Problems

1. The grinding machine occasionally breaks down and needs such services as replacement of a worn part or lubrication or resetting the speed of the grinding wheels. These break downs do not include the replacement of the grinding wheels, which is accomplished routinely by the operator. Here is the frequency of breakdowns observed over a period of 40 months. On the average it costs $200 to service the grinder. Routine maintenance, performed monthly, would cost $75.00 and would virtually eliminate breakdowns.

1. What is the mean number of breakdowns per month?
2. What is the average cost of repairing breakdowns per month?
3. Should the company repair the breakdowns or perform preventive maintenance?

2. (One step beyond) The robot welder is a temperamental machine and is prone to breaking down and going out of adjustment. The table contains records of the service calls for a 12 month period. (J = January, etc.) Preventive maintenance, performed monthly, would cost $750 each time, which would virtually eliminate breakdowns.

1. Construct the frequency distribution for calls per month.
2. What is the mean number of calls per month?
3. What is the mean cost per call?
4. Should the company service the welder when it breaks down or perform preventive maintenance?

3. (Similar to Example #S-2 in your textbook) The time between breakdowns of the stamping machine is normally distributed with a mean time between breakdowns of 13 weeks and a standard deviation of 1 week. The output of the stamping machine is indispensable to subsequent work stations, so that management assigns a high cost to any breakdown of $4,000.00. the cost of performing preventive maintenance is $1,200.00.

1. What is the area under the normal curve?
2. What is the value of z?
3. How often should preventive maintenance be performed?

SOLUTIONS

1. a. Mean: <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image299::/sites/dl/free/0072443901/24520/Image299.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image299 (1.0K)</a>Image299 breakdowns per month.
b Mean cost of services = .55(200) = $110 per month.
c. It would be cheaper to perform preventive maintenance at a cost of $75.00 per month.

2. a.

3. a. Preventive Cost = $1,200.00.
Breakdown Cost = $4,000.00.
Area = P. C. /(P. C. + B. C.) = 1200/(1200+4000) = 0.23.

b. z = -0.74 (from Table B in the appendix).
c. Mean + z <a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::Image302::/sites/dl/free/0072443901/24520/Image302.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif">Image302 (0.0K)</a>Image302 = 13 ? (0.74)(1) = 12.26 weeks.