Capacity Problems

7/27/2019 Capacity Problems

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Capacity Planning

Problems


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A manufacturing company has a product line consisting of

five work stations in series. The individual workstation

capacities are given. The actual output of the line is 500units per shift.

Calculate (i) System capacity

(ii) Efficiency of the production line

Workstation. No. A B C D E

Capacity/shift 600 650 650 550 600


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(i) The capacity of the system is decided by the workstation

with minimum capacity/shift, i.e., the bottleneck.

In the given example, the work station 'D' is having acapacity of 550 units/ shift which is a minimum.

Therefore, the system capacity = 550 units/shift

(ii) The actual output of the line = 500 units/shift.

Therefore, the system efficiency = Actual output X 100

System capacity

= (500) x 100

550

= 90.91 %


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A company intends to buy a machine having a capacity to

produce 1,70,000 good parts per annum. The machine

constitutes a part of the total product line. The system

efficiency of the product line is 85%.

(i) Find the system capacity.

(ii) The time required to produce each part is 100

seconds and the machine works for 2000 hours peryear. If the utilisation of the machine is 60% and the

efficiency of the machine is 90%, compute the output

of the machine.

(iii) Calculate the number of machines required?


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(i) System capacity = Actual output / annum

System efficiency

= 1,70,000 = 2,00,000 units / annum0.85

= 2,00,000 = 100 units / hours

2,000

(ii) output per annum = Unit capacity x % utilisation x efficiency

Unit capacity = 60 x 60 sec = 36 units

100 sec per unit

output per hour = 36 x 0.6 x 0.9 = 19.44 units = 20 units

(iii) Number of machines required = system capacity

output per hour

= 100 = 5 machines

20


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The following activities constitute a work cycle.

(i) Find the total time,

(ii) Theoretical output obtained from the machine.

(ii) Calculate the number of machines required to produce the three

components from the information given below.

Sr. No. Activity Time

(min)

1. Unloading 0.252. Inspection 0.35

3. Loading job on machine table 0.40

4. Machine operation time 0.90

Components A B C

1. Setup time per batch 25 min 55 min 45 min

2. Operation time (min/piece) 1.75 3.0 2.1

3. Batch size 350 550 575

4. Production per month 2450 4400 2875


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(i) Total cycle time (T) = 0.25 + 0.35 + 0.40 + 0.90 =1.90 min.

(ii) Output of the machine

Output = 60 = 31.5 = 31 units.1.9

(iii) Number of machines required

Assume that the plant works on the single shift basis per

day of 8 hours each.The total time required for the processing the components

is given by

Total time required = Setup time + operation time.

For component A,Total time required = Setup time + operation time

= Production quantity x Setup time + Operation time

Batch size Batch

= [2450/350] x [25/60] + [2450 x 1.75/60] = 74.374 hrs.


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For component A,

Total time required = Setup time + operation time

= Production quantity x Setup time + Operation time

Batch size Batch= [2450/350] x [25/60] + [2450 x 1.75/60] = 74.374 hrs.

For component B,

Total time required = [4400/550] x [55/60] + [4400 x 3/60]

= 227.33 hrs.

For component C,

Total time required = [2875/575] x [45/60] + [2875 x 21/60]

=104.375 hrs

Total time (hrs) required to process' all the three components=74.374 + 227.33 + 104.375 = 406.079 hrs.


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Total number of hours available (assuming 25 working

days) per month =8 x 25 = 200 hrs.

Number of machines required= Total number of machine hours required

Total number of hours available

= 406.079 = 2.030 = 2 machines

200

Assuming a machine efficiency of 85% and operator efficiency

of 75%, the number of machines required are:

Total hours required per month = 406.079 = 636-98 hrs.

0.85 x 0.75

Number of machines required = 636.98 = 3.18 = 4 machines200


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Three components are to be manufactured on three machines i.e. Center

lathe, Milling machine and Cylindrical grinding machine.

(i) Calculate the number of machines required of each kind to produce the

components if the plant works for 48 hours per week.(ii) Calculate the number of machines required assuming the machine

efficiency of 75%.

(iii) How do you reduce the number of machines.

The following information is given:

Machines Component A Component B Component C

setup operation setup operation setup operation

1.Center lathe 30 min 2 min 55 min 2.5 min 40 min 1.5 min

2.Milling machine 45 min 8 min 30 min 4 min ----------------

3.Cylindrical grinding 50 min 10 min 60 min 8 min 60 min 10 min

Other details

Lot size 350 400 600

Quantity demanded/month 1750 4000 3000


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Demand Forecasting

Weighted Moving Average

The past data on the load on the weaving machines is shown below:

a) Compute the load on the weaving machine centre using 5th moving average for the month of

December 1996

b) B) Compute a weighted 3 month moving average for Dec 96 where the weights are 0.5 for the latet

month,, 0.3 and 0.2 for the other months respectively

Month Load (Hrs)

May 96 -

June 585

July 610

Aug 675

Sep 750

Oct 860

Nov 970


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a) Five months moving average forecast for Dec 96

= (DNov+DOct+Dsept+DAug+DJuly)/5

= (970+860+750+675+610)/5

= 773 hrs.

b) A three months weighted moving average forecast for Dec96

= (WNov.x DNov) + (WOct x DOct) + (WSept x DSept)

=(0.5 x 970) + (0.3 x 860) + (0.2 x 750)

= 947.8 hrs


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Estimate the sales forecast for the year 2000, using exponential smoothing

forecaster. Take =0.5 and the forecast for the year 1995 as 160 x 105 units.

Compare the forecast with least square method.

Year 1995 1996 1997 1998 1999

Sales Rs (x105) 180 168 159 170 188


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Sol.: The demand for the year 2000 can be computed using least square

method as follows:

YearSales

(x 105

)

Deviation

(x)

x2 xy

1995 180 -2 4 -360

1996 168 -1 1 -168

1997 159 0 0 0

1998 170 +1 1 170

1999 188 +2 4 376

N=5 y=505 x=0 x2 =10 xy=18

Substituting these values in the equationsa=y/N = 505/5 = 101 b=xy/ x2 =18/10 = 1.8

The regression equation is y=100 + 1.8x

The demand for year 2000 = Rs.164.4 x 105


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The forecast demand using the smoothing constant =0.5 is computed as

shown in the table using the formula

Ft = .Dt-1 + (1- ).Ft-1

Period Demand Forecast demand for

=0.5

Ft

1995 180 160 (Given)

1996 168 0.5 x 180 + (1-0.5) x 160 170.0

1997 159 0.5 x 168 + (1-0.5) x 170 169.0

1998 170 0.5 x 159 + (1-0.5) x 169 164.0

1999 188 0.5 x 170 + (1-0.5) x 164 167.0

2000 0.5 x 188 + (1-0.5) x 167 177.5

The forecast demand for the year 2000 with =0.5 is 177.5


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A dealer for electrical appliances forecasts demand for the Geyser at the rate of

500 per month for the next three months. The actual demands turned out to be

400, 560 and 700.

Calculate the forecast error and bias, comment on the same

Solution:

MAD =nj=1 (Forecasted demand Actual demand) /3

= [(500 400) + (500 560) + (500 700)]/3

= (100 + 60 + 200)/3 = 120 units

Bias = [(500-400) + (500 560) + (500 700)]/3

= [(100 60 200)]/3

= -0.53 units

In this case MAD is 120 units and bias is -0.53 units

Since MAD increases absolute error, we can conclude that it has very high degree

of average error i.e 24% of the forecasted value of Geysers

The Bias The forecasting method has a tendency to under estimate the forecast

since bias is ve. i.e it is about 9.6% under forecast.


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Q.1 Given below are the details of a fan assembly line. Suppose we want to assemble 750

fans per day. What would our cycle time have to be? Draw the Precedence Diagram.

What is the theoretical minimum number of workstations for this problem? Find the

efficiency of this line. Suppose the demand rose from 750 to 1000 units per day. Whatwould you do?

Task Description Time (sec) Predecessor(s)

A Assemble Frame 20 none

B Mount switch 7 A

C Assemble motor housing 20 B

D Mount motor housing in frame 22 BE Attach blade 15 C

F Assemble and Attach safety grill 10 D

G Attach cord 16 E, F

H Test 8 G


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Example of Job Sequencing:

First-Come First-ServedJobs (in order Processing Due Date

of arrival) Time (days) (days hence)

A 4 5

B 7 10

C 3 6

D 1 4

Suppose you have the four

jobs to the right arrive for

processing on one machine.

What is the FCFS schedule?

Do all the jobs get done on time?


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Example of Job Sequencing:

First-Come First-Served

Jobs (in order Processing Due Date Flow Time

of arrival) Time (days) (days hence) (days)

A 4 5 4

B 7 10 11

C 3 6 14

D 1 4 15

Answer: FCFS Schedule

Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)

A 4 5

B 7 10

C 3 6

D 1 4

Suppose you have the four

jobs to the right arrive for

processing on one machine.

What is the FCFS schedule?

Do all the jobs get done on time?No, Jobs B, C,

and D are going

to be late.


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Capacity Problems

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