· web viewextra credit page 1103, problem 1, chapter 20 section 9 problem statement: a laundromat...

IE 417 Winter 2011Team 6: Robert Delgado, Adi Wijaya, Min PyonDr. Parisay’s comments are in red.

Extra Credit103, Problem 1, chapter 20 section 9

Problem statement: A laundromat has 5 washing machines. A typical machine breaks down once every 5 days. A repairer can repair a machine in an average of 2.5 days. Currently, three repairers are on duty. The owner of the laundromat has the option of replacing them with a superworker, who can

repair a machine in an average of 56 day. The salary of the superworker equals the pay of the

three regular employees. Breakdown and service times are exponential. Should the laundromat replace the three repairers with the superworker? (Solve manually.)

Added questions:1- Solve using WinQSB2- What if the numbers of machines decrease from 5 to 4, what will happen to the decision?3- Perform sensitivity analysis for the effect of changes in the number of queue capacity.

(not suitable)4- Analysis of best solution and report to manager

Manual solution for original problem:

For the regular workers: K = 5, R = 3, λ = 15= 0.2, µ=

12.5= 0.4

1

State No. of good machines Repair Queue No. of repair workers busy

0 GGGGG 01 GGGG 12 GGG 23 GG 34 G B 35 BB 3

2

5λ 4λ 3λ 2λ λ

µ 2µ 3µ 3µ 3µ

0 51 2 3 4

ρ = λ/µ = (15)/(

12.5 ) = 0.5

π1 = (51) (0.5)1π0 = 2.5π0

π2 =(52) (0.5)2π0 = 2.5 π0

π3 =(53) (0.5)3 π0 = 1.25 π0

π4 = (54) (0.5)4 4 !

3!3(4−3 )π 0 = 0.417π0

π5 = (55) (0.5)5 5!3!3(5−3) π0 = 0.0694 π0

π0 (1 + 2.5 + 2.5 + 1.25 + 0.417 + 0.0694) = 1

π0 = 0.129, π1 = 0.323, π2 = 0.323, π3 = 0.161, π4 = 0.054, π5 = 0.009

The expected number of machines in good condition is K - L

L = ∑j=0

j=5

j (πj)= 0(0.129) + 1(0.323) + 2(0.323) + 3(0.161) + 4(0.054) + 5(0.009) = 1.713

K - L = 5 – 1.713 = 3.287 machines are in good condition

Average Downtime for machines that needs repairs, W = L/λ

λ=¿ λ(K – L) = (15)(3.287) = 0.6574 machines per day

Since L = 1.713, W = 1.713

0.6574 = 2.61 days

Fraction of time a particular repair worker is idle

= π0 + (3−1 )π 1

3 +

(3−2 ) π23

= 0.129 + (23¿(0.323) + (

13)(0.323) = 0.452

For the Super worker: K = 5, R = 1, λ = 15 , µ=

65

3


0 GGGGG 01 GGGG 12 GGG B 13 GG BB 14 G BBB 15 BBBB 1

4

ρ = λ/µ = (15)/(

65 ) =

16

π1 = (51) (16 )1π0 =

56π0

π2 =(52) (16 )2 2!

1!1(2−1) π 0 = 0.556 π0

π3 =(53) (16 )3 3 !

1!1(3−1 )π 0 = 0.278 π0

π4 = (54) (1

6 )4 4 !1!1(4−1) π0 = 0.093π0

π5 = (55) (16 )5 5 !

1!1(5−1 )π 0 = 0.015 π0

π0 (1 + 56 + 0.556 + 0.278 + 0.093 + 0.015) = 1

π0 = 0.360, π1 = 0.300, π2 = 0.200, π3 = 0.100, π4 = 0.033, π5 = 0.005


L = ∑j=0

j=5

j(π j)=¿¿ 0(0.360) + 1(0.300) + 2(0.200) + 3(0.100) + 4(0.033) + 5(0.005) = 1.157



5

5λ 4λ 3λ 2λ λ

µ µ µ µ µ

10 2 3 4 5


Since L = 1.157, W = 1.1570.769 = 1.505 days

Fraction of time a particular repair worker is idle = π0 = 0.360

Summary Table

K =5, R=3 K=5, R=1 desiredExpected numbers of good condition machines

3.287 machines 3.843 higher

Average downtime for machines that needs repairs

2.61 days 1.51 lower


0.452 0.36 lower

Conclusion:Since the super worker has more expected good machines, lesser idle time and smaller average downtime for machines, we should replace the three workers with the super worker.

Solution using WinQSB:Notice the queue capacity value entered to WinQSB.

WinQSB input information for regular workers, K = 5

6

WinQSB input information for super worker, K = 5

Notice the queue capacity value entered to WinQSB.

7

Probability Summary

8

For regular workers

For super worker

What if the numbers of machines decrease from 5 to 4, what will happen to the decision?

For the regular workers:

K = 4, R = 3, λ = 15 , µ=

12.5


0 GGGGG 01 GGGG 1

9

2 GGG 23 GG 34 G B 3

ρ = λ/µ = (15)/(

12.5 ) = 0.5

π1 = (41) (0.5)1π0 = 2π0

π2 =(42) (0.5)2π0 = 1.5 π0

π3 =(43) (0.5)3 π0 = 0.5 π0

π4 = (44) (0.5)4 4 !

3!3(4−3 )π 0 = 0.0833π0

π0 (1 + 2 + 1.5 + 0.5 + 0.0833) = 1

π0 = 0.1967π1 = 0.3934π2 = 0.2951π3 = 0.0984π4 = 0.0164


L = ∑j=0

j=4

j(π j)=¿¿ 0(0.1967) + 1(0.3934) + 2(0.2951) + 3(0.0984) + 4(0.0164) = 1.3444




Since L = 1.3444, W = 1.34440.5311 = 2.531 days


= π0 + (3−1 )π 1

3 +

(3−2 ) π23

= 0.1967 + (23¿(0.3934) + (

13)(0.2951) = 0.5573

For the super workers:

K = 4, R = 1, λ = 15 , µ=

65

10


0 GGGGG 01 GGGG 12 GGG B 13 GG BB 14 G BBB 1

ρ = λ/µ = (15)/(

65 ) =

16

π1 = (41) (1

6 )1π0 = 0.667π0

π2 =(42) (1

6 )2 2!1!1(2−1)

π 0 = 0.3333 π0

π3 =(43) (1

6 )3 3 !1!1(3−1 ) π 0 = 0.1111 π0

π4 = (44) (1

6 )4 4 !1!1(4−1) π0 = 0.0185π0

π0 (1 + 0.667 + 0.3333 + 0.1111 + 0.0185) = 1

π0 = 0.4695π1 = 0.3132π2 = 0.1565π3 = 0.0522π4 = 0.0087


L = ∑j=0

j=4

j(π j)=¿¿ 0(0.4695) + 1(0.3132) + 2(0.1565) + 3(0.0522) + 4(0.0087) = 0.8176



λ=¿ λ(K – L) = (15)(3.1824) = 0.6365machines per day

Since L = 0.8176, W = 0.81760.6365 = 1.28 days

Fraction of time a particular repair worker is idle = π0 = 0.4695

Sensitivity analysis

11

The motivation for the sensitivity analysis is that, we were trying to find out if the decision changes if the number of queue capacity is lower. (This is not a good selection for sensitivity analysis from practical point of view. In repair shop situations we have the queue capacity that we need. It was better to perform SA on number of servers, or service rate, or arrival rate.)

The range of queue capacity can never be more than five because we only have 5 machines. It does not make sense if the queue capacity goes beyond five.

For regular workers:

12

For Super Worker

13

Analysis of best solution and report to manager:

Performance measure Observed Acceptance Value Comments/ SuggestionsL 1.16 ≤2 machines OkLq .52 ≤1 machine OkW 1.5 days ≤2 days OkWq .68 day ≤1 day Ok

Utilization 64 % 80-90% *Underutilizedπc=πi 36 % ≤50% Ok

Λb 0 ≤1 Ok

Mention motivation for acceptable values.

*underutilized: we can assigned other tasks (subjective).

Report to manager

Based on the calculated data, hiring a super worker will benefit the Laundromat more. With the super worker, the broken machines in the Laundromat will be on average of 1.6 with an average of .52 machines waiting in line for service. On the other hand, the machines will spend on average 1.5 days broken and on average .68 of a day waiting to get fixed, so a little over half of a work day. With that, the super worker would have a utilization of 64% and only a 36% of idle time.

The Acceptable values we expect for our average machine present in the Laundromat would be less than 2 machines, the average number of machines being in line to be service would be less than one machine, we expect the machine to be broken on average no more than 2 days, and on average one day waiting to get serviced or fixed. We expect a performance of 80 to 90% and can be flexible by accepting 5% having customer walk away. Furthermore, other than the utilization, the current system does satisfy most of our acceptance values of course these acceptance values are subjective.

We did a sensitivity analysis on the queue capacity. We did from a range of 1 to 5 and we found out that the decision changes when the queue capacity is at 3. When the queue capacity is at 3, we will have a change in the average number of broken machines to 1.14 machines, average number of machines waiting for service will change to .50 machines, the average time a machine spends broken .65 of a day, the average time a machine spends waiting for the service 1.4 of a day. Therefore we found out that the turning point of the decision happens when the queue capacity is at 3. (needs rewrite.)

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· web viewextra credit page 1103, problem 1, chapter 20 section 9 problem statement: a laundromat...

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