Consider twomachine m 2 station waiting line system with me

Consider two-machine (m = 2) station [waiting line system] with mean effective process time t_e per job and a fast single-machine (m = 1) station with mean effective process time t_e/2. Let CT_q & CT represent, respectively, average waiting time in queue & queue + machine. Let WIP_q, WIP represent the corresponding average inventory. In comparing a single-machine station to the two-machine station with the same average customer inter-arrival time “t_a”, station utilization, u, and CVs, which of the following are true (pick ALL that apply):

The two-machine system will be better than the single-machine system in terms of the four waiting line metrics, CT_q, CT, WIP_q, and WIP.

The single-machine system will be busier on average than the two-machine system.

Both systems will satisfy WIP = (1/ t_a) CT, WIP_q = (1/ t_a) CT_q.

Average number of customers in the waiting line WIP_q will be the same in both systems.

The comparison between CT in the two systems will depend on the variability factor, C= (Ca^2 +Ce2)/2

If the variability factor V > 0, then the two-machine system will have lower WIP_q.

If the variability factor is 1 (i.e., all CVs are 1), the single-machine system will have lower WIP.

If the variability factor is 1 (i.e., all CVs are 1), then CT will be higher in the two-machine system, but CT_q will be higher in the single-machine system

Solution

solution:

1)for given queuing model with two machine and single machine we have to make choices to select four parameter and select which system is better

2)firstly,for two machine system

mean waiting time=k=1/inter arrival time=1/ta

mean service time=s=1/process time=1/te

waiting time is given by

Ct=k/s(s-k)

here for queue only waiting time is

Ctq=[(te/ta(1/te-1/ta))]=te^2/(ta-te)

with machine

Ct=Ctq+process time=te^2/(ta-te)+te

where waiting inventory is given by

in queue

WIPq=k*Ctq=(1/ta)[te^2/(ta-te)]

for including machine

WIP=K*Ct=(1/ta)*Ct=[te^2/(ta-te)+te]

3)for single machine system

mean waiting time=k=1/inter arrival time=1/ta

mean service time=s=1/process time=2/te

waiting time is given by

Ct=k/s(s-k)

here for queue only waiting time is

Ctq=[(te/ta*2*(2/te-1/ta))]=te^2/2*(2*ta-te)

with machine

Ct=Ctq+process time=te^2/2*(2ta-te)+te/2

where waiting inventory is given by

in queue

WIPq=k*Ctq=(1/ta)[te^2/2*(2ta-te)]

for including machine

WIP=K*Ct=(1/ta)*Ct=[te^2/2*(2*ta-te)+te/2]

4)if we take te=2 and ta=4 sec then we get that

for two machine system

Ctq=2 sec

Ct=4 sec

WIPq=.5

WIP=1

for single amchine system

Ctq=.33

Ct=1.33

WIPq=.0825

WIP=.3325

for both of them variability factor is 1 and single machine system has lower WIP

5)hence it is observed that

two machine system

system busy=k/s=te/ta

for single machine

system busy=k/s=te/2ta

6)hence it is obeserved that

1)TRUE,system with two machine has more value of all waiting metric like Ctq,Ct,WIPq,WIP than single machine system,hence first parameter is true,but it is bad from point of service being provided.

2)false,system with two machine is more busy than system with one machine

3)TRUE,both system are satisfying waiting time equation

4)FALSE,average number of customer in queue is different and less for single machine due to fast process time