15
C H A P T E R
Short-Term Scheduling
DISCUSSION Q UESTIONS 1. Scheduling’s objective is to optimize the use of resources so that production objectives are met. 2. Four criteria for scheduling are: minimizing completion time, maximizing maximizing utilizat utilization, ion, minimizi minimizing ng work-in-proce work-in-process ss inventory, inventory, and minimizing minimizing customer customer waiting time. There is a one-to-one one-to-one correspondence between minimizing completion time and minimizing flowtime.
10. Four effectiveness measures for dispatching rules: average completion time, average number of jobs in the system, average job lateness, and utilization. utilization. 11. The assignment involves adding and subtracti subtracting ng assignment method involves appropriate numbers in the problem’s table in order to find the lowest opportunity cost for each assignment. The four steps are detailed in the chapter. 12.
1. 2. 3. 4.
Identify Identify the constraint constraints. s. Develop a plan for overcoming overcoming the identified constraints. constraints. Focus resource resourcess on accomplishin accomplishing g step 2. Reduce Reduce the effects effects of the constraint constraintss by off loading loading work or by expanding capacity. Make sure that the constraints are recognized by all those who can have impact upon them. 5. Once a set of constrai constraints nts is overcome, overcome, go back back to step 1 and identify new constraints.
3. Loading is the assignment of jobs to work processing centers. Work centers can be loaded by capacity or by assigning specific jobs to specific work centers. Gantt charts and the assignment method are loading techniques. 4.
Five priority sequencing rules are:
First First come, come, first first served served (FCFS) (FCFS);; or First First in, first out (FIFO): Jobs are sequenced in the order in which they arrive at the workstation. Earliest due date (EDD): Jobs are sequenced in the order in which they are due for delivery to the customer. Shortest Shortest processing processing time (SPT): Jobs are sequenced sequenced in order of the processing time required at the workstation, with with the job requir requiring ing the least processi processing ng time time at the workstation scheduled first. Longest Longest processing processing time (LPT): Jobs are sequenced sequenced in order of the processing time required at the workstation, with the job requiring the longest processing time at the workstation scheduled first. Criti Critica call rati ratio o (CR): (CR): Jobs Jobs are are sequ sequen ence ced d in orde orderr of increasing critical ratio (the ratio of time required by work left to be done to time left to do the work ).
5. SPT minimizes the average flow time, average lateness, and average number of jobs in the system. It maximizes the number of jobs completed at any point. The disadvantage is that long jobs are pushed back in the schedule. 6. A due date may range from a (meaningless) promise to a contractual obligation. It is a target with or without penalties. 7. Flow time is the length of time a job is in the system; lateness is completion time minus due date. 8. Most students will go for EDD, to gain minimum lateness. Others will go for SPT, on the grounds that the team can’t afford to tackle a job with an early due date and a long processing time. Interesting to see student assumption about sequence, damage, etc. 9. Johnson’s rule is used to sequence several jobs through two work centers.
200
Theory of constraints steps:
13.
Advantages of level material use are:
14.
Lower inventory costs Faster product throughput (shorter lead times) Improved component and product quality Reduced floor space requirement Improved Improved communicat communication ion between between employees employees because because they are closer together A smoother smoother production process because because large lots have no hidden problems
Techniques to deal with bottlenecks: 1. Increasin Increasing g capacity capacity of the constraint constraint 2. Ensuring Ensuring that well-trained well-trained and cross-trained cross-trained employees employees are available to operate and maintain the work center causing the constraint 3. Developing Developing alternate alternate routings, routings, processing processing procedures, procedures, or subcontractor 4. Moving inspect inspections ions and tests tests to a position just just before the constraint 5. Sche Schedul dulin ing g thro through ughpu putt to matc match h the the capa capaci city ty of the the bottleneck
The last technique may involve off-loading work from the bottleneck, scheduling less, meaning that the bottleneck itself gained no capacity, and no more throughput was accomplished through the bottleneck. 15. Input/output control keeps track of planned versus actual inputs and outputs, highlighting deviations and indicating bottlenecks. bottlenecks.
201
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ETHICA$ D I$E##A Students may not be aware of the issue from an employee’s or employer’s perspective, but scheduling the “graveyard” shift and achieving performance on the graveyard shift can be a problem under the best of circumstances. Rotating shifts is very difficult and the shorter the rotation cycle the worse—a one week cycle is a killer. Given the box the manager has gotten into, the manager may want to let the empowered employees move to a one-week rotation and simultaneously begin an education effort. The combination should result in the employees soon voting for a change to a more reasonable and intelligent schedule. As long as the manager is doing his best to educate the employees and the short-term rotation does not last long, we see no significant ethical problem. Not making employees aware of the literature and acquiescing to a one-week schedule does suggest an unethical position.
ACTI&E # ODE$ E 'ECISE
(b) Gantt load chart:
15.2
ACTIVE MODEL 15.1: Job Shop Sequencing 1. Which schedule (rule) minimizes the average completion time, maximizes the utilization and minimizes the average number of jobs in the system for this example? SPT 2. Use the scrollbar to change the processing time for job C and use the scrollbar to modify the due date for job C. Does the same rule always minimize the average completion time? Yes—SPT always minimizes the average completion time, maximizes utilization, and minimizes the average number of jobs in the system. 3. Which schedule (rule) minimizes the average lateness for this example? EDD
15.3
Original problem: Site*Cu+tome r
A
)
C
D
1 2 3 4
7 5 6 8
3 4 7 6
4 6 9 7
8 5 6 4
4. Use the scrollbar to change the due date for job C. Does the same rule always minimize the average lateness? No—if the due date for job C is 5, then FCFS and SPT minimize the average lateness.
Row subtraction is done next: A
)
C
D
END-O(-CHAPTE P O)$E#S
Site*Cu+tome r 1 2 3 4
4 1 0 4
0 0 1 2
1 2 3 3
5 1 0 0
15.1 (a) Gantt scheduling chart:
Column subtraction is done next: Site*Cu+tome r
A
)
C
D
1 2 3 4
4 1 0 4
0 0 1 2
0 1 2 2
5 1 0 0
Cover zeros with lines:
Optimal assignment: Taxi at post 1 to customer C
CHAPTE 1!
Taxi at post 2 to customer B Taxi at post 3 to customer A Taxi at post 4 to customer D Total distance traveled = 4 + 4 + 6 + 4 = 18 miles. 15.4
Cover zeros with lines:
(a) Assign: o
to
#,chine
1 2 3 4
# A C
Optimal assignment: Squad 1 to case C Squad 2 to case D Squad 3 to case B
(b) Total production: 40 [= 10 + 10 + 9 + 11] 15.5
15.6
A++ignment
,ting
C53 at %&ant 1 C81 at %&ant 3 #5 at %&ant 4 #44 at %&ant 2
10 !ents 4 !ents 30 !ents 14 !ents T'ta& (anu)a!turin* !'st 58
Convert the minutes into $: #,retin g
(in,nc e
O.er,tio n+
Hum,n e+ource+
Chris Steve
$80
$120
$125
$140
$20
$115
$145
$160
Juana
$40
$100
$85
$45
$65
$35
$25
$75
Ree!! a
The minimum-cost solution = Chris Steve Juana Rebecca 15.7
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Finance $120 Marketing $20 Human Resources $45 Operations $25 $210
Original problem: Su,d3C, +e
A
)
C
D
E
1 2 3 4 5
14 20 10 8 13
7 7 3 12 25
3 12 4 7 24
7 6 5 12 26
27 30 21 21 8
(a) Row subtraction is done next: Su,d3C, +e
A
)
C
D
E
1 2 3 4 5
11 14 7 1 5
4 1 0 5 17
0 6 1 0 16
4 0 2 5 18
24 24 18 14 0
Column subtraction is done next: Su,d3C, +e
A
)
C
D
E
1 2 3 4 5
10 13 6 0 4
4 1 0 5 17
0 6 1 0 16
4 0 2 5 18
24 24 18 14 0
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202
204
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Squad 4 to case A Squad 5 to case E 3+
Total person-days projected using this assignment 8 + 8 = 28 days.
=
3+ 6+
15.9 Because this is a maximization problem, each number is subtracted from 95. The problem is then solved using the minimization algorithm.
(a)
A++ignment +isher/)inan!e '&har/e!'n'(i!s Hu*/statisti!s Rusta*i/(ana*e(ent T'ta& ratin*
(b) We can avoid the assignment of squad 5 to case E occurring by assigning a very high value to that combination. In this case, we assign 50. Problem: Su,d3C,+ e
A
)
C
D
E
1 2 3 4 5
14 20 10 8 13
7 7 3 12 25
3 12 4 7 24
7 6 5 12 26
27 30 21 21 50
A
)
C
D
E
1 2 3 4 5
11 14 7 1 0
4 1 0 5 12
0 6 1 0 11
4 0 2 5 13
24 24 18 14 37
Column subtraction is done next:
15.10 o
A
)
C
D
E
1 2 3 4 5
11 14 7 1 0
4 1 0 5 12
0 6 1 0 11
4 0 2 5 13
10 10 4 0 23
Cover zeros with lines:
Due D,te
Dur,tion 9D,8+:
313 312 325 314 314
8 16 40 5 3
A C # E
(a) FCFS (first come, first served): Because all jobs arrived on day 275, and presumably in the order given, the FCFS sequence is: A, B, C, D, E (b)
Su,d3C, +e
EDD (earliest due date): o Seuence
Due D,te
A # E C
312 313 314 314 325
(c) SPT (shortest processing time): o Seuence E # A C
Optimal assignment: Squad 1 to case C Squad 2 to case D Squad 3 to case B Squad 4 to case E Squad 5 to case A Total person-days projected using this assignment 3 + 21 + 13 = 46. 15.8
95 75 85 80 335
(b) Since Fisher is not teaching statistics; the answer does not change. Total rating remains 335.
Row subtraction is done next: Su,d3C,+ e
,ting
Proce++ing Time 3 5 8 16 40
(d) LPT (longest processing time): o Seuence
=
C A # E
3+ 6+
The best pairs are assigned as follows: Ajay—Jackie Jack—Barbara Gray—Stella Raul—Dona Total compatibility score (overall) = 90 + 70 + 50 + 20 = 230
40 16 8 5 3
Scheduling A5er,ge ule +C+S E## SPT .PT
Proce++ing Time
A5er,ge
T,rdine+ (lo6 Time + 14,8 4,24,220,8
47,4 34,6 26,260,2
A5er,ge Numer o7 o+ in S8+tem 3,3 2,4 1,84,2
CHAPTE 1!
* = Best 15.11
o o
103 205 309 412 517
Due D,te
em,ining D,8+
214 223 217 219 217
Critic,l ,tio
10 7 11 5
1,40 3,29 1,55 3,80
15
1,13
Jobs should be scheduled in the sequence 517, 103, 309, 205, 412 if scheduled by the critical ratio scheduling rule. 15.12 o
em,ining Proce++ing Time
Due d,te
A C #
212 209 208 210
6 3 3 8
Minimize total lateness. Comparing the scheduling efficiency of the several algorithms presented in terms of lateness:
C # A
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Time
D,te
St,rt
End
3 3 8 6
208 209 210 212
205 208 211 219
207 210 218 224
(e)
Critical ratio:
o
Due D,te
em,ining Proce++ing Time
A
212
6
209
3
C
208
3
#
210
8
20;
D,8+ $,te
0 1 8 1 2 T'ta& 21 as
Critic,l ,tio 212 2056 1,17 209 2053 1,33 208 2053 1,00 210 2058 0,63
Need date − today s date ,
(a)
First come, first served (FCFS) :
Proce++ing o Time A C #
6 3 3 8
Critical ratio
Due D,te
St,rt
End
212 209 208 210
205 211 214 217
210 213 216 224
o C A #
St,rt
End
3 3 6 8
209 208 212 210
205 208 211 217
207 210 216 224
0,63 1,00 1,17 1,33
Critical ratio:
Shortest processing time (SPT): Due D,te
Critic,l ,tio
# C A
0 4 8 1 4 T'ta& 26 as
Proce++ing Time
Days required to complete job
o Seuence
D,8+ $,te
o
(b)
=
D,8+ $,te
0 2 4 1 4 T'ta& 20 as
# C A
Proce++ing Due Time D,te St,rt 8 3 6 3
210 208 212 209
205 213 216 222
End D,8+ $,te 212 215 221 224
2 7 9 1 5 T'ta& 33 as
A minimum total lateness of 20 days seems to be about the least we may achieve.
A5er,geA5er,ge o Seuence (lo6 Time Utili<,tion NumerNumer Schedulin A5er,ge A5er,ge ule o7 o+ o7 o+ g in C:RS;#ER ule 385 37,6< 2,66 S8+tem 10 End Avera*e D,8+ $,te $,tene++ (lo6 Time 212 .ateE##2 +C+S 375 6,5 11,8 2,4 SPT 38,6< 2,59 12 218 6 RC:S;#ER SPT 5,0 10,25 2,1 .PT R#ES;C:R 495 29,3< 3,41 44 221 13 +C+S C:R#ES;R .PT 390 37,2< 9,0 14,8 2,69 3,0 12 224 1 E## 5,25 10,8 2,2 Starting day 5 number: 241 (i.e., work can be done on day 241) Criti!a& rati' 8,3 14,0 2,8 T'ta& 36 Method: SPT—Shortest processing time as SPT is best on all criteria. Com.letion Proce++ing15.13 Due(a) D,te Order (lo6 Time Time $,te Time Di+.,tchin
(c) Longest processing time (LPT): g o # A C
Proce++ing Time
Due D,te
St,rt
8 6 3 3
210 212 208 209
205 213 219 222
(d) Earliest due date (EDD): Proce++ ing
Due
C:=01 R=02 #E=06 S;=11 R=05
25 15 35 30 40
270 300 320 310 360
2 1 4 3 5
40 15 105 70 14
280 255 345 310 385
10 0 25 0 25
20!
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Method: LPT—Longest processing time Proce++ing Time
Due D,te
Order
(lo6 Time
25 15 35 30 40 145
270 300 320 310 360
4 5 2 3 1
130 145 75 105 40 495 99
C:=01 R=02 #E=06 S;=11 R=05 T'ta& Avera*e
Com.letion Time
$,te
370 385 315 345 280
100 85 0 35 0 220 44
Se?uen!e R=05@#E=06@S;=11@C:=01@R=02@ Avera*e in sste( 3,414 495145
Method: Earliest due date (EDD); earliest to latest date Proce++ing Time
Due D,te
Sl,c
Order
25 15 35 30 4 0 145
270 300 320 310 360
0 0 0 0 0
1 2 4 3 5
C:=01 R=02 #E=06 S;=11 R=05
(lo6 Time
25 40 105 70 14 5 T'ta& 385 Avera*e 77 Se?uen!e C:=01@R=02@S;=11@#E=06@R=05 Avera*e in sste( 2,655 385145
Com.letion Time
$,te
265 280 345 310 385
0 0 25 0 25 50 10
Method: First come, first served (FCFS) Proce++ing Time C:=01 R=02 #E=06 S;=11 R=05
Due D,te
Sl,c
Order
(lo6 Time
Com.letion Time
$,te
270 300 320 310 360
0 0 0 0 0
1 2 3 4 5
25 40 75 105 14
265 280 315 345 385
0 0 0 35 25
25 15 35 30 4
(b)
The best flowtime is SPT; (c) best utilization is SPT;
(d)
best lateness is EDD. (e) Students could support either of these choices. LPT scores poorly on all three criteria.
15.14 o
D,te Production Order D,8+ ecei5ed Needed
> A r e r # u
110
20
D,te
180
(b)
EDD (earliest due date): o Seuence C A E #
Due D,te 175 180 200 210 230
(c) SPT (shortest processing time): o Seuence
(a) FCFS (first come, first served): o Seuence
D,te Order ecei5ed
A C # E
110 120 122 125 130
C # E A
Proce++ing Time 10 16 18 20 30
CHAPTE 1!
(d) LPT (longest processing time): o Seuence
Proce++ing Time
A E # C
30 20 18 16 10
Scheduling ule
A5er,ge T,rdine++
A5er,ge (lo6 Time
A5er,ge Numer o7 o+ in S8+tem
+C+S E## SPT .PT
5,4 0,0 7,2 9,6
60,0 54,4 47,6 65,2
3,2 2,9 2,5 3,5
EDD is best for average Lateness and SPT for the other two measures. 15.15
Johnson’s Rule finishes in 21 days, 2 days faster than the Firstcome, First-served schedule, which finishes in 23 days. 15.16 o
1 2 3 4 5
D,te Production o D,8+ ecei5ed Needed 215 220 225 240 250
30 20 40 50 20
D,te o Due
Critic,l ,tio
260 290 300 320 340
0,33 2,00 1,25 1,40 4,50
Jobs should be scheduled in the sequence 1, 3, 4, 2, 5 if scheduled by the critical ratio scheduling rule.
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20=
20>
15.17
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(a, c) The jobs should be processed in the sequence V–Y–U–Z–X–W–T, for a total time of 57. o Sho. Scheduling Printer )inder T D B : ; G
15 7 4 7 10 4 7
3 9 10 6 9 5 8
Order
(lo6 Time
Printer
)inder
54 15 4 39 32 8 22
57 28 14 51 45 19 3 6 57
seventh thir )irst siFth )i)th se!'n )'urth
Ti(e
(b, c)
Note: Y could also be placed first, with no change in total times. (d) Binding is idle from 0 to 4 and from 51 to 54 for a total of 7 hours. 15.18
o
O.er,tion 1 9Hour+:
A C # E +
O.er,tion 2 9Hour+:
10 7 5 3 2 4
5 4 7 8 6 3
Using Johnson’s rule, the optimal sequence is:
15.19
#ond,8 Tue+d,8?edne+d,8 Thur+d, (rid,8 S,turd, Sund,8 8 8 B'rer B'rer B'rer B'rer B'rer B'rer
1 2 3 4 5 6
6 5 4 3 2 1
5 4 3 3 2 1
5 4 3 3 2 2
5 4 3 2 1 1
B'rer 7
1
1
1
0
6 5 4 3 2 1
4 4 4 3 3 2
3 3 3 2 2 1
0
1
0
Note: Seven employees are needed; six have two consecutive days off. The 7th worker has two consecutive days off but only works 4 days. Days off are circled.
CHAPTE 1!
15.20
#ond, Tue+d, ?edne+d, Thur+d, (rid, S,turd, Sund, 8 8 8 8 8 8 8 B'rer 1
3
4
4
5
6
7
4
B'rer B'rer B'rer B'rer B'rer
2 3 4 5 6
3 3 2 1 1
3 3 3 2 1
3 2 2 2 1
4 3 2 2 1
5 4 3 2 1
6 5 4 3 2
4 3 2 1 1
B'rer 7
1
0
0
0
0
1
1
Note: Seven employees are needed: six have two consecutive days off. The 7th worker has 4 days off and only works 3.
15.21 Period P&anne in%ut A!tua& in%ut #eviati'n P&anne 'ut%ut A!tua& 'ut%ut #eviati'n a!&'* 30
1
2
4
;
Tot,l
80
80
100
100
360
85 5 90 85
85 5 90 85
85 15 90 80
85 15 90 80
340 20 360 330
5 30
5 30
10 35
10 40
30
Analysis: The completed input/output report shows that the Grinding work center did not process all the jobs that were available during the four periods; therefore, the desired output rate was not achieved. Also, rather than reducing the backlog, the backlog increased to 40 units.
S HOT- TE#
S CHEDU$IN%
20@
