Question

UPMC Hospital has been under recent pressure from stakeholders to improve cost efficiency and customer service. In response, the hospital calls in the Health Service Administration (HSA) Consulting Team of Robert Morris & Company (RMC). After initial analysis, we (HSA@RMC) decided to target the X-ray service process. We study the X-ray service process to recommend improvements.

We identified the point of entry into the process as the instant that a patient leaves the physician’s office to walk to the X-ray lab. The point of exit is defined as the instant that both the patient and the completed X-ray film are ready to enter the physician’s office for diagnosis. The unit of flow is a patient. We examined the entire process in detail and broke it down into the 11 constituent activities identified as in the table of the next page. Note that a, c and b denote the optimistic time, the most likely time and the pessimistic time. The flow chart is as follows:

Activity /Event	Description
Start	Patient leaves the physician’s office.
1	Patient walks to the X-ray lab.
2	The X-ray request travels to the X-ray lab by a messenger.
3	An X-ray technician fills out a standard form based on the information supplied by the physician.
4	The receptionist receives from the patient information concerning insurance, prepares and signs a claim form, and sends to the insurer.
5	Patient undressed in preparation for X-ray
6	A lab technician takes X-rays.
7	A darkroom technician develops X-rays.
8	A lab technician prepares X-rays for transfer.
9	Patient puts on clothes and gets ready to leave lab.
10	Patient walks back to the physician’s office.
11	The X-rays are transferred to the physician by a messenger.
End	Patient and X-rays arrive at the physician’s office.

1.1 (2 points) The duration of each activity is of Beta Distribution. Note that it is a BETA DISTRIBUTION. Fill in the following table.

Activity/event	a	c	b	Mean	Variance
Start
1	5	15	30
2	5	15	25
3	10	20	30
4	5	10	15
5	10	20	40
6	10	15	30
7	10	30	40
8	15	40	55
9	5	10	30
10	3	7	10
11	15	20	40
End

1.2 (2 points) There are 4 paths as follows:

            Path 1: Start → 1 → 4 → 5 → 6 → 7 → 8 → 9 → 10 → End

            Path 2: Start → 2 → 3 → 4 → 5 → 6 → 7 → 8 → 9 → 10 → End

            Path 3: Start → 1 → 4 → 5 → 6 → 7 → 8 → 11 → End

            Path 4: Start → 2 → 3 → 4 → 5 → 6 → 7 → 8 → 11 → End

What are the mean and the variance of the duration of each path? What is the longest path in mean time? Fill in the following table and identify the longest path in mean time.

	Mean	Variance
Path 1
Path 2
Path 3
Path 4

Answer: The longest path in mean time is                                                                .

1.3 (2 points) What is the probability of completing the longest path within 185 minutes? Assume that the duration of the longest path is normally distributed.

Answer:                                                                .

1.4 (2 points) What is the time T for which the probability to complete the longest path is 95%? Assume that the duration of the longest path is normally distributed.

Answer 1

1.1

Activity/event	a	c	b	Mean	Variance
Start
1	5	15	30	16.6666667	158.333333
2	5	15	25	15	100
3	10	20	30	20	100
4	5	10	15	10	25
5	10	20	40	23.3333333	233.333333
6	10	15	30	18.3333333	108.333333
7	10	30	40	26.6666667	233.333333
8	15	40	55	36.6666667	408.333333
9	5	10	30	15	175
10	3	7	10	6.66666667	12.3333333
11	15	20	40	25	175
End

1.2

		Variance
	Mean
Path 1	169.25	16381.1667
Path 2	19.0740741	15291.4383
Path 3	22.3809524	14421.2963
Path 4	172.916667	14201.3889

Answer: The longest path in mean time is Path 4.

1.3

z = (185 - 172.916667)/14201.3889 = 0.10

The probability is 0.5404.

1.4 The z-score for p-value = 0.95 is 1.645.

1.645 = (T- 172.916667)/14201.3889

T = 368.933183

Please give me a thumbs-up if this helps you out. Thank you!