Endless hours are wasted in most manufacturing operations arguing about the nameplate rating of the process. Weibull analysis will help resolve this important issue. The question about nameplate capacity will be answered by use of actual production quantities.
The May 1997 problem of the month showed daily production quantities. Look at the daily production output described in tons per day from a continuous processing plant where the production is "sold out", and every ton is needed for meeting customer demand. The data is shown in a frequency table format with tonnage * occurrence.
Daily Production Output (tons)
Mean = 897.48, Standard Deviation. = 254.69; Median = 925; Mode = 900
The data is not a simple distribution and is comprised of several problem areas as shown in Figure 1 below made by WinSMITH Weibull probability software. Notice how most of the data falls on a straight line with a beta (slope) = 5.5. Also note the X-axis is the daily output volume from the process.
The process reliability is defined at the point where the straight line is lost. This defines the failure in reliability terms. In short, the process should continue to respond up and down the green line, but deterioration in output results in a production gap quantifying the losses.
Many processes have characteristic production patterns when observed with Weibull probability plots-some are desirable and some are undesirable as seen in Figure 1.
What shape curve is desirable? The curve of actual production data should be steep and free of cutbacks in output which cause the production gaps of lost production. Good processes have steep curves with few production gaps. Less desirable processes have flat curves and many production losses.
The question is: What is the nameplate capacity for this plant based on the daily output and a Weibull plot?
The probability plot in Figure 1 shows a highly inclined curve for the normal production line with about 90.78% of the data falling on the green line. However, this slope has a b = 5.5 which suggest the process has an undesirable coefficient of variation. This is a troubled process because of the flat slope! It has many opportunities for improvement by reducing hidden factory losses from efficiency and utilization losses explained for the nameplate ratings and for this case, a good process with tight control would show a coefficient of variation of 5% with a line slope b = 24.4. The line slope with b = 24.4 will help define the nameplate capacity.
If the preferred slope is b = 24.4 based on tight control (and perhaps on benchmark data from similar processes in other operations) then where should the line be drawn? Figure 2 shows the nameplate line drawn through an actual data point achieved by production. Where the nameplate line cuts the characteristic value at 36.8% reliability (63.2% CDF) defines the point estimate of the nameplate capacity. Notice that the nameplate capacity will always lie to the right of the demonstrated production line. This technique for finding nameplate capacity is factual and objective, and it avoids wasting time in useless arguments.
The gap between the demonstrated production line and the nameplate production line is usually due to efficiency and utilization problems. This zone represents the typical hidden factory.
Hidden factories operate in stealth mode and for the assets installed, they rob the productive efforts from making saleable products.
Hidden factories are true burdens on the enterprise. Best in class companies work hard to eliminate the wasteful operations as capital is generally not required to correct the waste—usually the problems respond to well known methods of ferreting out the waste and making the assets productive. Least in class build more factories with the same “genetic” problems to increase output with terrible burdens on their return on capital key indicators—you need to replicate success and avoid replication of “genetic” problem plants.
The efficiency and utilization gap between the nameplate line and the demonstrated production line is 125,180 tons per year! That’s 125 days of typical output (1000 tons/day) wasted. This hidden plant is substantial! Convert this production loss into money assuming the lost gross margin is $200/ton—this problem is worth 125,180 tons/yr * $200/ton = $25,000,000/yr. Anytime you have a lost money opportunity of $25 million dollars, it’s very important!
Sumarizing the losses found by process reliability
analysis we have the following from the May 97 problem of the month and from
this problem of the month:
The Pareto distribution of problems in this plant:
Efficiency and utilization losses = 125,180 tons/year <--Problem #1 (92.5%) worth $25million!
Severe cutbacks due to crash and burn problems = 8,786 tons/year <--Problem #2 (6.5%)
Minor cutback problems = 1,397 tons/ year <--Problem #3 (1.0%)
Total losses = 135,363 tons/year
Even if you corrected all the reliability problems you would have solved 7.5% of the opportunities. Therefore in this plant, the emphasis needs to be on the #1 problem on the Pareto distribution!
Other pages you may want to visit concerning similar issue are:
· Papers On Process Reliability As PDF Files For No-charge Downloads
Every plant needs a Pareto distribution of problems measured in money. For this plant, the #1 problem is worth $25
million. Everyone needs to attack the
hidden factory as the first priority!
Keep plants running to generate gross margins.
Refer to the caveats on the Problem Of The Month Page about the limitations of the following solution. Maybe you have a better idea on how to solve the problem. Maybe you find where I've screwed-up the solution and you can point out my errors as you check my calculations. E-mail your comments, criticism, and corrections to: Paul Barringer by clicking here.
Technical tools are only interesting toys for engineers until results are converted into a business solution involving money and time. Complete your analysis with a bottom line which converts $'s and time so you have answers that will interest your management team!