the demand for trial 1? 4) What is the mean profit? 5) What is the maximum profit? 5) What is the probability of loss (in %)? 6) What is the value at risk (rounded to whole $)? Remember to give the absolute value. 7) Is this more or less risky than the original example solved in class? (8) Trial Profit $10 1 Sanotronics Simulation with 100 Trials 2 3 Data 4. Selling Price per Unit $249 5 Adm/Adv Cost $1 000 000 6 7 Direct Labor Cost Parts Cost (Normal Distribution) 8 Mean $90 Lower End Upper End of 9 Probability of Interval Interval Cost per Unit Standrard Deviation 10 0,1 0,0 0,1 $43 11 0,21 0,1 0,3 $44 12 0,4 0,3 0.7 $45 Demand (Uniform Distribution) 13 0,2 0,7 0.9 $46 Lower Bound 7500 14 0,1 0.9 1,0 $47 Upper Bound 30000 15 Expected Direct Labor Cost = $ 45,00 16 17 Simulation Trials 18 19 (1) (2) (3) (4) (5) (6) (7) Radom Direct Labor Radom First-Year 20 Number Cost per Unit Radom Number Parts Cost per Unit Number Demand 21 1 0,385 0,909 0,5201 22 2 0,528 0,195 0,049 23 3 0,037 0,574 0,963 24 4 0,587 0,028 0,554 25 5 0,087 0,848 0,782 115 95 0.572 0,756 0,066 116 96 0,098 0,753 0,545 117 97 0,024 0,822 0,562 118 98 0,392 0,703 0,335 119 99 0,670 0,560 0,507 120 100 0,381 0,137 0,831 121 Average This is the Sanotronics simulation with 100 trials that was covered in the lecture. Here, we want to illustrate how the results can change when different distributions are used for some of the random variables. Suppose the following changes occur in the assumptions. - Parts cost follows a normal distribution with mean of $90 and standard deviation of $10. - Demand now follows a uniform distribution between 7,500 and 30,000 units. The appropriate changes have been made in the Excel sheet. Do not change the given random numbers. Round all answers up to 2 decimal places. 1) What is the direct labor cost per unit for trial 1? 2) What is the parts cost per unit for trial 1? 3) What is the demand for trial 1? 4) What is the mean profit? 5) What is the maximum profit? 5) What is the probability of loss (in %)? 6) What is the value at risk (rounded to whole $)? Remember to give the absolute value. 7) Is this more or less risky than the original example solved in class? (8) Trial Profit
