project management in healthcare: a case for evaluating patient flow of an emergency department with...

YILDIZ TECHNICAL UNIVERSITY MECHANICAL FACULTY

INDUSTRIAL ENGINEERING DEPARTMENT

PROJECT MANAGEMENT IN HEALTHCARE: A CASE FOR

EVALUATING PATIENT FLOW OF AN EMERGENCY DEPARTMENT

WITH FUZZY CPM AND FUZZY PERT

GİZEM GÜNEŞ 11061007

ADVISOR

Assoc. Prof. Ali Fuat GÜNERİ

ISTANBUL, 2015

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PROJECT MANAGEMENT IN HEALTHCARE: A CASE FOR

EVALUATING PATIENT FLOW OF AN EMERGENCY

DEPARTMENT WITH FUZZY CPM AND FUZZY PERT

GİZEM GÜNEŞ

ADVISOR

Assoc. Prof. Ali Fuat GÜNERİ

Abstract

Nowadays projects are more complex than being in the past and complexibility increases

uncertainty. Fuzzy numbers are more effective for high uncertainty situations. In that study,

fuzzy numbers will be explained and then, how you can use fuzzy numbers in a project will

be explained with Fuzzy CPM (Critical Path Method) and Fuzzy PERT (Program Evaluation and

Review Technique). These tools provides to analyze flows and forecast project completing

time. In healthcare sector, FCPM and FPERT provide to standardize care, improve quality and

reduce patient’s staying time despite uncertainty.

Yıldız Technical University, Industrial Engineering Department, Istanbul, 2015

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CONTENTS

1. Introduction …………………………………………………………………………………………….……………………… 3

2. Fuzzy Numbers ……………………………………………………………………………………………………………….. 5

2.1. Arithmethic Operations Between Fuzzy Numbers ……………………………………………..………………………… 6

3. Fuzzy Critical Path Method ……..………………………………………………………………………………………. 6

3.1. Notation ……………………………………………………………………………………………………………………………………… 6

3.2. Steps of Fuzzy CPM ……………………………………………………………………………………………………………………… 7

4. A Case of Critical Path Method ………..……..………………………………………………………………………. 8

5. Fuzzy Program Evaluation and Review Technique ( FPERT ) ………..……..…………………………. 16

6. Conclusion ..…………………………………………………………………………………………………………………… 18

7. References ……………………………………………………………………………………………………………………. 19

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1. INTRODUCTION

In the past, uncertainity was less because projects were less complex. But now,

globalization improves day by day so, when anything is changed companies want to know

how it can affect result. Nowadays, information is power, so if the companies want to be

successful they have to have their past data to forecast their future. If the companies

analyze their past data they will grab the information; but, if they don’t they will look at the

flying information. If we specialize the issue for managing a project in healthcare sector

results are more important than other projects’. Because time is important but, in hospitals

time is the most important.

Healthcare Project Management Literature : L.C. Lee et al. [6] used PERT/CPM tools on

pulmonary lobectomy patients’ processes. provides to standardize care, reduce the lenghth

of stay and apply quality control processes to improve patients care with PERT/CPM. V.R.

Girija and M.S. Bhat [7] collected data with sample of 100 patients randomly selected from a

population of 460 and they researched to find critical path, expected completion time,

variance of the path and different slack timings associated with the emergency care

provided in the emergency department. U.K.M. Teichgräber et al. [8] used critical pathway

method to design and improve work flow in computed tomography.

While a Project is planning, you can use different kinds of data, these are : deterministic or

probabilistic, in this study fuzzy concept will be used for project management. Firstly, fuzzy

numbers will be explained and then, how you can use fuzzy numbers in a project while you

are planning, managing and controlling a project will be explained. While a project is

planned and managed, some tools are used such as CPM (Critical Path Method) and PERT

(Program Evaluation and Review Technique). These tools are effective and simple methods

to analyze flows and forecast ending date of a project. But, before these tools are used, right

collected database and right method of analyzing data has to be prepared.

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Fuzzy Project Management Literature : Zadeh [9] researched relationship between

possibility and fuzzy numbers. G.S. Liang and T.C. Han [1] used an algorithm that is related to

decision maker’s attitude (pessimistic/optimistic); Shankar et al. [2] used a defuzzification

algorithm; N.R. Shankar et al. [3] proposed a metric distance method for trapezoidal fuzzy

numbers and found float time for each activity to find the critical path in the fuzzy project

network. S.P. Chen and Y.J. Hsueh [10] proposed a method that based on linear

programming formulation and fuzzy number ranking method. M.F. El-Santaway and M. Abd-

Allah [4] used a linear programming approach which incorporates the concept of α-cuts into

two (primal and dual) models to identfy the critical path. V. Sireesha et al [5] proposed a

graphical method of solving fuzzy interval CPM problem. M. Sharafi et al. [12] used a linear

programming model for triangular fuzzy numbers to find earliest time, latest time and slack

times of a project. N.S. Pour [13] studied a method that doesn’t use any defuzzification

technique to find the final processing time.

In that study, fuzzy sets will be explained and then, how you can use fuzzy numbers in a

project will be explained with Fuzzy CPM (Critical Path Method) and Fuzzy PERT (Program

Evaluation and Review Technique) tools with a case. The case is about processes of patients

in emergency department. Sample size is 1500 patients. In the case, earliest and latest fuzzy

times and total slack time will be found with using FCPM and FPERT tools. In healthcare

sector, FCPM and FPERT provide to standardize care, improve quality and reduce patient’s

staying time. In addition, slack times between related activities can be found with

FCPM/FPERT tools and they provide to analyze space time and improve productivity in the

project.

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2. FUZZY NUMBERS

A fuzzy number A in R (Real Line) is a trapezoidal fuzzy number is its membership function

FA (x) : R → [0, 1] is

with - ∞ ≤ a ≤ b ≤ c ≤ d ≤ ∞, the trapezoidal fuzzy number A can be represented by (a, b, c,

d) (Liang and Han, 2004). Using this function, it is possible to assign a membership degree to

each of the element in the universe of discourse X. It is important to note the fact that

membership grades are not probabilities. (Shankar, Sireesha, Rao and Vani, 2010) It is

shown in Figure 1.

A fuzzy number represented by (a, b, c,

d). a means minimum number of data set;

b and c means range of average of data

set; d means maximum number of data

set. For example, we think a fuzzy number

is (12, 16, 18, 20). It can be minimum at 12

and maximum at 20. It can be

approximately between 16 and 18. It is

shown in Figure 2.

Figure 1. Crisp Set and Fuzzy Set (Shankar, Sireesha, Rao and Vani, 2010)

Figure 2. Trapezoidal Fuzzy Number (12, 16, 18, 20)

a = 12 b = 16 c = 18 d = 20

X

1

0

µA (X)

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If b equals to c that means it can be

approximately number of b = c. For

example (2, 4, 4, 6) means it can be

minimum at 2 and maximum at 6; and it can

be approximately 4. It is shown in Figure 3.

2.1. Arithmetic Operations Between Fuzzy Numbers

For A1 = (a1, b1, c1, d1) and A2 = (a2, b2, c2, d2),

Addition : A1 ⃝ A2

A1 ⃝ A2 = (a1, b1, c1, d1) ⃝ (a2, b2, c2, d2) = (a1 + a2, b1 + b2, c1 + c2, d1 + d2)

Subtraction : A1 ⃝ A2

A1 ⃝ A2 = (a1, b1, c1, d1) ⃝ (a2, b2, c2, d2) = (a1 - d2, b1 - c2, c1 - b2, d1 - a2)

Examples

Let A1 and A2 be fuzzy numbers which are (8, 9, 10, 11) and (3, 4, 5, 6).

A1 ⃝ A2 = (8, 9, 10, 11) ⃝ (3, 4, 5, 6) = (8 + 3, 9 + 4, 10 + 5, 11 + 6) = (11, 13, 15, 17)

A1 ⃝ A2 = (8, 9, 10, 11) ⃝ (3, 4, 5, 6) = (8 – 6, 9 – 5, 10 – 4, 11 – 3) = (2, 4, 6, 8)

3. FUZZY CRITICAL PATH METHOD ( FCPM )

3.1. Notation

N : All nodes in a project network

Aij : The activity between two nodes (from i to j)

FATij : Fuzzy activity time of Aij

EFT : Earliest fuzzy time

LFT : Latest fuzzy time

SFTij : Total slack fuzzy time of Aij

FCPM (Pn) : Fuzzy completion time of n th path

t : Number of activities in a project network

X

a = 2 b = c = 4 d = 6

1

0

µA (X)

Figure 3. Triangular Fuzzy Number (2, 4, 4, 6)

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- -

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3.2. Steps of Fuzzy CPM

Step 1. Accept EFT1 equals to (0, 0, 0, 0).

Step 2. Calculate β that means risk factor for each Aij with these formulation :

β = ∑ ∑ 𝐛𝐢𝐣−𝐚𝐢𝐣

(𝐛𝐢𝐣−𝐚𝐢𝐣)+(𝐝𝐢𝐣−𝐜𝐢𝐣) / t

Step 3. Calculate earliest fuzzy time (EFT) for each node.

EFTj = EFTi ⃝ FATij

Step 4. Compare EFTj s where intersection nodes and accept maximum number for EFTj for

each node.

EFTj = max { EFTi ⃝ FATij }

Step 5. Calculate latest fuzzy time (LFT) for each node.

LFTj = EFTk ⃝ FATjk

Step 6. Compare LFTj s where intersection nodes and accept minimum number as LFTj for

each node.

LFTj = min { EFTk ⃝ FATjk }

+

+

EFTj = max {(ax, bx, cx, dx), (ay, by, cy, dy)}

Step 4.1. Find x1 and x2 values with these equations :

x1 = min { ax, bx, cx, dx, ay, by, cy, dy} , x2 = max { ax, bx, cx, dx, ay, by, cy, dy}

Step 4.2. Calculate R ((ax, bx, cx, dx)) and R((ay, by, cy, dy)) values with these equation :

R ((ai, bi, ci, di)) = β [(di - x1)/( x2 - x1 - ci + di)] + (1 – β) [1 – (x2 - ai) / ( x2 - x1 + bi - ai)]

Step 4.3. Compare results of R ((ai, bi, ci, di)) and accept as EFTj which one is greater.

-

-

( ) i j

Note : β ˂ 0,5 , it’s fewer risky situation

If β = 0,5 , it’s neutral situation

β ˃ 0,5 , it’s risky situation

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Step 7. Calculate Total Slack Fuzzy Time (SFT) for each activity.

SFTij = LFTj ⃝ ( EFTi ⃝ FATij)

Step 8. Find all possible path and calculate Fuzzy Completion Time of Path (FCPM) for each

one. FCPM can be calculated with sum of activities between possible path nodes.

FCPM (Pn) = ∑ SFTij

Step 9. Compare FCPM s and accept minimum number as FCPM just like Step 6 calculation.

FCPM (Pn) = min { FCPM (Pi) | i = 1, 2, 3 … , n }

Step 10. Which FCPM of path is fewer, accept the path as critical path.

4. A CASE OF FUZZY CRITICAL PATH METHOD

This example is about patient’s flow in emergency department. The patient arrives the

emergency department by ambulance or patient’s private vehicle and patient’s is recorded

before being examined by doctor. If patient arrives by ambulance the patient passes

recording operation, so the patient goes to near of doctor directly. After doctor examines

the patient who transfers the patient to available part of emergency department depends

on patient’s situation. A sample is taken from the patient and performing lab tests. In same

time, the patient is got under control by nurse and the doctor. The patient can exit with 3

different ways depends on result of patient’s lab tests and doctor recommendation which

are : 1. the patient can leave hospital, 2. the patient can be transfered to another hospital, 3.

the patient can be shifted to appropriate department depends on patient’s situation. In the

case, sample size is 1500 patients.

LFTj = min {(ax, bx, cx, dx), (ay, by, cy, dy)}

Step 6.1. Find x1 and x2 values with these equations :

x1 = min { ax, bx, cx, dx, ay, by, cy, dy} , x2 = max { ax, bx, cx, dx, ay, by, cy, dy}

Step 6.2. Calculate R ((ax, bx, cx, dx)) and R((ay, by, cy, dy)) values with these equation :

R ((ai, bi, ci, di)) = β [(di - x1)/( x2 - x1 - ci + di) + (1 – β) (1 – (x2 - ai) / ( x2 - x1 + bi - ai)

Step 6.3. Compare results of R ((ai, bi, ci, di)) and accept as EFTj which one is fewer.

- +

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Patient’s activities in emergency department is shown in Table 1, project network is shown

in Figure 4 and activities’ times are shown in Table 2.

Symbol Activity

A Shifting of the patient from ambulance to emergency department

B Shifting of the patient from private vehicle to emergency department

C Recording of the patient

D Examining of the patient by doctor

E Sending of the patient to available part of the emergency department

F Checking of the patient by doctor and nurse

G Performing lab tests

H Shifting of the patient to appropriate department of the hospital depends on patient's situation

I Registration

J Shifting of the patient to ambulance to transfer another hospital

K Billing

Table 1. Activities of Patient in Emergency Department

Table 2. Activities’ Times of Patient in Emergency Department

1

2

A 3 4

B C

D

6

5E

F

G 7

H

8J

9K

10I

Figure 4. Project Network of Patient’s Activities in Emergency Department

Activity Predecessor a b c d

A - 0,5 1,2 1,5 4

B - 1 1,8 2,1 4

C B 0,7 1 1 2,5

D A, C 2,5 5 7 9

E D 0,4 0,5 0,8 1,4

F E 8 15 28 134

G E 1 10 10 125

H G, F 5 6 8 9

I G, F 5 8 8 14

J I 1 2,1 2,5 5

K I 5 6 6 8

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Step 1. EFT1 = (0, 0, 0, 0)

Step 2. β = 0,2715 is found with that equation:

Step 3. Calculate EFT s for each node with that equation:

EFTj = EFTi ⃝ FATij

Step 4. Compare EFT s for each node and accept maximum one as EFT.

EFT2 = (0, 0, 0, 0) + (1, 1,8 , 2,1 , 4) = (1, 1,8 , 2,1 , 4)

EFT3 = max {(EFT2 ⃝ FAT23) , (EFT1 ⃝ FAT13)}

EFT3 = max {((1, 1,8 , 2,1 , 4) ⃝ (0,7 , 1, 1, 2,5)), ((0, 0, 0, 0) ⃝ (0,5 , 1,2 , 1,5 , 4))}

EFT3 = max {(1,7 , 2,8 , 3,1 , 6,5) , (0,5 , 1,2 , 1,5 , 4)}

R((1,7 , 2,8 , 3,1 , 6,5)) = 0,4093, R((0,5 , 1,2 , 1,5 , 4)) = 0,1879

EFT3 = (1,7 , 2,8 , 3,1 , 6,5)

EFT4 = EFT3 ⃝ FAT34

EFT4 = (1,7 , 2,8, 3,1 , 6,5) ⃝ (2,5 , 5, 7, 9)

EFT4 = (4,2 , 7,8 , 10,1 , 15,5)

EFT5 = (4,2 , 7,8 , 10,1 , 15,5) ⃝ (0,4 , 0,5 , 0,8 , 1,4)

EFT5 = (4,6 , 8,3 , 10,9 , 16,9)

EFT6 = EFT5 = (4,6 , 8,3 , 10,9 , 16,9)

EFT7 = max {(EFT6 ⃝ FAT67) , (EFT5 ⃝ FAT57)}

EFT7 = max {((4,6 , 8,3 , 10,9 , 16,9) ⃝ (8, 15, 28, 134)) , ((4,6 , 8,3 , 10,9 , 16,9) ⃝ (1,

10, 10, 125))}

EFT7 = max {(12,6 , 23,3 , 38,9 ,150,9) , (5,6 , 18,3 , 20,9 , 141,9)}

R((12,6 , 23,3 , 38,9 ,150,9)) = 1,9933, R((5,6 , 18,3 , 20,9 , 141,9)) = 1,0911

EFT7 = (12,6 , 23,3 , 38,9 , 150,9)

EFT8 = EFT7 ⃝ FAT78

EFT8 = (12,6 , 23,3 , 38,9 , 150,9) ⃝ (5, 8, 8, 14)

EFT8 = (17,6 , 31,3 , 46,9 , 164,9)

EFT9 = EFT8 = (17,6 , 31,3 , 46,9 , 164,9)

EFT10 = max {(EFT7 ⃝ FAT710) , (EFT8 ⃝ FAT810) , (EFT9 ⃝ FAT910)}

EFT10 = max {((12,6 , 23,3 , 38,9 , 150,9) ⃝ (5, 6, 8, 9)) , ((17,6 , 31,3 , 46,9 , 164,9) ⃝

(1, 2,1 , 2,5 , 5)) , ((17,6 , 31,3 , 46,9 , 164,9) ⃝ (5, 6, 6, 8))}

EFT10 = max {(17,6 , 29,3 , 46,9 , 159,9) , (18,6 , 33,4 , 49,4 , 169,9) , (22,6 , 37,3 , 52,9

, 172,9)}

R((17,6 , 29,3 , 46,9 , 159,9)) = 1,5490 , R((18,6 , 33,4 , 49,4 , 169,9)) = 1,5159 ,

R((22,6 , 37,3 , 52,9 , 172,9)) = 1,6666

EFT10 = (22,6 , 37,3 , 52,9 , 172,9)

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Step 5. Calculate LFT s for each node with that equation:

LFTj = EFTk ⃝ FATjk

Step 6. Compare LFT s for each node and accept minimum one as EFT.

LFT10 = (22,6 , 37,3 , 52,9 , 172,9)

LFT9 = LFT10 ⃝ FAT910

LFT9 = (22,6 , 37,3 , 52,9 , 172,9) ⃝ (5, 6, 6, 8)

LFT9 = (14,6 , 31,3 , 46,9 , 167,9)

LFT8 = min { (LFT9) , (LFT10 ⃝ FAT810 ) }

LFT8 = min { (14,6 , 31,3 , 46,9 , 167,9) , [(22,6 , 37,3 , 52,9 , 172,9) ⃝ (1, 2,1 , 2,5 , 5)]}

LFT8 = min { (14,6 , 31,3 , 46,9 , 167,9) , (17,6 , 34,8 , 50,8 , 171,9) }

R((14,6 , 31,3 , 46,9 , 167,9)) = 0,2195 , R((17,6 , 34,8 , 50,8 , 171,9)) = 0,2377

LFT8 = (14,6 , 31,3 , 46,9 , 167,9)

LFT7 = min { (LFT10 ⃝ FAT710) , (LFT8 ⃝ FAT78 ) }

LFT7 = min { [(22,6 , 37,3 , 52,9 , 172,9) ⃝ (5, 6, 8, 9)] , [(14,6 , 31,3 , 46,9 , 167,9) ⃝

(5, 8, 8, 14)] }

LFT7 = min { (13,6 , 29,3 , 46,9 ,167,9) , (0,6 , 23,3 ,38,9 , 162,9) }

R((13,6 , 29,3 , 46,9 ,167,9)) = 0,2718 , R((0,6 , 23,3, 38,9 , 162,9)) = 0,2383

LFT7 = (0,6 , 23,3 ,38,9 , 162,9)


LFT6 = (0,6 , 23,3 ,38,9 , 162,9) ⃝ (8, 15, 28, 134)

LFT6 = (-133,4 , -4,7, 30,9 , 154,9)

LFT5 = min { (LFT6) , (LFT7 ⃝ FAT57 ) }

LFT5 = min { (-133,4 , -4,7, 30,9 , 154,9), [(0,6 , 23,3 ,38,9 , 162,9) ⃝ (1, 10, 10, 125)] }

LFT5 = min { (-133,4 , -4,7, 30,9 , 154,9) , (-124,4 , 13,3 , 28,9 , 161,9) }

R((-133,4 , -4,7, 30,9 , 154,9)) = 0,4048 , R((-124,4 , 13,3 , 28,9 , 161,9)) = 0,4340

LFT5 = (-133,4 , -4,7, 23,9 , 154,9)


LFT4 = (-133,4 , -4,7, 23,9 , 154,9) ⃝ (0,4 , 0,5 , 0,8 , 1,4)

LFT4 = (-134,8 , -5,5 , 23,4 , 154,5)


LFT3 = (-134,8 , -5,5 , 23,4 , 154,5) ⃝ (2,5 , 5, 7, 9)

LFT3 = (-143,8 , -12,5 , 18,4 , 152)


LFT2 = (-143,8 , -12,5 , 18,4 , 152) ⃝ (0,7 , 1, 1, 2,5)

LFT2 = (-146,3 , -13,5, 17,4 , 151,3)

LFT1 = min { (LFT2 ⃝ FAT12) , (LFT3 ⃝ FAT13 ) }

LFT1 = min { [(-146,3 , -13,5, 17,4 , 151,3) ⃝ (1, 1,8 , 2,1 , 4)] , [(-143,8 , -12,5 , 18,4 ,

152) ⃝ (0,5 , 1,2 , 1,5 , 4)] }

LFT1 = min { (-150,3, -15,6, 15,6 , 150,3) , (-147,8 , -14, 17,2 , 151,5) }

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-

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-

-

-

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12

R((-150,3, -15,6, 15,6 , 150,3)) = 0,4118 , R((-147,8 , -14, 17,2 , 151,5)) = 0,4159

LFT1 = (-150,3, -15,6, 15,6 , 150,3)

EFT and LFT values of each node are shown in Table 3.

Step 7. Calculate Total Slack Fuzzy Time (SFT) for each activity with that equation :

SFTij = LFTj ⃝ ( EFTi ⃝ FATij)

SFT13 = LFT3 ⃝ (EFT1 ⃝ FAT13)

SFT13 = (-143,8 , -12,5 , 18,4 , 152) ⃝ [(0, 0, 0, 0) ⃝ (0,5 , 1,2 , 1,5 , 4)]

SFT13 = (-147,8 , -14 , 17,2 , 151,5)


SFT12 = (-146,3 , -13,5 , 17,4 , 151,3) ⃝ [(0, 0, 0, 0) ⃝ (1, 1,8 , 2,1 , 4)]

SFT12 = (-150,3 , -15,6 , 15,6 , 150,3)


SFT23 = (-143,8 , -12,5 , 18,4 , 152) ⃝ [(1, 1,8 , 2,1 , 4) ⃝ (0,7 , 1, 1, 2,5)]

SFT23 = (-150,3 , -15,6 , 15,6 , 150,3)


SFT34 = (-134,8 , -5,5 , 23,4 , 154,5) ⃝ [(1,7 , 2,8 , 3,1 , 6,5) ⃝ (2,5 , 5, 7, 9)]

SFT34 = (-150,3 , -15,6 , 15,6 , 150,3)


SFT45 = (-133,4 , -4,7 , 23,9 , 154,9) ⃝ [(4,2 , 7,8 , 10,1 , 15,5) ⃝ (0,4 , 0,5 , 0,8 , 1,4)]

SFT45 = (-150,3 , -15,6 , 15,6 , 150,3)

SFT56 = LFT6 ⃝ EFT5

SFT56 = (-133,4 , -4,7 , 23,9 , 154,9 ) ⃝ (4,6 , 8,3 , 10,9 , 16,9)

a b c d a b c d

1 0 0 0 0 -150,3 -15,6 15,6 150,3

2 1 1,8 2,1 4 -146,3 -13,5 17,4 151,3

3 1,7 2,8 3,1 6,5 -143,8 -12,5 18,4 152

4 4,2 7,8 10,1 15,5 -134,8 -5,5 23,4 154,5

5 4,6 8,3 10,9 16,9 -133,4 -4,7 23,9 154,9

6 4,6 8,3 10,9 16,9 -133,4 -4,7 23,9 154,9

7 12,6 23,3 38,9 150,9 0,6 23,3 38,9 162,9

8 17,6 31,3 46,9 164,9 14,6 31,3 46,9 167,9

9 17,6 31,3 46,9 164,9 14,6 31,3 46,9 167,9

10 22,6 37,3 52,9 172,9 22,6 37,3 52,9 172,9

EFT LFT

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+

-

Table 3. EFT s and LFT s values of each node

13

SFT56 = (-150,3 , -15,6 , 15,6 , 150,3)


SFT67 = (0,6 , 23,3 , 38,9 , 162,9) ⃝ [(4,6 , 8,3 , 10,9 , 16,9) ⃝ (8, 15, 28, 134)]

SFT67 = (-150,3 , -15,6 , 15,6 , 150,3)


SFT57 = (0,6 , 23,3 , 38,9 , 162,9) ⃝ [(4,6 , 8,3 , 10,9 , 16,9) ⃝ (1, 10, 10, 125)]

SFT57 = (-141,3 , 2,4 , 20,6 , 157,3)

SFT710 = LFT10 ⃝ (EFT7 ⃝ FAT710)

SFT710 = (22,6 , 37,3 , 52,9 , 172,9) ⃝ [(12,6 , 23,3 , 38,9 , 150,9) ⃝ (5, 6, 8, 9)]

SFT710 = (-137,3 ,-9,6 , 23,6 , 155,3)


SFT78 = (14,6 , 31,3 , 46,9 , 167,9) ⃝ [(12,6 , 23,3 , 38,9 , 150,9) ⃝ (5, 8, 8, 14)]

SFT78 = (-150,3 , -15,6 , 15,6 , 150,3)

SFT89 = LFT9 ⃝ EFT8

SFT89 = (14,6 , 31,3 , 46,9 , 167,9) ⃝ (17,6 , 31,3 , 46,9 , 164,9 )

SFT89 = (-150,3 , -15,6 , 15,6 , 150,3)

SFT810 = LFT10 ⃝ (EFT8 ⃝ FAT810)

SFT810 = (22,6 , 37,3 , 52,9 , 172,9) ⃝ [(17,6 , 31,3 , 46,9 , 164,9 ) ⃝ (1 , 2,1 , 2,5 , 5)]

SFT810 = (-147,3 , -12,1 , 19,5 , 154,3)

SFT910 = LFT10 ⃝ (EFT9 ⃝ FAT910)

SFT910 = (22,6 , 37,3 , 52,9 , 172,9) ⃝ [(17,6 , 31,3 , 46,9 , 164,9 ) ⃝ (5, 6, 6, 8)]

SFT910 = (-153,3 , -15,6 , 15,6 , 153,3)

Step 8. Find all possible path and calculate Fuzzy Completion Time of Path (FCPM) for each

one. FCPM can be calculated with sum of activities between possible path nodes.

Let path P1 = (1 – 3 – 4 – 5 – 6 – 7 – 8 – 10) .

FCPM (P1) = SFT13 + SFT34 + SFT45 + SFT56 + SFT67 + SFT78 + SFT810

FCPM (P1) = (-1046,6 , -104,1 , 114,7 , 1057,3)

Let path P2 = (1 – 3 – 4 – 5 – 6 – 7 – 10) .

FCPM (P2) = SFT13 + SFT34 + SFT45 + SFT56 + SFT67 + SFT710

FCPM (P2) = (-886,3 , -86 , 103,2 , 908)

Let path P3 = (1 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 10) .


FCPM (P3) = (-1202,9 , -123,2 , 126,4 , 1206,6)

Let path P4 = (1 – 3 – 4 – 5 – 7 – 10) .

FCPM (P4) = SFT13 + SFT34 + SFT45 + SFT57 + SFT710

FCPM (P4) = (-727 , -52,4 , 92,6 , 764,7)

-

-

-

-

-

-

-

-

-

-

- +

+

+

+

+

+

+

+

+

+

+

+

-

-

-

14

Let path P5 = (1 – 3 – 4 – 5 – 7 – 8 – 10) .


FCPM (P5) = (-887,3 , -70,5 , 104,1 , 914)

Let path P6 = (1 – 3 – 4 – 5 – 7 – 8 – 9 – 10) .

FCPM (P6) = SFT13 + SFT34 + SFT45 + SFT57 + SFT78 + SFT89 +SFT910

FCPM (P6) = (-1043,6 , -89,6 , 115,8 , 1063,3)

Let path P7 = (1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 –10) .

FCPM (P7) = SFT12 + SFT23 + SFT34 + SFT45 + SFT56 + SFT67 + SFT78 + SFT810

FCPM (P7) = (-1199,4 , -121,3 , 128,7 , 1206,4)

Let path P8 = (1 – 2 – 3 – 4 – 5 – 6 – 7 – 10) .


FCPM (P8) = (-1039,1 , -103,2 , 117,2 , 1057,1)

Let path P9 = (1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 10) .


FCPM (P9) = (-1355,7 , -140,4 , 140,4 , 1355,7)

Let path P10 = (1 – 2 – 3 – 4 – 5 – 7 – 10) .


FCPM (P10) = (-879,8 , -69,6 , 106,6 , 913,8)

Let path P11 = (1 – 2 – 3 – 4 – 5 – 7 – 8 – 10) .


FCPM (P11) = (-1040,1 , -87,7 , 118,1 , 1063,1)

Let path P12 = (1 – 2 – 3 – 4 – 5 – 7 – 8 – 9 – 10) .

FCPM (P12) = SFT12 + SFT23 + SFT34 + SFT45 + SFT57 + SFT78 + SFT89 +SFT910

FCPM (P12) = (-1196,4 , -106,8 , 129,8 , 1212,4)

15

Step 9. Compare FCPM s and accept minimum number as FCPM.

FCPM and R (FCPM (Pi)) values of each possible path is shown in Table 4.

X1 = -1355,7 , X2 = 1355,7 , β = 0,2715

Step 10. Which FCPM of path is fewer, accept the path as critical path.

In a conclusion, R (FCPM (P9)) is minimum value of all possible paths’ FCPM s. So, the

critical path is 1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 10 in that case. Length of treatment period is

approximately between 37,3 and 52,9 minutes i.e. (22,6 , 37,3 , 52,9 ,172,9).

a b c d

(1 – 3 – 4 – 5 – 6 – 7 – 8 – 10) -1046,6 -104,1 114,7 1057,3 0,4288

(1 – 3 – 4 – 5 – 6 – 7 – 10) -886,3 -86 103,2 908 0,4382

(1 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 10) -1202,9 -123,2 126,4 1206,6 0,4203

(1 – 3 – 4 – 5 – 7 – 10) -727 -52,4 92,6 764,7 0,4506

(1 – 3 – 4 – 5 – 7 – 8 – 10) -887,3 -70,5 104,1 914 0,4404

(1 – 3 – 4 – 5 – 7 – 8 – 9 –10) -1043,6 -89,6 115,8 1063,3 0,4311

(1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 10) -1199,4 -121,3 128,7 1206,4 0,4209

(1 – 2 – 3 – 4 – 5 – 6 – 7 – 10) -1039,1 -103,2 117,2 1057,1 0,4296

(1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9 – 10) -1355,7 -140,4 140,4 1355,7 0,4130

(1 – 2 – 3 – 4 – 5 – 7 – 10) -879,8 -69,6 106,6 913,8 0,4412

(1 – 2 – 3 – 4 – 5 – 7 – 8 – 10) -1040,1 -87,7 118,1 1063,1 0,4317

(1 – 2 – 3 – 4 – 5 – 7 – 8 – 9 –10) -1196,4 -106,8 129,8 1212,4 0,4232

FCPM (Pi)R (FCPM(Pi))

Table 4. FCPM and R (FCPM (Pi)) values of each possible path

16

5. FUZZY PROGRAM EVALUATION AND REVIEW TECHNIQUE ( FPERT )

In that issue will be explained with previous case.

Step 1. Compute Mean for each activity with that equation :

m = ( a + b + c + d ) / 4

All means of activities are shown in Table 5.

Step 2. Compute Variance for each activity with that equation :

σ2 = [ 3 ( a2 + b2 + c2 + d2 ) – 2 ( ab + ac + ad + bc + bd + cd ) ] / 36

All variances of activities are shown in Table 6.

Activity a b c d Mean

A 0,5 1,2 1,5 4 1,800

B 1 1,8 2,1 4 2,225

C 0,7 1 1 2,5 1,300

D 2,5 5 7 9 5,875

E 0,4 0,5 0,8 1,4 0,775

F 8 15 28 134 46,250

G 1 10 10 125 36,500

H 5 6 8 9 7,000

I 5 8 8 14 8,750

J 1 2,1 2,5 5 2,650

K 5 6 6 8 6,250

Table 5. Mean Values of Each Activity

Table 6. Variance Values of Each Activity

Activity a b c d Mean Variance

A 0,5 1,2 1,5 4 1,800 0,7756

B 1 1,8 2,1 4 2,225 0,5386

C 0,7 1 1 2,5 1,300 0,2200

D 2,5 5 7 9 5,875 2,5764

E 0,4 0,5 0,8 1,4 0,775 0,0675

F 8 15 28 134 46,250 1163,6389

G 1 10 10 125 36,500 1166,3333

H 5 6 8 9 7,000 1,1111

I 5 8 8 14 8,750 4,7500

J 1 2,1 2,5 5 2,650 0,9522

K 5 6 6 8 6,250 0,5278

119,375 2341,4914

17

Note : ij is in Step 3 and 4 that means activities are on the critical path which has known.

And that case critical path includes all activities.

Step 3. Compute Standart Deviation of the Project with sum of variance of activities are on

critical path :

σ = ( ∑ σ2 ij )

1/2

In that case σ = ( 2341,4914 ) 1/2 = 48,389

Step 4. Compute Expected Project Completing Time with sum of mean of activities are on

critical path :

E ( Project ) = ∑ m ij

In that case m = 119,375

Step 5. To find probability of the project completed time is less than X value , Central Limit

Theorem will be used for it. Trapezoidal distribution is accepted to approximate normal

distribution. Normal distribution has two variables are mean and standart deviation and they

have been computed previous steps.

Z = ( X - µ ) / σ

Step 6. After Z value is computed, find probability with z table.

In that case, some probability calculation results are shown in Table 7.

< 80 < 100 < 120 < 140 < 160 < 180 < 200 < 220 < 240 < 280

Z value -0,814 -0,400 0,013 0,426 0,840 1,253 1,666 2,080 2,493 3,319

Probability 0,209 0,345 0,504 0,663 0,800 0,894 0,953 0,981 0,994 1,000

Table 7. Probabilities of the Project Completed Time is Less than X Values

18

6. CONCLUSION

Nowadays projects are more complex than being in the past and complexibility increases

uncertainty. In a conclusion deterministic approaches can’t afford to complex projects so

fuzzy numbers are more effective for high uncertainty situations. In the hospitals, time is the

most important issue so management department has to manage, control and plan the

projects with right numbers and fuzzy numbers provides to us to do that part right. In that

study, fuzzy numbers were explained and then, how you can use fuzzy numbers in a project

were explained with Fuzzy CPM (Critical Path Method) and Fuzzy PERT (Program Evaluation

and Review Technique). These tools provides to analyze flows and forecast project

completing time, and in healthcare sector, FCPM/FPERT provides to standardize care and

quality, reduce patient’s staying time. In addition, slack times between related activities can

be found with FCPM/FPERT tools and it provides to analyze space time and improve

productivity in the project.

19

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