Question

Larry Ellison starts a company that manufactures high-end custom leather bags. He hires two employees. Each employee only begins working on a bag when a customer order has been received and then she makes the bag from beginning to end. The average production time of a bag is 1.8 days with a standard deviation of 2.7 days. Larry expects to receive one customer order per day on average. The interarrival times of orders have a coefficient of variation of 1. (Carry at least 4 decimal places in all intermediate calculations. Round your final answer to 2 decimal places.)

Answer 1

Answer:

12.55 days

Explanation:

Provided information

Number of employees 2

Average production time=1.8 days

Standard deviation=2.7 days

Inter-arrival time= 1 day

Coefficient of variation= 1 day

Standard deviation of inter-arrival time= 1 day

The coefficient of variations

Inter-arrival coefficient of variation

$C_{vi}=\frac {\sigma}{T}$ where $\sigma$ is standard deviation of inter-arrival time, T is inter-arrival time and $C_v$ is coefficient of variation of inter-arrival time

$C_{vi}=\frac {1 day}{1 day}=1$

Production time coefficient of variation

$C_{vp}=\frac {2.7}{1.8}=1.5$

Total utilization time

$U=\frac {T}{n*T_i}$ where T is the time of production, n is number of employees, U is utilization, $T_i$ is inter-arrival time

$U=\frac {1.8}{2*1}=0.9$

Therefore, utilization time by 2 employees is 0.9

Expected average waiting time

$T_e=(\frac {T}{n*T_i})*0.5(C_{vi}^{2}+C_{vp}^{2})*(\frac{U^{\sqrt{2(n+1)}-1}}{1-U})$

Where $T_e$ is expected average waiting time and the other symbols as already defined

Substituting 1.5 for $C_{vp}$, 1 for $C_{vi}$, 0.9 for U, 2 for n, 1 for $T_i$and 1.8 for T

$T_e=(\frac {1.8}{2*1})*0.5(1^{2}+1.5^{2})*(\frac{0.9^{\sqrt{2(2+1)}-1}}{1-0.9})$

$T_e=0.9*1.625*8.583709=12.55367 days$  and rounding off to 2 decimal places we obtain 12.55 days

Therefore, expected duration between order received and beginning of production is approximately 12.55 days