Q1. At valley hospital nurses
beginning a new shift report to a central area to receive their primary
assignments. not every nurses is as efficient as another with particular kinds
of patients. Given the following patient roster care levels, and time estimates, assign nurses to patients
to optimize efficiency. Also, determine how long it will take for nurses to
complete their routine tasks on the shift.
|
atpatient
|
Care level
|
Nurse1
|
Nurse2
|
Nurse3
|
Nurse4
|
|
Jones
|
A2
|
3
|
5
|
4
|
3
|
|
hathaway
|
B2
|
2
|
1
|
3
|
2
|
|
bryant
|
B1
|
3
|
4
|
2
|
2
|
|
sweeney
|
A1
|
4
|
3
|
3
|
4
|
ANS
|
|
Nurse1
|
Nurse2
|
Nurse3
|
Nurse4
|
|
A2
|
0
|
2
|
1
|
0
|
|
B2
|
1
|
0
|
2
|
1
|
|
B1
|
1
|
2
|
0
|
0
|
|
A1
|
1
|
0
|
0
|
1
|
|
B2
|
N2
|
|
A1
|
N3
|
|
B1
|
N4
|
|
A2
|
N1
|
Time
take
N1 take
3
N2 take
1
N3 take
3
N4 take
2
Q2. Valley hospital wants to focus on
customer perceptions of quality , so it has asked its patients to evaluate the
nursing staff and indicate preference for assignment. Reassign the nursing
staff to obtain the highest customer approval rating possible .
|
Patient
|
Nurse1
|
Nurse2
|
Nurse3
|
Nurse4
|
|
A Jones
|
89
|
95
|
83
|
84
|
|
B Hathaway
|
88
|
80
|
96
|
85
|
|
C Bryant
|
87
|
92
|
82
|
84
|
|
D Sweeney
|
93
|
82
|
86
|
94
|
ANS
|
|
N1
|
N2
|
N3
|
N4
|
|
A
|
6
|
1
|
13
|
12
|
|
B
|
8
|
16
|
0
|
11
|
|
C
|
9
|
4
|
14
|
12
|
|
D
|
3
|
14
|
10
|
2
|
|
|
N1
|
N2
|
N3
|
N4
|
|
A
|
5
|
0
|
12
|
11
|
|
B
|
8
|
16
|
0
|
11
|
|
C
|
5
|
0
|
10
|
8
|
|
D
|
1
|
12
|
8
|
0
|
|
|
N1
|
N2
|
N3
|
N4
|
|
A
|
4
|
0
|
12
|
11
|
|
B
|
7
|
16
|
0
|
11
|
|
C
|
4
|
0
|
10
|
8
|
|
D
|
0
|
12
|
8
|
0
|
N1=
A
N2=C
N3=B
Q.no 3-
Ans -
|
Item
|
1
|
2
|
3
|
4
|
5
|
|
Chocolate Mint
|
30
|
18
|
26
|
17
|
15
|
|
Peanut Butter
|
23
|
22
|
32
|
25
|
30
|
|
Short Bread
|
17
|
31
|
24
|
22
|
29
|
|
Fudge Delight
|
28
|
19
|
13
|
18
|
23
|
|
Macarooons
|
23
|
14
|
16
|
20
|
27
|
Matrix :
|
30
|
18
|
26
|
17
|
15
|
|
23
|
22
|
32
|
25
|
30
|
|
17
|
31
|
24
|
22
|
29
|
|
28
|
19
|
12
|
18
|
23
|
|
23
|
14
|
16
|
20
|
27
|
|
15
|
3
|
11
|
2
|
0
|
|
1
|
0
|
10
|
3
|
8
|
|
0
|
14
|
7
|
5
|
12
|
|
15
|
6
|
0
|
5
|
10
|
|
9
|
0
|
2
|
6
|
13
|
Crosses
line not equal to order of matrix so,
 15 |
3
|
11
|
0
|
0
|
|
1
|
0
|
10
|
1
|
8
|
 0 |
14
|
7
|
3
|
12
|
 15 |
6
|
0
|
3
|
10
|
|
9
|
0
|
2
|
4
|
13
|
 15 |
3
|
11
|
0
|
0
|
 0 |
0
|
10
|
0
|
7
|
 0 |
15
|
7
|
3
|
12
|
 15 |
7
|
0
|
3
|
10
|
 9 |
0
|
2
|
4
|
13
|
|
Item
|
|
|
Chocolate Mint
|
5
|
|
Peanut Butter
|
4
|
|
Short Bread
|
1
|
|
Fudge Delight
|
3
|
|
Macarooons
|
2
|
Q.
no -4
Answer
–
|
Product
|
A
|
B
|
C
|
D
|
E
|
|
1
|
17
|
10
|
15
|
16
|
20
|
|
2
|
12
|
9
|
16
|
9
|
14
|
|
3
|
11
|
16
|
14
|
15
|
12
|
|
4
|
14
|
10
|
10
|
18
|
17
|
|
5
|
13
|
12
|
9
|
15
|
11
|
Matrix
:
|
3
|
10
|
5
|
4
|
0
|
|
8
|
11
|
4
|
11
|
6
|
|
9
|
4
|
6
|
5
|
8
|
|
6
|
10
|
10
|
2
|
3
|
|
7
|
8
|
11
|
5
|
9
|
|
3
|
10
|
5
|
4
|
0
|
|
4
|
7
|
0
|
7
|
2
|
|
5
|
0
|
2
|
1
|
4
|
|
4
|
8
|
8
|
0
|
1
|
|
2
|
3
|
6
|
0
|
4
|
|
A
|
B
|
C
|
D
|
E
|
 1 |
10
|
5
|
4
|
0
|
 2 |
7
|
0
|
7
|
2
|
 3 |
0
|
2
|
1
|
4
|
 2 |
8
|
8
|
0
|
1
|
 0 |
3
|
6
|
0
|
4
|
|
E
|
1
|
|
C
|
2
|
|
B
|
3
|
|
D
|
4
|
|
A
|
5
|
|
|
|
7.
Answer
Questn 9
|
s.p.t.
|
typing
|
proofing
|
|
1
|
-30
|
20
|
|
2
|
90
|
25
|
|
3
|
60
|
15
|
|
4
|
-45
|
30
|
|
5
|
75
|
60
|
|
6
|
-20
|
30
|
Sol.
|
Chap
|
Typing
|
Proofing
|
|
|
6
|
20
|
30
|
20
|
|
1
|
30
|
20
|
50
|
|
4
|
45
|
30
|
95
|
|
6
|
60
|
15
|
155
|
|
5
|
75
|
60
|
230
|
|
2
|
90
|
25
|
320
|
Total completion time=320 days
Avg. Flow time=(20+50+95+155+230+320)/6
=145 days.
Avg. No. Of jobs each in same days=1.22days
Avg. Job lateness=(0+20+50+95+155+230)/6
=91.6 days
10.
|
|
|
|
|
1
|
8
|
5
|
|
2
|
15
|
10
|
|
3
|
6
|
5
|
|
4
|
5
|
10
|
|
5
|
17
|
15
|
|
6
|
10
|
10
|
|
7
|
5
|
3
|
|
8
|
4
|
15
|
|
9
|
12
|
10
|
|
10
|
20
|
3
|
|
|
|
|
|
|
8
|
4
|
4
|
15
|
|
4
|
5
|
9
|
10
|
|
7
|
5
|
14
|
3
|
|
3
|
6
|
20
|
5
|
|
1
|
8
|
28
|
5
|
|
6
|
10
|
38
|
10
|
|
9
|
12
|
50
|
10
|
|
2
|
15
|
65
|
10
|
|
5
|
17
|
52
|
15
|
|
10
|
20
|
92
|
3
|
time completion=92 days
avg. Flow time=40.2
avg. No. Of jobs in same day=41.2
avg. Job lateness=32.1
Question 11-jobs processed
through Percy machine shop pass through three operation milling, grinding and
turning. the hours required for each of these operation is as follow:-
Sequence the operation and
make the Gantt chart:-
|
Jobs
|
Milling
|
Grinding
|
Turning
|
|
A
|
5
|
1
|
4
|
|
B
|
2
|
2
|
5
|
|
C
|
3
|
2
|
1
|
|
D
|
0
|
3
|
0
|
|
E
|
4
|
1
|
2
|
Q 12- Given that all the
jobs listed below must be machined first, then polished, determine a sequence
that will minimize the time required to process all 6 jobs. Chart the schedule
on a gantt chart and indicate the makespan.
|
Jobs
|
MACHINING
|
POLISHING
|
|
|
|
|
A
|
5.0
|
4.0
|
|
|
|
|
B
|
7.0
|
3.0
|
|
|
|
|
C
|
3.0
|
2.0
|
|
|
|
|
D
|
2.0
|
1.0
|
|
|
|
|
E
|
2.0
|
3.0
|
|
|
|
F 6.0 5.0
Ans- Idle time for M1= 1 min
Idle time for M2= 4 min
ASSIGNMENT
QUESTIONS
Q13 .Given
the prep. And painting time in days for following five houses.Chart out your sequence on gantt
charts and calculate the makespan.
|
Houses
|
Prep.
|
Paint
|
|
Addison
|
4
|
2
|
|
Brown
|
3
|
6
|
|
Clayton
|
6
|
4
|
|
Daniels
|
5
|
6
|
|
Ebersole
|
1
|
4
|
ANS—
Q14. Sequence the houses in previous problem by SPT
and SLACK.How much time is saved with johnsons rule ?
ANS-- SLACK
Time saved= 27-23=4
SPT
Time saved=25-23=2
Q.15 Job A,B,C, & D must be processed
through the same machine centre .Sequence
the following Jobs by
a.SPT
b. SLACK
|
JOB
|
PROCESSING TIME
|
DUE DATE
|
|
A
|
20
|
20
|
|
B
|
10
|
15
|
|
C
|
30
|
50
|
|
D
|
15
|
30
|
Calculate the mean flow time ,mean tardiness
and max tardiness .which sequencing rule would you recommend? Why?
Ans:
According to
SPT
|
JOB
|
PROCESSING TIME
|
DUE DATE
|
FLOW TIME
|
|
B
|
10
|
15
|
1 0
|
|
D
|
15
|
30
|
25
|
|
A
|
20
|
20
|
45
|
|
C
|
30
|
50
|
75
|
According to slack rule
|
JOB
|
PROCESSING TIME
|
DUE DATE
|
FLOW TIME
|
SLACK
|
|
A
|
20
|
20
|
45
|
0
|
|
B
|
10
|
15
|
1 0
|
5
|
|
D
|
15
|
30
|
25
|
15
|
|
C
|
30
|
50
|
75
|
20
|
|
RULE
|
TOTAL COMPLETION TIME
|
AVG. TIME FLOW
|
AVG. NO. OF JOBS
|
AVG. JOB LATENESS
|
AVG. JOB EARLINESS
|
NO. OF TARDI JOBS
|
|
SPT
|
75
|
38.7
|
2.066
|
12.5
|
2.5
|
3
|
|
SLACK
|
75
|
38.7
|
2.23
|
13.75
|
0
|
3
|
Mean tardiness = 0.75
Maximum tardiness = 3
Shortest Path time sequencing is preferred
because in this case avg. Job lateness is less and avg. Job earliness is more.
Q.16 Sequence the following jobs by (a)SPT
(b)DDATE, and (c)SLACK .Calculate the mean flow time mean tardiness and max tardiness .which
sequencing rule would you recommend? Why?
|
JOB
|
PROCESSING TIME
|
DUE DATE
|
|
A
|
5
|
8
|
|
B
|
3
|
5
|
|
C
|
9
|
18
|
|
D
|
6
|
7
|
Sol.
SPT:
|
JOB
|
PROCESSING TIME
|
DUE DATE
|
FLOW TIME
|
|
B
|
3
|
5
|
3
|
|
A
|
5
|
8
|
8
|
|
D
|
6
|
7
|
14
|
|
C
|
9
|
9
|
23
|
Avg. Flow time = (3+8+14+23)/4 = 8.75
No. of tardi jobs = 2
SLACK:
|
JOB
|
PROCESSING TIME
|
DUE DATE
|
FLOW TIME
|
|
C
|
9
|
9
|
9
|
|
D
|
6
|
7
|
15
|
|
B
|
3
|
5
|
18
|
|
A
|
5
|
8
|
23
|
Avg. Flow time = (9+15+18+23)/4 = 16.25
No. of tardi jobs = 3
DUE DATE:
|
JOB
|
PROCESSING TIME
|
DUE DATE
|
FLOW TIME
|
|
B
|
3
|
5
|
3
|
|
D
|
6
|
7
|
9
|
|
A
|
5
|
8
|
14
|
|
C
|
9
|
18
|
23
|
Avg. Flow time = (3+9+14+23)/4 = 12.25
No. of tardi jobs = 2
|
JOB
|
AVG. FLOW TIME
|
NO. OF TARDI JOBS
|
MAX. TARDINESS
|
|
SPT
|
8.75
|
2
|
14
|
|
SLACK
|
16.25
|
3
|
15
|
|
DUE DATE
|
12.25
|
3
|
6
|
17 Complete the report
|
period
|
1
|
2
|
3
|
4
|
5
|
total
|
|
Planned
input
|
50
|
55
|
60
|
65
|
65
|
295
|
|
Actual
input
|
50
|
50
|
55
|
60
|
65
|
280
|
|
Deviation
|
0
|
-5
|
-5
|
-5
|
0
|
-15
|
|
Planned
output
|
65
|
65
|
65
|
65
|
65
|
325
|
|
Actual
output
|
60
|
60
|
60
|
60
|
60
|
300
|
|
Deviation
|
-5
|
-5
|
-5
|
-5
|
-5
|
-15
|
|
Backlog
30
|
20
|
10
|
5
|
5
|
10
|
|
18
complete the table
|
Period
|
1
|
2
|
3
|
4
|
5
|
total
|
|
Planned
input
|
50
|
55
|
60
|
65
|
65
|
295
|
|
Actual
input
|
40
|
50
|
55
|
60
|
65
|
270
|
|
Deviation
|
-10
|
-5
|
-5
|
-5
|
0
|
-25
|
|
Planned
output
|
50
|
55
|
60
|
65
|
65
|
295
|
|
Actual
output
|
50
|
50
|
55
|
60
|
65
|
280
|
|
Deviation
|
0
|
-5
|
-5
|
-5
|
0
|
-15
|
|
Backlog
10
|
0
|
0
|
0
|
0
|
0
|
|
Q: 19
Kim jonshan R.N. the charge nurse of
the antepartum ward of city hospital , need help in scheduling the nurse
workforce for next work.
(1) Create the employee
schedule that will meet the demand requirements &garuntee each nurse 2 days off per week.
ANS:
|
Day of
week
|
Mo
|
Tu
|
We
|
Th
|
Fr
|
St
|
su
|
|
No of
nurse
|
3
|
3
|
4
|
5
|
4
|
3
|
3
|
|
Kim
jonshan
|
O
|
X
|
X
|
X
|
X
|
O
|
x
|
|
Tom
swan
|
O
|
X
|
X
|
X
|
X
|
O
|
x
|
|
Flo
cology
|
X
|
O
|
X
|
X
|
X
|
X
|
o
|
|
Shelli
|
X
|
O
|
X
|
X
|
X
|
X
|
o
|
|
Troung
|
X
|
X
|
O
|
X
|
O
|
X
|
x
|
(2) Revise the schedule so
that 2 days off are consecutive.
ANS:
|
Day of
week
|
Mo
|
Tu
|
We
|
Th
|
Fr
|
St
|
su
|
|
No of
nurse
|
3
|
3
|
4
|
5
|
4
|
3
|
3
|
|
Kim
jonshan
|
o
|
o
|
x
|
x
|
x
|
x
|
x
|
|
Tom
swan
|
O
|
O
|
x
|
X
|
X
|
X
|
X
|
|
Flo
cology
|
X
|
X
|
O
|
X
|
O
|
X
|
X
|
|
Shelli
|
X
|
X
|
X
|
X
|
X
|
O
|
o
|
|
Troung
|
x
|
x
|
X
|
x
|
x
|
o
|
o
|
Here o = off days
X= working day
Flo cology not have
consecutive off days .
Q 20:
Rosemary needs help in
scheduling volunteers . create a work schedule that will meet the requirements
,given that a volunteer will only work 4 days per week .try to make the work
consecutive.
|
Days of weeks
|
Mo
|
Tu
|
We
|
Th
|
Fr
|
St
|
Su
|
|
No. of volunteers
|
4
|
3
|
2
|
3
|
6
|
4
|
2
|
|
Rosemary
|
|
|
|
|
|
|
|
|
Albert
|
|
|
|
|
|
|
|
|
Richard
|
|
|
|
|
|
|
|
|
Gail
|
|
|
|
|
|
|
|
|
Shelly
|
|
|
|
|
|
|
|
|
Karen
|
|
|
|
|
|
|
|
ANS:
|
Days of weeks
|
Mo
|
Tu
|
We
|
Th
|
Fr
|
St
|
Su
|
|
No. of volunteers
|
4
|
3
|
2
|
3
|
6
|
4
|
2
|
|
Rosemary
|
O
|
O
|
O
|
X
|
X
|
X
|
X
|
|
Albert
|
O
|
O
|
O
|
X
|
X
|
X
|
X
|
|
Richard
|
X
|
O
|
O
|
X
|
X
|
X
|
O
|
|
Gail
|
X
|
X
|
O
|
O
|
X
|
X
|
O
|
|
Shelly
|
X
|
X
|
X
|
O
|
X
|
O
|
O
|
|
Karen
|
X
|
x
|
x
|
O
|
x
|
O
|
O
|
3 Consecutive offdays
possible for only 2 volenteer &for
remaining 2 consecutive off days .
Q21
|
Job
|
1
|
2
|
3
|
4
|
5
|
6
|
|
M/C 1
|
25
|
30
|
60
|
20
|
35
|
45
|
|
M/C
2
|
25
|
15
|
40
|
25
|
30
|
70
|
Find idle time, flow time and total time using gantt
chart.
Ans 1
Sequence-- 4 6 3 5 1 2
Idle time – machine 1
–30
Machine 2 – 40
Total time -- 245
Q22
|
Job
|
M1
|
M2
|
M3
|
M4
|
|
A
|
7
|
11
|
2
|
14
|
|
B
|
18
|
15
|
18
|
13
|
|
C
|
4
|
11
|
14
|
18
|
|
D
|
14
|
27
|
32
|
21
|
|
E
|
6
|
16
|
14
|
16
|
Ans
2
M1-M4
M1+M2--M3+M4
M1+M2+M3—M2+M3+M4
Q.no 10-
What the function are typically performed
by a production control department ?
Ans –
Production control department responsibilities
consist of
1. Loading—checking the
availability of material, machines, and labour. The MRP system
plans for material availability.
CRP converts the material plan into machine and
labour requirements, and projects
resource overloads and under loads. Production
control assigns work to individual
workers or machines, and then attempts to
smooth out the load to make the
MRP schedule "doable." Smoothing the load is
called load levelling.
2. Sequencing—releasing work
orders to the shop and issuing dispatch lists for individual
machines. MRP recommends when
orders should be released (hence the name,
planned order releases). After
verifying their feasibility, production control actually
Releases the orders. When several
orders are released to one machine center, they must
be prioritized so that the worker
will know which ones to do first. The dispatch list
contains the sequence in which
jobs should be processed. This sequence is often based
on certain sequencing rules.
3. Monitoring—maintaining progress
reports on each job until it is completed. This is
important because items may need
to be rescheduled as changes occur in the system.
In addition to timely data collection,
it involves the use of Gantt charts and input/output
control charts.
Q.no 11 –
Explain Johnsons rule and its Procedure .
Ans-Since few factories consist of
just one process, we might wonder if techniques exist that will
produce an optimal sequence for any
number of jobs processed through more than one
machine or process. Johnson's
rule finds the fastest way to process a series of jobs through a
two-step system in which every job
follows the same sequence through two processes. Based
on a variation of the SPT rule, it
requires that the sequence be "mapped out" to determine
the final completion time, or makespan,
for the set of jobs. The procedure is as follows:
1. List the time required to
complete each job at each process. Set up a one-dimensional
matrix to represent the desired
sequence with the number of slots equal to the number
of jobs.
2. Select the smallest processing
time at either process. If that time occurs at process 1,
put the associated job as near to
the beginning of the sequence as possible.
3. If the smallest time occurs at
process 2, put the associated job as near to the end of the
sequence as possible.
4. Remove the job from the list.
5. Repeat steps 2-4 until all
slots in the matrix have been filled or all jobs have been
sequenced.
Q.no 3-
Ans -
|
Item
|
1
|
2
|
3
|
4
|
5
|
|
Chocolate
Mint
|
30
|
18
|
26
|
17
|
15
|
|
Peanut
Butter
|
23
|
22
|
32
|
25
|
30
|
|
Short
Bread
|
17
|
31
|
24
|
22
|
29
|
|
Fudge
Delight
|
28
|
19
|
13
|
18
|
23
|
|
Macarooons
|
23
|
14
|
16
|
20
|
27
|
Matrix :
|
30
|
18
|
26
|
17
|
15
|
|
23
|
22
|
32
|
25
|
30
|
|
17
|
31
|
24
|
22
|
29
|
|
28
|
19
|
12
|
18
|
23
|
|
23
|
14
|
16
|
20
|
27
|
|
15
|
3
|
11
|
2
|
0
|
|
1
|
0
|
10
|
3
|
8
|
|
0
|
14
|
7
|
5
|
12
|
|
15
|
6
|
0
|
5
|
10
|
|
9
|
0
|
2
|
6
|
13
|
Crosses
line not equal to order of matrix so,
 15 |
3
|
11
|
0
|
0
|
|
1
|
0
|
10
|
1
|
8
|
 0 |
14
|
7
|
3
|
12
|
 15 |
6
|
0
|
3
|
10
|
|
9
|
0
|
2
|
4
|
13
|
 15 |
3
|
11
|
0
|
0
|
 0 |
0
|
10
|
0
|
7
|
 0 |
15
|
7
|
3
|
12
|
 15 |
7
|
0
|
3
|
10
|
 9 |
0
|
2
|
4
|
13
|
|
Item
|
|
|
Chocolate
Mint
|
5
|
|
Peanut
Butter
|
4
|
|
Short
Bread
|
1
|
|
Fudge
Delight
|
3
|
|
Macarooons
|
2
|
Q. no -4
Answer –
|
Product
|
A
|
B
|
C
|
D
|
E
|
|
1
|
17
|
10
|
15
|
16
|
20
|
|
2
|
12
|
9
|
16
|
9
|
14
|
|
3
|
11
|
16
|
14
|
15
|
12
|
|
4
|
14
|
10
|
10
|
18
|
17
|
|
5
|
13
|
12
|
9
|
15
|
11
|
Matrix :
|
3
|
10
|
5
|
4
|
0
|
|
8
|
11
|
4
|
11
|
6
|
|
9
|
4
|
6
|
5
|
8
|
|
6
|
10
|
10
|
2
|
3
|
|
7
|
8
|
11
|
5
|
9
|
|
3
|
10
|
5
|
4
|
0
|
|
4
|
7
|
0
|
7
|
2
|
|
5
|
0
|
2
|
1
|
4
|
|
4
|
8
|
8
|
0
|
1
|
|
2
|
3
|
6
|
0
|
4
|
|
A
|
B
|
C
|
D
|
E
|
 1 |
10
|
5
|
4
|
0
|
 2 |
7
|
0
|
7
|
2
|
 3 |
0
|
2
|
1
|
4
|
 2 |
8
|
8
|
0
|
1
|
 0 |
3
|
6
|
0
|
4
|
|
E
|
1
|
|
C
|
2
|
|
B
|
3
|
|
D
|
4
|
|
A
|
5
|
No comments:
Post a Comment