Session – XI

Scheduling

Of Operations Activities

Once the planning process is completed, actual work is to be

performed by man power with the help of machine power. It is said

that one machine can do the work of 100 ordinary persons, but no

machine can do the work of one extra ordinary person (MBA

qualified). It means all repetitive work is done by machine

(automatic) and all intelligent work is done by MBAs. (software

designing, process planning, problem solving etc).

Thus, scheduling of operations is assigning of work to individuals,

who will operate some equipments for “specified time” and produce

desired product or service or part of it.

The capacity has been decided based on investment. Working days

in India is 300 (250 in Western world) and all industries work for 24

hrs (3 shifts) based on 8 hrs working per person per day as per

labour laws. Therefore per shift or per hour work target is pre-set

and the intension is to achieve optimum output, for which

scheduling is carried out.

Ex. Capacity is 9,00,000 per year, 300 working days and 3 shifts /

day. Thus per shifts, output is 1000 nos. or MT

In 8 hrs shift, there is lunch time and rest time of 30 min + 2 x 15

min, totalling 60 min and working time is 7 hrs or 420 minutes

approximately.

2

The time for doing each operation is also pre calculated (10/20

minutes) by experience and scheduling is time management of both

machine and worker.

Ex : Take a simple example.

Four pars are being manufactured by four person on four machines

and time taken is given as follows.

Find the minimum time for completing all parts.

Work Time of equipment as per capability

1 2 3 4

A

B

C

D

10

6

7

9

5

2

6

5 xx

6

4

5

4

10

6

6

10

Scheduling for each Transformations process.

1st Job Shop – Difficult, since which job will come and when ?

2nd Batch Process : Easy, since some process is used for all products

and products are smaller.

3rd Moving assembly line : Line Balancing for large products and no.

of components (mostly bought out0 are assembled on a moving line.

4th Continuous Process : Mostly automotive and goes on without any

rest or holiday. Machines are doing repetitive work and only

supervision is done.

Schedules are fixed by planning and efficient maintenance.

3

Objectives of Scheduling.

1. Achieve target output as per capacity planned.

2. Scheduling for maximising worker and machine time utilization.

3. And minimising idle time by better MRP & ERP.

4. Meeting the promised delivery of customer.

5. Good time scheduling increases the efficiency of worker by

proper loading of job on the machine, deciding the sequences

of work (or fixing priority) and then monitoring the work.

Sequencing or fixing priority for target results.

Rules of sequencing for job shops (one work centre).

1st Rule : First come first serve FCFS for all services – Q formation.

How to serve the customer at early as possible and minimum

waiting time, such as Banks, Reservations, Paying Bills, Hospitals,

Traffic, Restaurant.

Single Server

Multi Server

Rate of arrival A or

Rate of Service S or

Formulae : 1) Waiting time in Q = . (-)

2) Waiting time in System = 1 . -

3) Length of Q (no. of persons) = 2 . (-)

4) Length of persons (in no.) in system = . -

4

2nd Rule : Last come first serve LCFS for H.R. Deptt i.e . Last man

come, will be first served the notice (not used in operations area).

3rd Rule : Shortest operating time “SOT” rule. The work which takes shortest time to complete will be given first priority or sequence, used for maximum production.

4th Rule : Earliest delivery due EDD rule. Priority is given to, which is to be delivered earliest, for customer satisfaction (for customer focus industries).

5th Rule : Smallest Critical Ratio SCR rule. For projects under taken of large value.

C.R. is Promised delivery Process time

Smallest C.R. is given 1st priority

6th Rule : Minimum Slack time MST rule. Remaining means extra time remaining between delivery and production time. Minimum S.T. is given 1st probity.

Ex. to understand above five rules (LCFS is dropped).

Jobs Process Time Delivery Time A 5 10B 10 15C 2 5D 8 12E 6 8

Compare the average processing time and average delay time by 5 rules and give recommendations.

Solution :

1st Rule FCFS, means same sequence as given.

5

No. of Job

TimeStart Process Completio

nDue Delay

A 0 5 5 10 0B 5 10 15 15 0C 15 2 17 5 12D 17 8 25 12 13E 25 6 31 8 23

Total 93 48Average ( by 5) 18.60 9.6

2nd Rule : (dropped)

3rd Rule : SOT

Arrange sequence as per SOT i.e. C 1st, A 2nd, E 3rd, D 4th & B 5th

No. of Job

TimeStart Process Completio

nDue Delay

C 0 2 2 5 0A 2 5 7 10 0E 7 6 13 8 5D 13 8 21 12 9B 21 10 31 15 16

Total 74 30 Average ( by 5) 14.80

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4th Rule : Earliest delivery due i.e. arrange sequence as per EDD rule C 1st, E 2nd, A 3rd, D 4th, and B 5th

No. of Job

TimeStart Process Completio

nDue Delay

C 0 2 2 5 0E 2 6 8 8 0A 8 5 13 10 3D 13 8 21 12 9B 21 10 31 15 16

Total 75 28Average ( by 5) 15 5.6

6

5th Rule : Smallest critical ratio is promised delivery process time SCR for A is 10 = 2 Fourth

5B is 15 = 1.5 Third

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C is 5 = 2.5 Fifth 2

D is 12 = 1.5 Second 8

E is 8 = 1.3 First 6

No. of Job

TimeStart Process Completio

nDue Delay

E 0 6 6 8 0D 6 8 14 12 2B 14 10 24 15 9A 24 5 29 10 19C 29 2 31 5 26

Total 104 56Average ( by 5) 20.80 11.2

6th Rule : Minimum slack time remaining.

= Due date – Process lime For A 10 – 5 = 5 Fourth

B 15 - 10 = 5 Fifth

C 5 – 2 = 3 Second

D 12 – 8 = 4 Third

E 8 – 6 = 2 First

No. of Job

TimeStart Process Completio

nDue Delay

E 0 6 6 8 0C 6 2 8 5 3

7

D 8 8 16 12 4A 16 5 21 10 11B 21 10 31 15 16

Total 82 34Average ( by 5) 16.40 6.8

Summery or comparison of Rules

Rule Av. Process Av. Delay FCFS 18.60 9.60SOT 14.80 Best 6.00 2nd BestEDD 15.00 2nd Best 5.60 BestSCR 20.80 11.20MST 16.40 6.80

SOT & EDD give best results.

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Ex - 1:

25 Customers arrive at ATM Centre per hour and the ATM takes 2 minutes to serve one customer.

Find 1) Waiting time for customer 2) Waiting time in system 3) Length of Q 4) No. of persons waiting in the system 5) Utilization of ATM.

Solution

Arrival is = 25 per hour Service is = 60 = 30 per hour

2

1) Waiting time for customer in q is Wq = Arrival

Service (Service - Arrival)

= = 25 ( - ) 30 (30 – 25)

= 25 . = 1 hr or = 10 minutes 30 (5) 6

2) Waiting time in system is Ws

Or 1 = 1 . = 1 hr or 12 minutes (10+2) - 30 – 25 5

3) Length of queue = 2

9

(-)

= 25 x 25 0. 30 (30 – 25) = 4.17 persons or 4 persons

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4) No. of persons in system

Ls = = 25 . = 5 persons - 30 – 25

5) Utilization of system of ATM

= 25 x 100 = 83.34%30

means for 16.64% ATM is lying idle.

Ex – 2 :

In a bank, 15 customers arrive per hour () and the teller machine serves customers every 3 minutes / person.

Find 1. Waiting time in queue Wq 2. Waiting time in system Ws3. Length of queue Lq4. Length queue in system Ls5. Utilization of teller machine

Solution

= 15 and = 60 = 20 3

1. Wq = = 15 x 60 = 9 min. ( -) 20 (20 – 15)

2. Ws = 1 = 1 x 60 = 12 min (9+3) - 20 – 15

10

3. Lq = 2 = 15 x 15 = 9 = 2 persons ( -) 20 (20 – 15) 4

4. Ls = = 15 = 3 persons.

- 20 – 15

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5. Utilization ratio = 15 x 100 = 75% & 25% is idle time 20

Ex – 3 :

On Economics of waiting time i.e. Generating savings by increasing the service efficiency.

The average arrival of customers is 30 per hour and the server, serves the customer at the rate of 45 per hour. The processing system will save Rs 0.50 per minute per customer if service is fast. For this, some assistance is hired @ Rs 15 per hour and service rate is increased to 75 per hour. Find out if assistance be provided ?

Solution :

Arrival rate is = 30 per hour Service rate is = 45 per hour

waiting time in System WS =

1 . = 1 = 1 x 60 = 4 min - 45 - 30 15

When service rate is 75 per hour Then Ws = 1 = 1 x 60 = 4 min

75 – 30 45 3

Time saved per customer is 4 – 4 = 8 minutes 3 3

Savings per hour will be 8 x .50 x 30 = Rs 40

11

3

Amount paid to assistant is Rs 15, Net Saving is Rs 25/-.

Yes, provide assistance

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Ex-4

In a factory, the break down of machines is 3 nos. per hour.

The servicing of machine is done by two mechanics A & B, who take 15 minutes (A) and 10 minutes (B) resp. The wages are Rs 8/- for A and Rs 10/- for B per hour.

Cost of idle time per machine per hour is Rs 16/-.

Find economics of A & B.

Solution :

Machine breakdown for repairs is rate of arrival 3 nos. per hour =

Rate of service for A is 15 min. i.e. 4 machines per hour = no. of machine in system is Ls

= = 3 = 3 nos/hour - 4 - 3

Cost of idle time per hour is Rs 16/-.

Total cost for A is (16 x 3) + 8 = 56 in Rs

For B, rate of service is 10 min/machine Or 6 machines per hour

Ls for B is 3 = 1 machine 6 – 3

Total cost for B = (1 x 16) + 10 = 26

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For A Cost is Rs 56/-.For B Cost is Rs 26/-.

Saving is Rs 30/- in case of B

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