Using lab 7 with a two-day lead time (all original values), if we increased the min and max by 5% each (from original values), the total cost would be . However, if we increased the min and max by 10% each (from original values), the total cost would be x. Lastly, if we increased the min and max by 15% each (from original values), the total cost would be x. Therefore, increasing the min and max by x would be the best scenario to minimize the overall cost, while increasing the min and max by x would be the worst scenario. Using lab 7 with a three-day lead time (all original values), we will analyze six different scenarios to determine which would be the best scenario based on the total cost. Scenario A - Min is 3,500 and Max is 5,500. All other values remain the same as lab 7 original values. The total cost would be Scenario B - Min is 3,500 and Max is 6,000. All other values remain the same as lab 7 original values. The total cost would be Scenario C - Min is 4,000 and Max is 5,500. All other values remain the same as lab 7 original values. The total cost would be Scenario D - Min is 4,000 and Max is 6,000 . All other values remain the same as lab 7 original values. The total cost would be Scenario E - Min is 4,500 and Max is 5,500. All other values remain the same as lab 7 original values. The total cost would be Scenario F. Min is 4,500 and Max is 6,000 . All other values remain the same as lab 7 original values. The total cost would be Based on the above scenarios, we know that scenario x would cost us the most, therefore we wouldn't want to use those min and max values, On the flip side, scenario * ylelds us the best combination of min and max resulting in the least cost