operations management 4

Chapter 16

Scheduling

Teaching Notes

Scheduling techniques are designed to disaggregate the master production schedule into time-phased daily or hourly activities. A detailed production schedule must include when and where each activity must take place in order to meet the master schedule.

Scheduling involves the following major activities:

1. Short-run machine, manpower and production scheduling.

2. Short-run capacity planning.

3. Routing (determining where the work is going to be done).

4. Dispatching (issuing the order to begin work).

5. Expediting (speeding the progress of the work order) late, critical jobs.

6. Controlling the progress of orders and monitoring the process to determine that operations are running according to plan.

7. Determining the sequence of activities (determining when the work is to be done).

8. Revising the schedule based on changes in order status of jobs, material and/or capacity availability and various other reasons.

Elements of Scheduling Problems:

1. Job arrival patterns (static vs. dynamic). Dynamic arrival pattern means that more jobs will arrive in the system during the time those currently in the system are being processed. In a static system, all jobs that will ever enter the system are known. Most job shops are dynamic.

2. Ratio of workers to machines (machine limited vs. labor limited environment).

3. Priority rules for scheduling.

4. Flow patterns of jobs through the plant.

a. Flow shop: All jobs follow the same pattern of flow through the system. In a flow shop, routing is not typically a problem.

b. Job shop: Each job follows its own specified pattern.

c. Job shop is more difficult to analyze than the flow shop. It requires the storage of additional information.

5. Evaluation of the scheduling technique (maximization of service level, minimization of stockouts, WIP inventory, idle time, setup time, number of late jobs and the like). The evaluation process provides necessary feedback to the scheduler about the performance of the scheduling method.

6. Number and types of machines in the plant (as the number and types of machines increase, the scheduling environment gets more complicated.)

7. Facility layout (different layouts lead to different types of scheduling problems.)

Scheduling decision is based on the following critical factors:

1. Material availability.

2. Capacity availability (worker or machine).

3. Bottleneck vs. nonbottleneck operations.

4. Job priorities.

5. Queue of work before each work station.

A major portion of the chapter is devoted to production scheduling. I generally emphasize production scheduling because it has more “meat” than other areas of scheduling and because it enables students to get a better grasp on what scheduling involves and some of the tools available.

It is important to spend at least a few moments to establish the basic differences in scheduling needs among high volume, intermediate volume and job shop systems. It is also important to compare and contrast the scheduling needs of service operations and the scheduling needs of manufacturing operations (similarities as well as differences).

Answers to Discussion and Review Questions

1. Job shops are intended to handle a wide range of processing requirements; jobs tend to follow many different paths through the shop, and often differ significantly with respect to processing times. In addition, rush orders, changes in specs, and short planning horizons must be contended with. Consequently, scheduling can be exceedingly complex. Conversely, in continuous systems there is a high degree of uniformity which presents much less difficulty in terms of scheduling.

2. The main decision areas of job shop scheduling concern loading and sequencing.

3. Gantt charts are visual aids used by managers to plan and adjust facility and equipment loading. Advantages of Gantt charts include ease of manipulation, the fact that they provide a visual model of loading, and the ability to assist in trial and error changes. Disadvantages include the need to repeatedly update the chart, the inability to reveal costs associated with different alternatives and the inability to include other details (e.g., processing items which are dependent on the equipment being used rather than the same regardless of the equipment being used).

4. The assignment model assumes a one-for-one matching is possible, that costs for each combination are known and fixed, and that in general, each machine is capable of handling each job. (The last assumption is not strict, however. For example, certain combinations may be “undesirable” because certain jobs cannot be processed on certain pieces of equipment.)

5. a. FCFS: process jobs in order of arrival.

b. SPT: process jobs according to processing times, shortest ones first.

c. DD: process jobs by due date, earliest due dates first.

d. S/O: takes into account remaining processing time on all remaining operations for each job. Jobs with lowest slack per remaining operations are scheduled first.

e. Rush: Emergency or preferred customers first.

6. Priority rules allow for the fact that jobs are not equally important: different processing sequences will have different consequences for the organization. Priority rules enable an organization to emphasize those measures of effectiveness deemed most important.

7. Service systems must usually contend with random arrivals and variations in service times. Because of these factors, service systems loads tend to be uneven, making scheduling less effective.

8. Forward scheduling is when scheduling starts with a specific date and moves forward;

backward scheduling starts at a specific end date and moves backward. Forward scheduling shows the earliest finish date; backward scheduling shows the latest start date.

9. To the extent that scheduling efforts can achieve a balance in facility or equipment loading, the amount of output for a given amount of input will be higher than it otherwise would, and hence, productivity will be higher than otherwise. Conversely, poor scheduling will adversely affect productivity because it will not enable maximum use of resources.

10. Will throughput time decrease? Is it technically feasible to split a job? How disruptive will it be? What additional costs will be involved (e.g., setup, increased paperwork)?

11. Makespan is the total time needed to complete a group of jobs from the beginning of the first job to the completion of the last job.

Taking Stock

1. In sequencing jobs, we have to consider the trade-off between the customer service and operational efficiency. Sequencing job A before job B may result in higher customer service, because we could meet the due dates of both jobs, but this sequence may be inefficient due to the extra setup time involved.

Scheduling jobs, due to the variability in setup times, operation times, interruptions, cancellations and addition of orders, becomes a very difficult task to manage. Thus, a manager must constantly balance the trade-offs between customer service and operational efficiency. For example, scheduling the bottleneck operations before the non-bottleneck operations may improve throughput of the system, but may require additional labor and material movement and inefficient use of resources.

2. In large facilities, Production Planning and Control manager, or planners and schedulers are primarily responsible for scheduling. In small facilities, the plant manager may be involved in scheduling the jobs.

3. Technology, more specifically, the computers had a profound impact on scheduling. When the schedule is determined, it can be automatically released from manufacturing control to the shop floor, and the workers on the shop floor can get the machines and materials ready for production without wasting valuable time. In addition, the schedulers can automatically check to make sure that there is sufficient capacity and materials to produce the parts so that the schedule to produce the part can be released to the shop floor.

Critical Thinking Exercise

1.

a. The production bottleneck limits the production of the entire system due to insufficient capacity or excessive demand or some combination of both at a given work station. The throughput of the entire production system will depend on the output of the bottleneck workstation.

b. The small lot sizes will result in smaller transfer batches from one workstation to another. Since we can produce the smaller batch faster at a bottleneck workstation and transfer it to the next work station, the next work station can begin to work on the transfer batch sooner, experiencing less idle time. This process will eventually lead to higher throughput of the entire manufacturing system.

c. If we use small lot sizes for the purpose of reducing the bottleneck effect, it will cost us more due to more frequent setups, because we will have to produce additional batches. On the other hand, smaller lot sizes will result in less work-in process inventory and smaller transfer batches, which in turn will increase the throughput of the system.

d. As the bottlenecks shift from one workstation to another (bottleneck shiftiness), it becomes more difficult to identify and take appropriate action (reduce lot sizes, assign more resources, etc.) on bottleneck work stations. Therefore, it may be to our advantage to use smaller lot sizes in producing many different jobs at many different workstations because we simply do not know how long a bottleneck will remain a bottleneck. Also producing small batch sizes across many different machines and jobs rather than only on bottlenecks will improve the efficiency of the production process. However, on some occasions we may not want to reduce the batch size. The main reason for not using small batch sizes is a high setup cost.

2. Unless each customer or job is expected to have the same processing time requirements, scheduling at regular intervals will tend to create waiting lines. Ideally, appointments should be made with processing times taken into account. Obviously, it is not always possible to do this. One reason is the inability to closely estimate how much time each appointment will require. However, if appointments can be expected to all take about the same length of time (e.g., standard physical exams, driver’s test, eye exam), then scheduling at regular intervals is reasonable.

3. Student answers will vary.

Memo Writing Exercises

1. Job-shop scheduling is used in producing make-to-order products. These products are usually made in small batches, they have a wide variety of processing and setup requirements, material needs and processing sequences. The characteristics listed above make job shop scheduling very complex and difficult. None of these characteristics are present in the call-in service phone line.

The service phone line scheduling (employees answering technical questions about a product) is considered a scheduling issue for service operations. The scheduling of service operations is very different than job shop scheduling not only because a typical service operation does not have any of the job shop characteristics listed above, but also the service (customer phone calls) cannot be inventoried. Due to the nature of the service (requiring a response as soon as possible), appointment or reservation systems will not be appropriate. The company needs to estimate the number of service calls for different time periods. Based on the projected number of phone calls, an appropriate number of employees can be scheduled. The key issue is to balance the cost of waiting time and poor customer service with the cost of employee idle time.

2. The local approach to scheduling only considers the criteria and scheduling issues at a given workstation (department) without considering the priorities and scheduling issues at other workstations. The global approach to scheduling not only considers the scheduling priority at the workstation in question but also takes into account the priorities in other departments or workstations. The global approach to scheduling considers capacity limitations using finite loading. The local approach to scheduling uses infinite loading (assumes infinite capacity). An example of a local approach to scheduling would be using a single machine scheduling heuristic such as (earliest due date) in a job shop or flow shop environment.

Solutions:

Critical Ratio

Job A has the lowest critical ratio, therefore it is scheduled first and completed on day 14. After the completion of Job A, the revised critical ratios are:

Job C has the lowest critical ratio, therefore it is scheduled next and completed on day 21. After the completion of Job C, the revised critical ratios are:

Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27.

The critical ratio sequence is A–C–D–B and the makespan is 37 days.

jobs

67

.

2

37

99

jobs

of

number

Average

days

75

.

24

4

99

time

flow

Average

;

days

25

.

10

4

41

ness

Tardi

Average

=

=

=

=

=

=

	b.	times job time Flow center the at jobs of number Average jobs of Number tardy Days ness ardi t job Average jobs of Number time Flow time flow Average å = = =	FCFS	SPT	EDD	CR
			26.50	19.75	21.00	24.75
			11.0	6.00	6.00	9.25
			2.86	2.14	2.27	2.67

c. SPT is superior.

7.		FCFS: a–b–c–d–e
		SPT: c–b–a–e–d
		EDD: a–b–c–e–d
		CR: a–e–b–c–d

		FCFS:		Operation	Flow time	Due date	Hours
			Job	time (hr.)	(hr.)	(hr.)	tardy
			a	7	7	4	3
			b	4	11	10	1
			c	2	13	12	1
			d	11	24	20	4
			e	8	32	15	17
				32	87		26

		SPT:		Operation	Flow time	Due date	Hours
			Job	time (hr.)	(hr.)	(hr.)	tardy
			c	2	2	12	0
			b	4	6	10	0
			a	7	13	4	9
			e	8	21	15	6
			d	11	32	20	12
				32	74		27

		EDD:		Operation	Flow time	Due date	Hours
			Job	time (hr.)	(hr.)	(hr.)	tardy
			a	7	7	4	3
			b	4	11	10	1
			c	2	13	12	1
			e	8	21	15	6
			d	11	32	20	12
				32	84		23

Critical Ratio

Job	Processing Time (Hours)	Due Date	Critical Ratio Calculation
A	(.14 x 45) + .7 = 7	4	(4 – 0) / 7 = .57
B	(.25 x 14) + .5 = 4	10	(10 – 0) / 4 = 2.5
C	(.10 x 18) + .2 = 2	12	(12 – 0) / 2 = 6
D	(.25 x 40) + 1 = 11	20	(20 – 0) / 11 = 1.82
E	(.10 x 75) + .5 = 8	15	(15 – 0) / 8 = 1.88

Job A has the lowest critical ratio, therefore it is scheduled first and completed after 4 hours, the revised critical ratios are:

Job	Processing Time (Hrs.)	Due Date	Critical Ratio Calculation
A	–	–	–
B	4	10	(10 – 7) / 4 = 0.75
C	2	12	(12 – 7) / 2 = 2.50
D	11	20	(20 – 7) / 11 = 1.18
E	8	15	(15 – 7) / 8 = 1.00

Job B has the lowest critical ratio therefore it is scheduled next and it is completed after 11 hours (7 + 4). After the completion of Job B, the revised critical ratios are:

Job	Processing Time (Hours)	Due Date	Critical Ratio Calculation
A	–	–	–
B	–	–	–
C	2	12	(12 – 11) / 2 = 0.5
D	11	20	(20 – 11) / 11 = .81
E	8	15	(15 – 11) / 8 = .5

Job C and Job E are tied for the lowest critical ratio and Job C is arbitrarily selected and is scheduled next. Job C is completed in 2 hours bringing the total completion time to 11 + 2 = 13. After the completion of Job C, the revised critical ratios are:

Job	Processing Time (Hours)	Due Date	Critical Ratio Calculation
A	–	–	–
B		–	–
C	–	–	–
D	11	20	(20 – 13) / 11 = .63
E	8	15	(15 – 13) / 8 = .25

Job E has the lowest critical ratio therefore it is scheduled next. The critical ratio final sequence is A–B–C–E–D. Total completion of all six jobs (makespan) is 32 hours.

Critical Ratio sequence	Processing Time (Days)	Flow time	Due Date	Tardiness
A	7	7	4	3
B	4	11	10	1
C	2	13	12	1
E	8	21	15	6
D	11	32	20	12
(	32	84		23

jobs

63

.

2

32

84

jobs

of

number

Average

hours

8

.

16

5

84

time

flow

Average

;

6

.

4

5

23

ness

ardi

t

Average

=

=

=

=

=

=

		times job time Flow center the at jobs of number Average jobs of Number late ours H ness ardi t job Average jobs of Number time Flow time flow Average å = = =	FCFS	SPT	EDD	CR
			17.40	14.80	16.80	16.8
			5.20	5.4	4.60	4.6
			2.72	2.31	2.63	2.63

8.	a.	(1) FCFS: A–B–C–D–E
		(2) S/O: B–D–C–A–E OR D–B–C–A–E [see below]

			Time	Due date		Remaining number
		Job	(days)	(days)	Slack	of operations	Ratio	Rank
		A	8	20	12	2	6.0	4
		B	10	18	8	4	2.0	1,2 (tie)
		C	5	25	20	5	4.0	3
		D	11	17	6	3	2.0	1,2 (tie)
		E	9	35	26	4	6.5	5

b. S/O: [Assume B–D–C–A–E]

			Time	Flow time	Due date	Days
		Job	(days)	(days)	(days)	tardy
		B	10	10	18	0
		D	11	21	17	4
		C	5	26	25	1
		A	8	34	20	14
		E	9	43	35	8
			43	134		27

			Time	Flow time	Due date	Days
		Job	(days)	(days)	(days)	tardy
		A	8	8	20	0
		B	10	18	18	0
		C	5	23	25	0
		D	11	34	17	17
		E	9	43	35	8
			43	126		25

					FCFS	S/O
		Average flow time =	flow time:		25.20	26.80
			number of jobs
		Average number of jobs in the system =	flow time:	2.93	3.12
			(job times

9.

	Time (hr.)
Order	Step 1	Step 2
A	1.20	1.40
B	0.90	1.30
C	2.00	0.80
D	1.70	1.50
E	1.60	1.80
F	2.20	1.75
G	1.30	1.40

		Sequence of assignment:
		.80	[C]	last (i.e., 7th)
		.90	[B]	first
		1.20	[A]	2nd
		1.30	[G]	3rd
		1.60	[E]	4th
		1.50	[D]	6th
		1.75	[F]	5th

Thus, the sequence is b-a-g-e-f-d-c.

10.	a.	Job		Machine A			Machine B
		a		16			5	7
		b		3	2		13		Thus, the sequence is e–b–g–h–d–c–a–f.
		c		9			6	6
		d		8			7	5
		e		2	1		14
		f		12			4	8
		g		18			14	3
		h		20			11	4
	b.

0 2 5 23 43 51 60 76 88

e

B

g

h

d

c

a

f

e

b

g

h

d

c

a

f

0 2 16 29 43 54 61 67 76 81 88 92

c. Original idle time for B: 2 + 9 + 7 = 18 hrs., and original makespan is 92.

The last two tasks in the sequence are a and f. Splitting both of their completion times evenly, we get the following results.

Machine 1 Machine 2

a1 8 2.5

a2 8 2.5

f1 6 2

f2 6 2

After splitting, we get the following Gantt chart:

0 2 5 23 43 51 60 68 76 82 88

e

b

g

h

d

c

a1

a2

f1

f2

e

b

g

h

d

c

a1

a2

f1

f2

0 2 16 29 43 54 61 67 68 70.5 76 78.5 84 90

After splitting, idle time is: 2 + 1 + 5.5 + 3.5 + 4 = 16 hours, and the new makespan = 90.

There is a savings of 2 hr.

				Time (minutes)
11.	a.	Job		Center 1			Center 2
		A		20	2		27
		B		16	1		30		Thus, the sequence is B–A–C–E–F–D.
		C		43	3		51
		D		60			12	6
		E		35			28	4
		F		42			24	5
	b.

0 16 36 79 114 156 216

		B	A	C	E	F	D
			B	A		C	E	F		D

0 16 46 73 79 130 158 182 216 228

Idle time of 56 hours.

12.	a.	Job		Station A			Station B
		a		27	2		45
		b		18	1		33		Thus, the sequence is b–a–c–d–e.
		c		70			30	3
		d		26			24	4
		e		15			10	5

0 18 45 115 141 156

		b	a	c	d	e
			b	a		c	d	e

0 18 51 96 115 145 169 179

	b.	The Idle time for Station B is = 18 + 19 = 37 minutes.
	c.	Jobs B, A, C, D and E are candidates for splitting in order to reduce throughput time and idle time.

141 156

0 9 18 31.5 45 80 115 128 148.5

B

18

A

27

C 70

D

26

E

15

B 33

A 45

C 30

D 24

E 10

0 9 42 87 117 128 152 162

		Throughput time is 162 minutes, reducing this time by 17 minutes. The idle time for B of 20 minutes has decreased by 17 minutes.

13.

Determine job times from the schedule table, and then use Johnson’s Rule to sequence the jobs. The job times are:

		Job	A	B	C	D	E	F	G
		Cutting	2	4	5	4	2	3	1
		Polishing	3	3	2	5	3	1	4

Using Johnson’s Rule, we obtain:

			Cutting	Polishing
		Job	Start	Finish	Start	Finish
		G	0	1	1	5
		A	1	3	5	8
		E	3	5	8	11
		D	5	9	11	16
		B	9	13	16	19
		C	13	18	19	21
		F	18	20	21	22

Note: The order of Jobs A and E can be reversed with no effect on times.

14.	a.,b.
		SPT	Grinding	Deburring
		Job	Start	Finish	Start	Finish
		C	0	1	1	6
		B	1	3	6	10
		A	3	6	10	16
		D	6	10	16	19
		G	10	16	19	21
		F	16	24	24	31
		E	24	33	33	37
				93		140

The Grinding flow time is 93 hours and Deburring flow time is 140 hours. The Total time is 37 hours.

c. Johnson’s Rule

			Grinding	Deburring
		Job	Start	Finish	Start	Finish
		C	0	1	1	6
		B	1	3	6	10
		A	3	6	10	16
		F	6	14	16	23
		E	14	23	23	27
		D	23	27	27	30
		G	27	33	33	35
				107		147

The Grinding flow time is 107 hours and Deburring flow time is 140 hours. The Total time is 35 hours.

d. The tradeoff is between shorter flow time in the Grinding and Deburring departments and shorter total processing time. Ed would be indifferent if the benefit to be gained by shorter total processing time was equal to the cost of additional flow time in the Grinding and Deburring departments.

15.	a.	FCFS:						SPT:
			Job	Flow	Due	Days			Job	Flow	Due	Days
		Job	time	time	date	tardy		Job	time	time	date	tardy
		a	4.5	4.5	10	0		d	1.6	1.6	27	0
		b	6.0	10.5	17	0		e	2.8	4.4	18	0
		c	5.2	15.7	12	3.7		f	3.3	7.7	19	0
		d	1.6	17.3	27	0		a	4.5	12.2	10	2.2
		e	2.8	20.1	18	2.1		c	5.2	17.4	12	5.4
		f	3.3	23.4	19	4.4		b	6.0	23.4	17	6.4
			23.4	91.5		10.2			23.4	66.7		14.0
		EDD:						CR:
			Job	Flow	Due	Days			Job	Flow	Due	Days
		Job	time	time	date	tardy		Job	time	time	date	tardy
		a	4.5	4.5	10	0		a	4.5	4.5	10	0
		c	5.2	9.7	12	0		c	5.2	9.7	12	0
		b	6.0	15.7	17	0		b	6.0	15.7	17	0
		e	2.8	18.5	18	.5		e	2.8	18.5	18	0.5
		f	3.3	21.8	19	2.8		f	3.3	21.8	19	2.8
		d	1.6	23.4	27	0		d	1.6	23.4	27	0
			23.4	93.6		3.3			23.4	93.6		3.3

Critical Ratio

Job	Processing Time (Days)	Due Date	Critical Ratio Calculation
A	4.5	10	(10 – 0) / 4.5 = 2.22
B	6.0	17	(17 – 0) / 6.0 = 2.83
C	5.2	12	(12 – 0) / 5.2 = 2.31
D	1.6	27	(27 – 0) / 1.6 = 16.88
E	2.8	18	(18 – 0) / 2.8 = 6.43
F	3.3	19	(19 – 0) / 3.3 = 5.76

Job A has the lowest critical ratio, therefore it is scheduled first and completed after 4.5 days. The revised critical ratios are:

Job	Processing Time (Hrs.)	Due Date	Critical Ratio Calculation
A	–	–	–
B	6.0	17	(17 – 4.5) / 6.0 = 2.08
C	5.2	12	(12 – 4.5) / 5.2 = 1.44
D	1.6	27	(27 – 4.5) / 1.6 = 14.06
E	2.8	18	(18 – 4.5) / 2.8 = 4.82
F	3.3	19	(19 – 4.5) / 3.3 = 4.39

Job C has the lowest critical ratio, therefore it is scheduled next and it is completed after 9.7 days (4.5 + 5.2). After the completion of Job C, the revised critical ratios are:

Job	Processing Time (Hrs.)	Due Date	Critical Ratio Calculation
A	–	–	–
B	6.0	17	(17 – 9.7) / 6.0 = 1.22
C	–	–	–
D	1.6	27	(27 – 9.7) / 1.6 = 10.81
E	2.8	18	(18 – 9.7) / 2.8 = 2.96
F	3.3	19	(19 – 9.7) / 3.3 = 2.82

Job B has the lowest critical ratio, therefore it is scheduled next and it is completed after 15.7 days (9.7 + 6). After the completion of Job B, the revised critical ratios are:

Job	Processing Time (Hrs.)	Due Date	Critical Ratio Calculation
A	–	–	–
B	–	–	–
C	–	–	–
D	1.6	27	(27 – 15.7) / 1.6 = 7.06
E	2.8	18	(18 – 15.7) / 2.8 = 0.82
F	3.3	19	(19 – 15.7) / 3.3 = 1.0

Job E has the lowest critical ratio, therefore it is scheduled next and it is completed after 18.5 days (15.7 + 2.8). After the completion of Job E, the revised critical ratios are:

Job	Processing Time (Hrs.)	Due Date	Critical Ratio Calculation
A	–	–	–
B	–	–	–
C	–	–	–
D	1.6	27	(27 – 18.5) / 1.6 = 5.31
E	–	–	–
F	3.3	19	(19 – 18.5) / 3.3 = 0.15

Job F has the lowest critical ratio therefore it is scheduled next and it is completed after 21.8 days (18.5 + 3.3). The final critical ratio sequence is A–C–B–E–F–D. Total completion of all six jobs (makespan) is 23.4 days.

Critical Ratio sequence	Processing Time (Days)	Flow time	Due Date	Lateness	Tardiness
A	4.5	4.5	10	–5.5	0
C	5.2	9.7	12	–2.3	0
B	6.0	15.7	17	–1.3	0
E	2.8	18.5	18	0.5	0.5
F	3.3	21.8	19	2.8	2.8
D	1.6	23.4	27	–3.6	0
(	23.4	93.6		–9.4	3.3

Rule

Average

=

days late

Average flow

=

flow time

Average

=

flow time

tardiness

no. of jobs

time

no. of jobs

no. of jobs

job time

FCFS

10.2/6 =

1.7 days

91.5/6 =

15.25 days

91.5/23.4 =

3.91 jobs

SPT

14.0/6 =

2.33 days

66.7/6 =

11.117 days

66.7/23.4 =

2.85

EDD

3.3/6 =

0.55 days.

93.6/6 =

15.60 days

93.6/23.4 =

4.00

CR

3.3/6 =

.55 days

93.6/6 =

15.60 days

93.6/23.4 =

4.00

b. There are several ways to show this. One is to calculate the ratio of average flow time to average number of jobs for each rule and then observe that they are equal. Here the ratios are approximately 3.90. [Slight differences in ratios may arise due to rounding.]

c. S/O job sequence is a-c-b-e-d-f

16.
		Job	Remaining processing time	Due date	Slack	Remaining number of operations	Slack ( Remaining number of operations	Rank
		a	5	8	3	2	1.50	4
		b	6	5	–1	4	–0.25	1
		c	9	10	1	4	0.25	2
		d	7	12	5	3	1.67	5
		e	8	10	+2	2	+1.00	3
				Using the S/O rule, the sequence is B–C–E–A–D
17.		FCFS
			Job time	Due date	Flow time	Tardy
		Job	(hr.)	(hr.)	(hr.)	(hr.)
		a	3.5	7	3.5	0
		b	2.0	6	5.5	0
		c	4.5	18	10.0	0
		d	5.0	22	15.0	0
		e	2.5	4	17.5	13.5
		f	6.0	20	23.5	3.5
			23.5		75.0	17.0

		SPT
		Job	Job time	Flow time	Due date	Tardy
		b	2.0	2.0	6	0
		e	2.5	4.5	4	0.5
		a	3.5	8.0	7	1
		c	4.5	12.5	18	0
		d	5.0	17.5	22	0
		f	6.0	23.5	20	3.5
			23.5	68.0		5.0

		EDD
		Job	Job time	Flow time	Due date	Tardy
		e	2.5	2.5	4	0
		b	2.0	4.5	6	0
		a	3.5	8.0	7	1
		c	4.5	12.5	18	0
		f	6.0	18.5	20	0
		d	5.0	23.5	22	1.5
			23.5	69.5		2.5

Critical Ratio

Job	Processing Time (Days)	Due Date	Critical Ratio Calculation
A	3.5	7	(7 – 0) / 3.5 = 2.0
B	2.0	6	(6 – 0) / 2.0 = 3.0
C	4.5	18	(18 – 0) / 4.5 = 4.0
D	5.0	22	(22 – 0) / 5.0 = 4.4
E	2.5	4	(4 – 0) / 2.5 = 1.6
F	6.0	20	(20 – 0) / 6 = 3.33

Job E has the lowest critical ratio, therefore it is scheduled first and completed after 2.5 hours. The revised critical ratios are:

Job	Processing Time (Hrs.)	Due Date	Critical Ratio Calculation
A	3.5	7	(7 – 2.5) / 3.5 = 1.29
B	2.0	6	(6 – 2.5) / 2.0 = 1.75
C	4.5	18	(18 – 2.5) / 4.5 = 3.44
D	5.0	22	(22 – 2.5) / 5.0 = 3.90
E	–	–	–
F	6.0	20	(20 – 2.5) / 6 = 2.92

Job A is scheduled next because Job A has the lowest critical ratio. Job A will be completed after 6 hours (2.5 + 3.5). After the completion of Job A, the revised critical ratios are:

Job	Processing Time (Hrs.)	Due Date	Critical Ratio Calculation
A	–	–	–
B	2.0	6	(6 – 6) / 2.0 = 0
C	4.5	18	(18 – 6) / 4.5 = 2.67
D	5.0	22	(22 – 6) / 5.0 = 3.20
E	–	–	–
F	6.0	20	(20 – 6) / 6 = 2.33

Since Job B has the lowest critical ratio, it is scheduled next and it is completed after 8 hours (6 + 2). After the completion of Job B, the revised critical ratios are:

Job	Processing Time (Hrs.)	Due Date	Critical Ratio Calculation
A	–	–	–
B	–	–	–
C	4.5	18	(18 – 8) / 4.5 = 2.22
D	5.0	22	(22 – 8) / 5.0 = 2.80
E	–	–	–
F	6.0	20	(20 – 8) / 6 = 2.00

Since Job F has the lowest critical ratio, it is scheduled next and it will be completed after 14 hours (8 + 6). After the completion of Job F, the revised critical ratios are:

Job	Processing Time (Hrs.)	Due Date	Critical Ratio Calculation
A	–	–	–
B	–	–	–
C	4.5	18	(18 – 14) / 4.5 = 0.89
D	5.0	22	(22 – 14) / 5.0 = 1.60
E	–	–	–
F	–	–	–

Since Job C has the lowest critical ratio, it is scheduled next. Job C will be completed after 18.5 hours. The final critical ratio sequence of all jobs is E–A–B–F–C–D. Total completion of all six jobs (makespan) is 23.5 hours.

		Job	Critical ratio	Job time	Flow time	Due date	Tardy
		e	1.6	2.5	2.5	4	0
		a	2.0	3.5	6.0	7	0
		b	3.0	2.0	8.0	6	2
		f	3.3	6.0	14.0	20	0
		c	4.0	4.5	18.5	18	.5
		d	4.4	5.0	23.5	22	1.5
				23.5	72.5		4.0

hours

08

.

12

6

5

.

72

time

flow

Average

hours

67

.

6

4

ardiness

t

Average

=

=

=

=

			FCFS	SPT	EDD	CR
		Average flow time	12.5	11.33	11.58	12.08
		Average job tardiness	2.83	0.83	0.42	.67

18.	a.
		Order		Job time
		A		16 x	4 =	64
		B		6 x	12 =	72
		C		10 x	3 =	30
		D		8 x	10 =	80
		E		4 x	1 =	4

		DD
		Job	Job time	Flow time	Due date	Tardiness
		A	64	64	160	0
		C	30	94	180	0
		D	80	174	190	0
		B	72	246	200	46
		E	4	250	220	30
			250	828		76

b. Average job tardiness = 76/5 = 15.2 minutes

c. Average number of jobs in the system = 828/250 = 3.31

	d.	SPT
		Job	Job time	Flow time	Due date	Tardiness
		E	4	4	220	0
		C	30	34	180	0
		A	64	98	160	0
		B	72	170	200	0
		D	80	250	190	60
						60

Average job tardiness = 60/5 = 12 minutes

19.		Sequence	Setup times	Total
		A–B–C	2 + 3 + 2 =	7 (best)
		A–C–B	2 + 5 + 3 =	10
		B–A–C	3 + 8 + 5 =	16
		B–C–A	3 + 2 + 4 =	9
		C–A–B	2 + 4 + 3 =	9
		C–B–A	2 + 3 + 8 =	13

20.		Sequence	Setup times	Total
		A–B–C	2.4 + 1.8 + 1.4 =	5.6
		A–C–B	2.4 + 2.2 + 1.3 =	5.9
		B–A–C	3.2 + 0.8 + 2.2 =	6.2
		B–C–A	3.2 + 1.4 + 2.6 =	7.2
		C–A–B	2.0 + 2.6 + 1.8 =	6.4
		C–B–A	2.0 + 1.3 + 0.8 =	4.1 (best)

21.		Sequence	Setup times	Total
		A–B–C–D	2 + 5 + 3 + 2 =	12
		A–B–D–C	2 + 5 + 2 + 6 =	15
		A–D–B–C	2 + 4 + 3 + 3 =	12
		A–D–C–B	2 + 4 + 6 + 2 =	14
		B–A–D–C	1 + 7 + 4 + 6 =	18
		B–C–D–A	1 + 3 + 2 + 4 =	10 (best)
		C–B–A–D	3 + 2 + 7 + 4 =	16
		C–B–D–A	3 + 2 + 2 + 4 =	11
		C–D–A–B	3 + 2 + 4 + 5 =	14
		C–D–B–A	3 + 2 + 3 + 7 =	15
		D–A–B–C	2 + 4 + 5 + 3 =	14
		D–C–B–A	2 + 6 + 2 + 7 =	17

22. Each period’s backlog is equal to actual input – actual output. That amount is added to (or subtracted from) the previous backlog to obtain the current (shown) backlog for the period.

					Period
	Input		1	2	3	4	5
		Planned	24	24	24	24	20
		Actual	25	27	20	22	24
	Output	Planned	24	24	24	24	23
		Actual	24	22	23	24	24
	Backlog	12	13	18	15	13	13

23. Period

		1	2	3	4	5	6
Input	Planned	200	200	180	190	190	200
	Actual	210	200	179	195	193	194
	Deviation	+10	0	-1	+5	+3	-6
	Cum. Dev.	+10	+10	+9	+14	+17	+11

Period

		1	2	3	4	5	6
Output	Planned	200	200	180	190	190	200
	Actual	205	194	177	195	193	200
	Deviation	+5	-6	-3	+5	+3	0
	Cum. Dev.	+5	-1	-4	+1	+4	+4

Backlog

7

12

18

20

20

20

14

24. Day Mon Tue Wed Thu Fri Sat

Staff needed 2 3 1 2 4 3

Worker 1 2 3 1 2 4 3

Worker 2 1 2 1 2 3 2 (tie)

Worker 3 0 2 1 1 2 1

Worker 4 0 1 0 0 1 1 Part-time worker

No. working: 2 3 1 2 4 3

25. Day Mon Tue Wed Thu Fri Sat

Staff needed 3 4 2 3 4 5

Worker 1 3 4 2 3 4 5

Worker 2 2 3 2 3 3 4 (tie)

Worker 3 1 3 2 2 2 3 (tie)

Worker 4 0 2 1 2 2 2

Worker 5 0 2 0 1 1 1 (part-time worker)

Worker 6 0 1 0 1 0 0 (tie) (part-time worker)

No. working: 3 4 2 3 4 5

26.

Day Mon Tue Wed Thu Fri Sat

Staff needed 4 4 5 6 7 8

Worker 1 4 4 5 6 7 8

Worker 2 4 4 4 5 6 7 (tie)

Worker 3 3 4 4 4 5 6

Worker 4 3 4 3 3 4 5

Worker 5 2 3 3 3 3 4

Worker 6 2 3 2 2 2 3 (tie)

Worker 7 1 2 2 2 1 2 (tie)

Worker 8 0 1 1 1 1 2

Worker 9 0 1 0 0 0 1 (tie) Part-time worker

No. working: 4 4 5 6 7 8

Case: Hi-Ho Yo-Yo Case Study Grading Guide

Technical

Did student add setup time to the production time for each order and change due dates to days

From the beginning of schedule?

Date Order Setup Production Due

Job Received Time Time w/ Setup Date

A 6/4 2 hrs. 6.25 days 7

B 6/7 4 hrs. 2.5 days 5

C 6/12 2 hrs. 8.25 days 19

D 6/14 4 hrs. 3.5 days 14

E 6/15 4 hrs. 9.5 days 23

Did the student use all the heuristics available in the templates to evaluate the sequences?

Did the student properly evaluate the results?

SPT yields the lowest average flow time and number of jobs in the system.

DD yields the lowest average job tardiness.

FCFS and CR are tied for all three measures.

Rule	Sequence	Average flow time	Average tardiness	Average no. of jobs late
FCFS SPT DD CR	A-B-C-D-E B-D-A-C-E A-B-C-D-E A-B-C-D-E	16.50 14.25 14.80 16.50	3.45 2.75 2.05 3.45	2.75 2.38 2.47 2.75

Did the student discuss tradeoffs, and make and justify a recommendation?

Managerial/Editorial

Was report organized professionally?

Was report appropriate for intended reader (i.e., Jeff Baker)?

Were there more than a few typographical, spelling, and grammatical errors?

Were technical aspects (methods/results/etc.) explained appropriately?

Were outside sources (if used) appropriately cited?

Enrichment Module: Runout Time Method

Make-to-stock companies produce different products on a common machine or an operation. For example, a paint manufacturing company may decide to mix different colors of paint using the same “mixer”. In this scenario, the plant manager has to decide how much of different colors of paint to produce in each batch and the sequence of production. This decision is generally made based on the current level of inventory, production rate associated with a particular product and the rate of demand. The optimal lot size can be determined using the production lot size model covered in the inventory management chapter which balances the trade-off between carrying cost and the setup cost. However, when several products share common machinery for production, the batch sizes my need to be modified because the product sequencing may also affect cost. If the setups are sequence dependent, sequence of the items may be important in making the sequencing decision at the common work center or machine. In addition, the economic production model covered in the inventory chapter does not take into account the level of inventory. Therefore, using the economic production model may result in too much inventory in for some items and not enough inventory for other items. The economic production model also is not designed for scheduling specific work centers. The “runout time” method of scheduling can be utilized to determine production runs for a group of items that share the same operation or work center. The basic objective of this method is to balance the utilization of the machine or equipment across all products that share this facility or machine such that the demand is satisfied and the inventory is minimized. In other words, the production is balanced across all products using the common facility or the machinery rather than concentrating on a few products while the other products are ignored.

There are two versions of runout method available:

1. Aggregate runout method: This method is used if the lot size is variable.

2. Individual runout time method: This method is used if there are fixed lot sizes.

Aggregate Runout Method

First, we will illustrate the aggregate runout time method with the following example.

Kim Reuter starts her own company producing computer diskettes, CD-ROMs, DVDs and cassette tapes. All of these products are processed by the “Blue Monster”, an automated assembly line to produce these types of products. Kim has 100 machine hours available for production each week. Kim feels that her responsibility is to balance the utilization of the automated assembly line across all four products such that the demand for all products is satisfied and the inventory for each product is minimized. After consulting with an operations management professor at the local university, Kim decides to utilize the aggregate runout time method to schedule these four parts. Given the following data, determine the schedule that would achieve her objective using the runout time method.

Item	Inventory	Production time (hours/unit)	Forecast in units (per week)
Diskette	94	.10	85
CD-ROM	50	.15	150
Cassette	117	.30	60
DVD	150	.60	120

In order to solve this problem, first we must convert all of the inventory and forecasted demand into machine hours. Once we have the inventory and the forecasted demand in machine hour terms, then we need to sum the machine hours for inventory as well as the forecasted demand. After getting these totals we determine the aggregate runout time. The basic principle behind the aggregate runout time formula is that we need to manufacture a sufficient quantity of all products so that they will last until a future time period called the “runout time” where theoretically we run out of all the parts at the same time. The aggregate runout time is based on the inventory and the production capacity using the following formula:

Hours

Machine

in

Demand

Forecasted

Aggregate

Hours

Machine

in

Capacity

Production

Hours

Machine

in

Inventory

Aggregate

Time

Runout

Aggregate

+

=

In order to determine the aggregate runout time, we will utilize the following table.

Item	Production time/unit (hours)	Inventory (units)	Forecasted demand (units)	Inventory (hours)	Forecasted demand (hours)	Runout time (in weeks)
Diskette	.10	94	85	9.4	8.5	2
CD-ROM	.15	50	150	7.5	22.5	2
Cassette	.30	117	60	35.1	18	2
DVD	.60	150	120	90	72	2
Totals				142	121

weeks

2

121

100

142

e

runout tim

Aggregate

=

+

=

Since the aggregate runout time is two weeks, we need to have two weeks supply of each item, given our current inventory levels. The following table illustrates the calculation to determine the production quantities of each item. In these calculations, we start by multiplying the aggregate runout time with forecasted demand for each individual product to determine how many units would be needed for the duration of the runout time period. After getting how many units of each product are needed for the runout time period, then we subtract the current inventory level from the “needed quantity” and this becomes our production quantity for the week. In the last column of the table, we multiply the production quantity with the production time per unit to determine the length of the production run for each item. The sum of the hours or time periods in the last column must add up to the production capacity of the machine. Therefore, the sum of the values in the last column can serve as a “check” value. If the sum does not equal the production capacity specified, then we made a mistake and need to go back and correct it.

Item	Production Time (in hrs.) (1)	Inventory (in units) (2)	Forecast (in units) (3)	ROT (in weeks) (4)	# of items needed (5) = (3)x(4)	Production schedule (in units) (6) = (5) – (2)	Production schedule (in machine hrs). (7) = (1)x(6)
Diskette	.10	94	85	2	170	76	7.6
CD-ROM	.15	50	150	2	300	250	37.5
Cassette	.30	117	60	2	120	3	.90
DVD	.60	150	120	2	240	90	54
Totals							100

Individual Runout Time Method

In many situations, the aggregate runout time method described above is inadequate because the company produces products in fixed lot sizes. Note that in the table given above, we are scheduled to manufacture only three cassettes. This may not be an acceptable lot size—especially if the setup cost is significant. If the company has fixed lot sizes, we can utilize the individual runout time method. Instead of calculating an aggregate runout time for all items, we will calculate a runout time for each item. The individual runout time (ROT) is: ROT = inventory level /forecasted demand. In other words, the individual runout time is the amount of time inventory will be available to satisfy demand. Let’s use the following example to demonstrate this method.

Using the same problem scenario, we now assume that Kim has decided to produce all four products in fixed lot sizes and she determined the following fixed lot sizes:

Diskettes = 200 units

CD-ROMs = 260 units

Cassettes = 150 units

DVDs = 60 units

Use the runout method to determine the production schedule. Remember that Kim has 100 machine hours available per week.

If we multiply the production time per unit with the lot sizes we get the following results:

Diskettes = (200 units) (.10 hrs.) = 20 hours

CD-ROMs = (260 units)(.15 hrs.) = 39 hours

Cassettes = (150 units) (.30 hrs.) = 45 hours

DVDs = (60 units) (.60 hrs.) = 36 hours

Summing the lot-size hours (20 + 39 + 45 +36) = 140, we see that the production capacity is less than the required hours for all products. Therefore, since not all products can be manufactured within a week, we need to sequence the products in ascending order of their individual runout times. This way, the product that we are running out of the fastest is scheduled to be manufactured first. Using this logic, the runout times for the four products are:

weeks

ROT

weeks

ROT

weeks

ROT

weeks

ROT

DVD

cassette

ROM

CD

diskette

125

.

120

150

;

95

.

1

60

117

333

.

150

50

;

11

.

1

85

94

=

=

=

=

=

=

=

=

-

Based on the above calculations, the sequence of production in ascending order of runout times is:

1. CD-ROMs, 2. Diskettes, 3. DVDs, 4. Cassettes. Based on this order, the week’s production schedule is summarized in the following table:

Sequence	Machine hours required	Cumulative Machine hrs	Week’s production schedule in units
CD-ROM	39	39	260 units
Diskettes	20	59	200 units
DVD	36	95	60 units
Cassette	45	140	16 units
Totals	140

Note that we will exceed the weekly capacity after scheduling DVDs for production. The question is, how many cassettes can we manufacture within the remaining time of the week? After producing DVDs, we have 5 hours left (100 – 95) to manufacture the cassettes. Since each cassette takes .30 hours to manufacture, the estimated number of cassettes we can manufacture during this week are 5 / .3 = 16.667 or 16 units.

We would like to demonstrate the continuous nature of this process by showing the transition from one week to another. Let’s assume that the time has passed, and the week has gone by. Kim checks the company records and determines that the company has sold the following quantities of each item.

Diskettes = 100 units

CD-ROMs = 170 units

Cassettes = 50 units

DVDs = 140 units

Update the inventory records for the next week and determine the new runout time.

Using the following formula, we can update the inventory records.

New (ending) Inventory = Old (beginning) inventory + production – actual demand

Ending inventory for Diskettes = 94 + 200 – 100 = 194 units

Ending inventory for CD-ROMs = 50 + 260 – 170 = 140 units

Ending inventory for Cassettes = 117 + 16 – 50 = 83 units

Ending inventory for DVDs = 150 + 60 – 140 = 70 units

Item	Production time per unit	Inventory in units	Forecasted demand in units	Lot size in units	Runout time in weeks
Diskette	.10	194	85	200	2.28 weeks
CD-ROM	.15	140	150	260	0.933 weeks
Cassette	.30	83	60	150	1.383 weeks
DVD	.60	70	120	60	0.583 weeks

Based on these updated runout times, the new sequence is as follows:

1. Cassettes (Even though cassettes do not have the lowest ROT, we have to finish the lot that was started last week.)

2. DVDs

3. CD-ROMs

4. Cassettes

5. Diskettes

Sequence	Machine hours required	Cumulative Machine hours	Week’s production schedule in units
Cassettes	40*	40	150 – 16 = 134#
DVDs	36	76	60
CD-ROMs	39	115 > 100	(100 – 76) /.15 = 160

*Since we started batch of cassette tapes last week and produced 16 cassettes in 5 hours, we have 40 hours left for this week because the lot for cassettes takes 45 hours to manufacture.

# Since a batch of cassettes consist of 150 units, and we have completed 16 units last week, we will complete the remaining 134 units this week.

Problems

1. Given the following data and 95 hours available per week for production, use the individual runout time method of scheduling and determine the runout time sequence, batch production times and scheduled production quantity for each product.

Product	Production time/unit	Lot size	Inventory on-hand	Forecasted demand
A	.25	160	120	100
B	.10	120	140	120
C	.40	100	150	80
D	.30	80	100	70

2. Given the following data and 91 hours available per week for production, use the aggregate runout time method of scheduling and determine the aggregate runout time, batch production times and scheduled production quantity for each product.

Product	Production time/unit	Inventory on-hand	Forecasted demand
A	.25	120	100
B	.10	140	120
C	.40	150	80
D	.30	100	70

Solutions to problems

1.

weeks

ROT

weeks

ROT

weeks

ROT

weeks

ROT

D

C

B

A

429

.

1

70

100

;

875

.

1

80

150

16

.

1

120

140

;

2

.

1

100

120

=

=

=

=

=

=

=

=

Sequence	Machine hours required	Cumulative Machine hrs	Week’s production schedule in units
B	12	12	120 units
A	40	52	160 units
D	24	76	80 units
C	40	116	47* units
Totals	116

*

units

47

5

.

47

4

.

76

95

C

size

Lot

»

=

-

=

2.

Item	Production time/unit (hours)	Inventory (units)	Forecasted demand (units)	Inventory (hours)	Forecasted demand (hours)	Runout time (in weeks)
A	.25	120	100	30	25	2.5
B	.10	140	120	14	12	2.5
C	.40	150	80	60	32	2.5
D	.30	100	70	30	21	2.5
Totals				134	90

weeks

5

.

2

90

91

134

e

runout tim

Aggregate

=

+

=

Solutions to problems

2. continued

Item	Production Time (in hrs.) (1)	Inventory (in units) (2)	Forecast (in units) (3)	ROT (in weeks) (4)	# of items needed (5) = (3) x (4)	Production schedule (in units) (6) = (5) – (2)	Production schedule (in machine hrs). (7) = (1) x (6)
A	.25	120	100	2.5	250	130	32.5
B	.10	140	120	2.5	300	160	16
C	.40	150	80	2.5	200	50	20
D	.30	100	70	2.5	175	75	22.5
Totals							91

1

16-40

16-1

_1128630044.unknown

_1128633820.unknown

_1134069646.unknown

_1128633931.unknown

_1128630553.unknown

_1128633739.unknown

_1128630277.unknown

_1128337455.unknown

_1128369464.unknown

_1128369737.unknown

_1050076567.unknown