max to = 1.5 min/part.
The operational cycle time of the flow-line is then
tcf = 0.05 min/part + 1.5 min/part = 1.55 min/part.
The cycle rate is then:
Rc = 1/1.55 min/part = 0.645 part/min or 38.7 parts/hr.
2.3.2 Other Mathematical Quantifying Concepts
Although productivity is of primary importance in justifying automation, other quantifying concepts come into play as well. These include production capacity, utilization, availability, and manufacturing lead-time.
Production capacity is the maximum rate of output of a particular product for a manufacturing system over a specified time period. The time can be expressed in days, weeks, months, or years. The system under consideration could be the whole plant, a production line, or a manufacturing cell. The calculation takes the production rate of the system under consideration and multiplies it by the number of hours worked during the specified time interval and the number of subsystems producing at that production rate. The general form of the equation is:
Pc = Rnmhrsw,
where
| R | = | production rate of system (Rp, Rpq, or Rc) in parts/hr |
| n m | = | number of machines or work centers producing at that rate |
| hrs w | = | hours worked during the specified time interval. |
Example 2.8
Calculate the monthly production capacity (Pc) of a product produced by the injection molding process described in Example 2.4. Assume the plant uses 3 injection molding machines and molds to produce the part. Also, assume the plant operates in three 8-hour shifts per day, 5 days per week. Suggest a plan by which the plant may increase production capacity in the short term. How could it do so in the long term?
Solution
The governing equation is
Pc = Rnmhrsw,
where
| R pq | = | 179.64 parts/hr (calculated in Example 2.6) |
| n m | = | 3 |
| hrs w | = | (3 shifts)(8 hr/shift)(5 days/week)(4 weeks/month) = 480 hr/month. |
Thus,
| P c | = | (179.64 parts/hr)(3)(480 hr/month) = 258,681 parts/month. |
To increase production in the short term the plant could run the injection presses on weekends. Doing so would increase hours worked (hrsw). Long term solutions might include building more molds to run on more injection molding machines (increase nm) and/or increase the production rate (R). This could be accomplished by building molds with more cavities or decreasing the process’s operational cycle time (tc).
Production capacity is a theoretical value. In practice, actual production may be significantly less due to lack of orders, lack of supplies, processing problems, or labor issues. Thus, management will often evaluate the utilization of a manufacturing system. Utilization U is defined as the ratio of the actual number of products divided by production capacity. Thus,
U = 100Q /Pc,
where
| Q | = | actual production over specified time frame |
| P c | = | production capacity over specified time frame. |
Note that U is expressed as a percentage.
Example 2.9
Calculate the utilization of the injection molding process described in Example 2.8 if actual production in the previous month was 175,000 parts.
Solution
The governing equation is
U = 100Q/Pc,
where
| Q | = | 175,000 parts |
| P c | = | 258,681 parts/month. |
Thus,
U = (100)(175,000 parts)/258,681 parts = 67.7%.
Additionally, a manufacturing system under repair may not be fully used. Thus, the availability of a system, expressed as a percentage, can be calculated. It is determined by the equation
A = 100(tmtbf − tmtbr)/tmtbf
where
| A | = | availability |
| t mtbf | = | mean time between failures (hr) |
| t mtbr | = | mean time to repair (hr). |
These two measures provide solid insight into a manufacturing system and can also help in identifying automation opportunities. Additionally, if utilization and availability information is known within a facility, realistic actual production values can be calculated. Consider the following example.
Example 2.10
A manufacturing system has a theoretical production capacity of 100,000 parts/ month. Typical utilization of the system is 80% and availability is 93%. What is the anticipated actual monthly production of the system?
Solution
Rearranging the equation for utilization and factoring in the availability of the system yields the following equation:
Q = UPcA.
Thus,
Q = (80%)(100,000 parts/month)(93%) = 74,400 parts/month.
Another important quantifying