Question

A firm uses a serial assembly system and needs answers to the following:

a. An output of 900 units per shift (7.5 hours) is desired for a new processing system. The system requires the product to pass through four stations where the work content at each station is 30 seconds. What is the required cycle time for such a system?

b. How efficient is your system with the cycle time you calculated?

c. Station 3 changes and now requires 45 seconds to complete. What will need to be done to meet demand (assume only 7.5 hours are available)? What is the efficiency of the new system?

Short answer

1. The cycle time is 30 seconds.
2. The efficiency is 100%.
3. The efficiency of the new system is 75%.

Step-by-step solution

Step-by-Step Solution

Step 1: (a) Cycle Time

$\begin{array}{c}\text{Cycle\hspace{0.17em}time\hspace{0.17em}=\hspace{0.17em}}\frac{\text{Available\hspace{0.17em}production\hspace{0.17em}time}}{\text{Desired\hspace{0.17em}output}}\\ \\ \text{Available\hspace{0.17em}production\hspace{0.17em}time\hspace{0.17em}=\hspace{0.17em}7.5hours\hspace{0.17em}×60minutes\hspace{0.17em}×60seconds}\\ \\ \text{=\hspace{0.17em}27000seconds}\\ \\ \text{Desired\hspace{0.17em}output\hspace{0.17em}=\hspace{0.17em}900\hspace{0.17em}units}\end{array}$

Now,

role="math" localid="1651292487719" $\begin{array}{c}\text{Cycle\hspace{0.17em}time\hspace{0.17em}=\hspace{0.17em}}\frac{\text{Available\hspace{0.17em}production\hspace{0.17em}time}}{\text{Desired\hspace{0.17em}output}}\\ \\ \text{=\hspace{0.17em}}\frac{\text{27000}}{\text{900}}\\ \\ \text{=\hspace{0.17em}30\hspace{0.17em}seconds}\end{array}$

The cycle time is30 seconds.

(b) Efficiency

$\begin{array}{c}\text{Sum\hspace{0.17em}of\hspace{0.17em}task\hspace{0.17em}times\hspace{0.17em}Ti\hspace{0.17em}=\hspace{0.17em}30\hspace{0.17em}seconds\hspace{0.17em}×\hspace{0.17em}4\hspace{0.17em}workstations}\\ \\ \text{=\hspace{0.17em}120\hspace{0.17em}seconds}\\ \\ \text{Number\hspace{0.17em}of\hspace{0.17em}workstations\hspace{0.17em}N\hspace{0.17em}=\hspace{0.17em}4}\\ \\ \text{Cycle\hspace{0.17em}time\hspace{0.17em}C\hspace{0.17em}=\hspace{0.17em}30\hspace{0.17em}seconds}\end{array}$

We can find the efficiency by using the given equation:

$\begin{array}{c}\text{Efficiency\hspace{0.17em}=\hspace{0.17em}}\frac{\text{Ti}}{\text{N\hspace{0.17em}×\hspace{0.17em}C}}\\ \\ \text{=\hspace{0.17em}}\frac{\text{120}}{\text{4\hspace{0.17em}×\hspace{0.17em}30}}\text{\hspace{0.17em}×100}\\ \\ \text{=\hspace{0.17em}100\%}\end{array}$

The efficiency is 100%.

(c) Efficiency of the new system

Since the third station currently requires 45 seconds to finish and the accessible production is the same at 7.5 hours, the accompanying things should be possible to satisfy the need on time and keep up with efficiency:

1. More talented specialists could be employed for station 3 to accelerate the interaction.
2. Laborers could be approached to stay at work longer than required.
3. Some upgrading in the item could likewise assist with efficiency.
4. Updated hardware and machine could be introduced at the third station to limit the process duration.

For the efficiency of the new system,

$\begin{array}{c}\text{Sum\hspace{0.17em}of\hspace{0.17em}task\hspace{0.17em}times\hspace{0.17em}Ti=\hspace{0.17em}}\left(\text{30seconds\hspace{0.17em}×\hspace{0.17em}3\hspace{0.17em}workstations}\right)\text{+}\left(\text{45seconds\hspace{0.17em}×\hspace{0.17em}1\hspace{0.17em}workstation}\right)\\ \\ \text{=\hspace{0.17em}135\hspace{0.17em}seconds}\end{array}$

For this situation, the longest workstation time will be viewed as the cycle time. Accordingly, the cycle time will currently be 45 seconds. For efficiency:

role="math" localid="1651292885944" $\begin{array}{c}\text{Efficiency\hspace{0.17em}=\hspace{0.17em}}\frac{\text{Ti}}{\text{N\hspace{0.17em}×\hspace{0.17em}C}}\\ \\ \text{=\hspace{0.17em}}\frac{\text{135}}{\text{4\hspace{0.17em}×\hspace{0.17em}}45}\text{\hspace{0.17em}×100}\\ \\ \text{=\hspace{0.17em}75\%}\end{array}$

The efficiency of the new system is 75%.