Training Industrial Engineering Queuing Theory & Work Measurement
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Queuing Theory & Work Measurement

24 min Industrial Engineering

Queuing Theory & Work Measurement

Queuing theory models waiting lines (service systems). Work measurement determines standard times for operations.

M/M/1 Queue (Single Server)

Arrival rate: $\lambda$ (customers/hr), Service rate: $\mu$ (customers/hr). Utilization:

$$\rho = \frac{\lambda}{\mu} < 1 \text{ for stability}$$

Average number in system: $L = \frac{\rho}{1 - \rho}$. Average time in system: $W = \frac{1}{\mu - \lambda}$.

Little's Law

$$L = \lambda W$$

Average number in system = arrival rate × average time in system. Works for any stable queue.

Learning Curve

$$T_n = T_1 \cdot n^b \quad \text{where} \quad b = \frac{\ln(\text{learning rate})}{\ln 2}$$

An 80% learning curve means each doubling of production reduces time per unit to 80%.

Example 1

Customers arrive at $\lambda = 20$/hr, service rate $\mu = 25$/hr. Find utilization, average number in system, and average wait time.

$\rho = 20/25 = 0.8$

$L = 0.8/(1-0.8) = 4$ customers

$W = 1/(25-20) = 0.2$ hrs $= 12$ minutes

$W_q = W - 1/\mu = 0.2 - 0.04 = 0.16$ hrs $= 9.6$ min (wait in queue only)

Example 2

Verify Little's Law for Example 1.

$L = \lambda W = 20 \times 0.2 = 4$ ✓

Example 3

The first unit takes 100 hours. With an 80% learning curve, find the time for the 8th unit.

$b = \ln(0.8)/\ln(2) = -0.2231/0.6931 = -0.3219$

$$T_8 = 100 \times 8^{-0.3219} = 100 \times 0.512 = 51.2 \text{ hours}$$

Note: $8 = 2^3$, so three doublings: $100 \times 0.8^3 = 51.2$ ✓

Practice Problems

1. $\lambda = 10$/hr, $\mu = 15$/hr. Find $\rho$, $L$, and $W$.
2. Verify Little's Law for #1.
3. $\lambda = 30$/hr, $\mu = 32$/hr. Find the average number waiting in queue $L_q = \rho^2/(1-\rho)$.
4. If we add a second server (M/M/2), utilization becomes $\rho = \lambda/(2\mu)$. For $\lambda = 30$, $\mu = 20$, find $\rho$.
5. 90% learning curve: first unit takes 50 hrs. Find time for unit 4.
6. 75% learning curve: $T_1 = 200$ hrs. Find $T_{16}$.
7. Normal time $= $ observed time $\times$ performance rating. If observed $= 5$ min at 110% rating, find normal time.
8. Standard time $= $ normal time $\times (1 + $ allowance$)$. If normal $= 5.5$ min, allowance $= 15\%$, find standard time.
9. A process has 5 operations with standard times: 2, 3, 4, 2, 3 minutes. Find the cycle time and production rate per hour.
10. Line balancing: the bottleneck operation is the longest at 4 min. If the line has 5 stations, find the efficiency.
11. $\lambda = 8$/hr, $\mu = 10$/hr. Find the probability of zero customers in the system: $P_0 = 1 - \rho$.
12. If waiting cost is \$20/hr per customer and adding a server costs \$15/hr, should we add the server? (Use #1 data.)
Show Answer Key

1. $\rho = 0.667$; $L = 2$; $W = 1/5 = 0.2$ hrs $= 12$ min

2. $L = 10 \times 0.2 = 2$ ✓

3. $\rho = 30/32 = 0.9375$; $L_q = 0.9375^2/0.0625 = 14.06$ customers

4. $\rho = 30/40 = 0.75$

5. $b = \ln(0.9)/\ln 2 = -0.152$; $T_4 = 50 \times 4^{-0.152} = 50 \times 0.81 = 40.5$ hrs

6. $16 = 2^4$; $T_{16} = 200 \times 0.75^4 = 200 \times 0.3164 = 63.3$ hrs

7. $NT = 5 \times 1.10 = 5.5$ min

8. $ST = 5.5 \times 1.15 = 6.325$ min

9. Cycle time $= 4$ min (bottleneck); rate $= 60/4 = 15$ units/hr

10. Efficiency $= \sum t_i/(n \times c) = 14/(5 \times 4) = 70\%$

11. $P_0 = 1 - 0.8 = 0.2 = 20\%$

12. Waiting cost: $L \times \$20 = 2 \times 20 = \$40$/hr. Server costs \$15. Net saving = \$25 — yes, add the server.

Inventory Models — EOQ & Reorder Point Placement Test Practice — Industrial Engineering