Question

1. the arrival rate to a machining center is 15 parts per hour and the machining center takes an average of 2 minutes to process each part. assume part arrivals and processing times are variable with $c_a^2 = 1$ and $c_e^2 = 2$.

(a) what is average utilization? will a queue form in front of the center and why?
(b) what is the average length of the queue? what is the overall average level of work-in-process inventory?
(c) what is the average cycle time?
(d) what is average throughput? what is the maximum feasible throughput?
(e) the machining center feeds into an assembly workstation as described in the following figure. what is the coefficient of variation in the arrival process of the assembly workstation?

Explanation:

Step1: Define key parameters

Arrival rate $\lambda = 15$ parts/hour, processing time $t_e = \frac{2}{60} = \frac{1}{30}$ hours/part, so service rate $\mu = \frac{1}{t_e} = 30$ parts/hour. $c_a^2=1$, $c_e^2=2$.

---

Part (a)

Step1: Calculate utilization

Utilization $u = \frac{\lambda}{\mu}$
$u = \frac{15}{30} = 0.5$

Step2: Analyze queue formation

Queue forms if $u>0$ and variability exists. Here $u=0.5>0$, and $c_a^2,c_e^2>0$, so queue forms due to variable arrivals/processing.

---

Part (b)

Step1: Calculate queue length $L_q$

Use queue formula: $L_q = \frac{c_a^2 + c_e^2}{2} \cdot \frac{u^2}{1-u}$
$L_q = \frac{1+2}{2} \cdot \frac{0.5^2}{1-0.5} = \frac{3}{2} \cdot \frac{0.25}{0.5} = 0.75$

Step2: Calculate WIP $L$

$L = L_q + u$
$L = 0.75 + 0.5 = 1.25$

---

Part (c)

Step1: Calculate cycle time $T$

$T = \frac{L}{\lambda}$
$T = \frac{1.25}{15} = \frac{1}{12}$ hours $= 5$ minutes

---

Part (d)

Step1: Average throughput

Average throughput equals arrival rate: $TH = \lambda = 15$ parts/hour

Step2: Maximum feasible throughput

Maximum throughput is service rate: $TH_{max} = \mu = 30$ parts/hour

---

Part (e)

Step1: Calculate arrival CV for assembly

Use formula for output CV: $c_{a,out}^2 = c_a^2(1-u^2) + c_e^2 u^2$
$c_{a,out}^2 = 1(1-0.5^2) + 2(0.5^2) = 1(0.75) + 2(0.25) = 0.75 + 0.5 = 1.25$

Answer:

(a) Average utilization = 0.5 (50%). A queue will form because the center has positive utilization and there is variability in both arrival and processing times.
(b) Average queue length = 0.75 parts; Overall average WIP inventory = 1.25 parts
(c) Average cycle time = 5 minutes (or $\frac{1}{12}$ hours)
(d) Average throughput = 15 parts per hour; Maximum feasible throughput = 30 parts per hour
(e) Coefficient of variation squared for assembly arrivals = 1.25, so coefficient of variation = $\sqrt{1.25} \approx 1.118$