Problem (1)

Bottlenecks

A Granite Quarry has to perform following operations:-

1. Blast the granite hill;2. Load the boulders into trucks;3. Truck the boulders to the crusher;4. Crush the boulders into granite stones;5. Sieve the granite stones.

Data.

No. of shovels = 3 All three shovels can operate at the same time Time taken by the each shovel to load the truck = 10 minutes No. of crushers = 3 Crushers operational at a given time = 2 Maximum output from each crusher = 520 tones / hr The ‘octopus’ can process 1200 tones/hr Total loading time for loading a truck = 10 minutes Time for a trip from pit to crusher = 15 minutes Time for unloading at the crusher = 10 minutes Time for a trip from crusher to the pit = 10 minutes Each truck can carry 80 tones of ore Operation time for company = 24 hours a day, 7 days a week Production of the company = 24,000 tones per day

a) Determine the minimum no of trucks required

Number of Trips / truck

Total cycle time for one truck to pick up ore from shovels and depositing them at the

crusher and back to the shovel = 10 + 15 + 10 + 10 = 45 minutes

Number of minutes in a day = 60 x 24 = 1440 minutes

Total Cycle time for one truck to pick up ore from shovels One cycle time

Therefore

1440 = 32 Trips 45

i

Number of tones a truck can carry / day

A tuck can carry 80 tones in one tripNumber of round up trips / truck can make in a day = 32

Number of tones in a day = load in each truck x Total trips in a day

= 80 x 32 = 2560 tones

Trucking Operation:

The company operates 24 hrs / day, 7 day a week to meet the signed contract to

supply the customer, i.e. 24,000 tones/ day, in addition the assumption we made that

there is no depletion or wastage of time and final product supplying to the customer,

therefore, amount extracted is equal to the amount of granite produced. Thus

Output of trucks operations = granite produced = 24,000 tones per day

Minimum number of Trucks required in a day. To find the minimum number of trucks, in a day, required to meet the daily output

requirements we divide the total output in a day by amount of tones carried by a truck

in a day i.e.

24,000 / 2560 = Minimum number of trucks required in a day ≈ 10 trucks

b) Percent Utilization

i. Percentage utilization of shovels:

Each shovel can load a truck in 10 minutes, and each truck can carry 80 tones of load.

As there are three shovels working, therefore load of 80 tones by one shovel can make it

in every 10 minute.

Thus, in one hour, 6 loads of 80 tones = 80 x 6 = 480 tones (Amount loaded by one

shovel)

ii

By three Shovels. In One Hour 3 shovels load = 480 x 3 = 1440 tones

In 24 Hours 3 shovels can load = 1440 * 24 = 34,560 tones (Maximum Utilization)

Percentage utilization of shovels.

= (24,000 ÷ 34,560) x 100 = 69.44 %

ii. Percentage utilization of Trucks:

Number of hours in a day = 24 hrs.Number of minutes in 24 hours = 24 x 60 = 1440 Mins.

Every 45 minutes a truck completes one whole cycle from the quarry to the pit.

Minimum No. of trucks in a day = 10

Every 45 minutes one load of 80 tones by one truckIn 1440 minutes 32 loads of 80 tones = 32 x 80 = 2560 tones

As there are 10 trucks working in a day:

In 1440 minutes (32 x 80) x 10 = 25,600 tones (Maximum Utilization)

Percentage utilization of trucks.

Percentage Utilization = (Actual utilization ÷ Maximum utilization) x 100

Where actual utilization = 24,000

Thus, = (24,000 ÷ 25,600) * 100 = 93.75%

iii. Percentage utilization of crusher:

No. of operational Crushers at any given time = 2Maximum output from both crushers in one hour = 2 x 520 = 1040 tones

In 24 hours output from crushers = 1040 x 24 = 24,960 (Maximum Utilization)

iii

Percentage utilization of crushers.

(Actual utilization ÷ Maximum utilization) * 100

Actual utilization = 24,000

Thus: = (24,000 ÷ 24,960) * 100 = 96.15%

iv. Percentage utilization of Octopus

Output of the octopus in hour = 1200 tonesTotal output in a day = 1200 x 24 = 28800 tones (maximum utilization)

Percentage Utilization

(Actual utilization ÷ Maximum utilization) * 100, where Actual utilization = 24,000

Thus: = (24,000 ÷ 28,800) x 100 = 83.33%

Shovels = 69.44%Trucks = 93.75%Crushers = 96.15%Octopus = 83.33%

c) Bottleneck Operation. The apparent bottleneck is the crushers it has highest maximum percent utilization

recorded as 96.15%.

d) Recommendation.

OPT (Optimized Production technology) cited by (Galloway, Rowbotham and

Azhashemi, 2000 P-251), which places the primary focus upon the bottleneck process or

processes, Suggest that all non-bottlenecks must be optimized to help relieve the

bottleneck. To relieve the bottleneck company may emphasize on Bottleneck allocation

methodology (BAM) attempts to blend the best features of other popular production

planning philosophies. As a result, BAM could offer a better-integrated schedule with

more flexible production planning results. In this case the crushers are creating

bottlenecks, part of the reason could be the third crushing lines which went for repair and

maintenance,

iv

lets say that third crushing line will be operational with in the time limit contract than the

following will be answer working will all crushing lines.

No. of operational Crushers at any given time = 3Maximum output from both crushers in one hour = 3 x 520 = 1560 tones

In 24 hours output from crushers = 1560 x 24 = 37,440 (Maximum Utilization)

(Actual utilization ÷ Maximum utilization) * 100

Actual utilization = 24,000

Thus: = (24,000 ÷ 37,440) * 100 = 64.10% utilization

As we can see the percent utilization will be reduced by staggering 32%, than again the

new bottleneck will be recoded in trucks percentage utilization i.e 93.75%, In that case

necessary investment is much needed to avoid creating any bottlenecks

v

Problem (2)

Line Balancing

a) Precedence Diagram

A11

D11

G7

B7

C12 E5 F13

H9 I 15

b) Calculating the minimum number of workstations. (Ignoring precedence requirement and idle time)

Total time requirement = 90 seconds

No. of workstations = (Number of outputs x total time) / Working time

No. of outputs = 1000 toys

Total time of tasks = 1.5 minutes.

Factory operates for 450 minutes per day (Counting all breaks)

Thus, = (1000 x 1.5) / 450 = 3.33 ≈ 4 workstations

C. Calculating the control cycle time for assembly process

C.C.T = (seconds in a minute x total minutes) / total output

= (60 x 450) / 1000 = 27 minutes

vi

d. Line Balancing.

A11 D11 B7 C12 E5 F13 H9 G7 I15

22 24 22 22

Workstation 1 Workstation 2 Workstation 4 Workstation 5

e. Calculating Idle Time.

i. The overall Percentage idle time for the process

An average it took 27 seconds to produced one toy

Available time = 450 x 4 workstations = 1800

Required time = 1000 x 90 = 90,000 seconds = 1500 minutes

Idle time = total time – required time = 1800 – 1500 = 300 minutes

Overall percentage idle time = (idle time / total time) x 100

(300 / 1800) x 100 = 16.66%

ii. The percentage idle time at each workstation

[(Process time – cycle time) / cycle time] x 100

Therefore,

Work station 1:

Process time = 22, Cycle time = 27

= [(22-27) / 27] x100 = 18.51%

Workstation 2:

Process time = 24, Cycle time = 27

= [(24-27) / 27] x 100 = 11.11%Workstation 3:

vii

Process time = 22, Cycle time = 27

= [(22-27) / 27] x 100 = 18.51%

Workstation 4:

Process time = 22, Cycle time = 27

= [(22-27) ÷ 27] x 100 = 18.51%

Problem (3)

viii

Statistical Process Control

Measurement of abandon call rates nation-wide taken every hour

Abandon call rates

HourAvg. of 9

measurementHour

Avg. of 9 measurement

HourAvg. of 9

measurement1 16.1% 5 16.5% 9 16.3%

2 16.8% 6 16.4% 10 14.8%

3 15.5% 7 15.2% 11 14.2%

4 16.5% 8 16.4% 12 17.3%

a)

µ = 16.1σ = 1

E( x ) = µ E( x ) = 16.1

σx = σ/√ n= 1/√12

σ x = 0.288

b) 3 σ Control limits

UCL x = µ + 3(σ/√ n)

= 16.1 + 3(0.288)

UCL x = 16.9

LCL x = µ - 3(σ/√ n) = 16.1 - 3(0.288)

16% 17% 18% 17% 13% 16% 16% 17% 15%

ix

LCL x = 15.2

c)

X = 16.1+16.8+15.5+16.5+16.5+16.4+15.2+16.4+16.3+14.8+14.2+17.312

X = 16

Graph shows 3 outlines

Sample no X

10 14.811 14.212 17.3

Out of control criteria due to one or more assignable causes

Problem no. (4)

x

Workers Method.

a. Construct a worker-machine bar chart and use it to determine the number of items that the worker can produce during an eight-hour day at 100% of standard.

@ 100% of standard, the total cycle time is calculated as 2.45 min and worker will work (8 hours) 480 min / per day.

Therefore, Capacity per day = 480 min / 2.45 min

= 195 units / day / worker

b. (i) 115% of Standard pay; (i.e 15% bonus)

If the bonus of 15% is to be paid, items must be shaped 100 / 115 of the standard time

S.no Worker Machine 1 Machine 2

1 Unloading M1 = 0.25 Unloading M1 = 0.25 Running time = 1.65

2 Reloading M1 = 0.30 Reloading M1 = 0.3

3 Internal time M1= 0.40 Running time M1 = 1.90

4 Internal time M2 = 0.52

5 Idle time = 0.30 Idle time = 0.12

6 Unloading M2 = 0.30Unloading M2 = 0.30

7 Reloading M2 = 0.38 Reloading M2 =0.38

Total cycle time = 2.45 Total cycle time = 2.45 Total cycle time = 2.45

% Utilization = 87.76

% Idle time = 12.24

% Utilization = 100

% Idle time = 0

% Utilization = 95.10

% Idle time = 4.90

xi

Therefore, new external time M1 = (100 / 115) x 2.45 Min

= 2.13 min

= 2.45-2.13 = 0.32

The new external time M1 = 2.13-1.90

= 0.23 min

Reduced from = 0.55 min

The new external time M2 = 2.13-1.65

= 0.48 min

Reduced from = 0.68

New Process Cycle 115% of Standard pay.

S.no Worker Machine 1 Machine 2

1Unloading

M1 = 0.25

Reducing

time by

0.23

Unloading

M1 = 0.25 Reduction

time by

0.23

Running time = 1.65

2Reloading

M1 = 0.30

Reloading

M1 = 0.3

3 Internal time M1= 0.40 Running time M1 = 1.90

4 Internal time M2 = 0.52

5 Idle time = 0.30

6Unloading

M2 = 0.30 Reducing in

time by 0.48

Unloading

M2 = 0.30Reduction

the time

by 0.487Reloading

M2 = 0.38

Reloading

M2 =0.38

8 Total cycle time = 2.45 Total cycle time = 2.45 Total cycle time = 2.45

(ii) 130% of Standard pay; (i.e 45% bonus)

xii

If the bonus of 30% is to be paid, items must be processed 100 / 130 of the standard time (2.45 min).

Therefore, new external time M1 = (100 / 130) x 2.45 Min

= 1.88- 1.90

Critical Path

xiii

a. Network Diagram

A2 C4 F4 H4 J3

D4 G7 L6

B4 E1 I6 K5

b. Critical Path

A2 + D4 +G7 + H4 + J3+ L6 = 26 weeks

c. Normal completion time

Because critical path is the longest part in the project and total calendar time for required

project, therefore normal completion time for the project is however 26 weeks to

complete this project

d. Over time Management

Manager signed the contract to complete the project in 45 weeks and the normal

completion time is 26 weeks which actually is two weeks over the normal completion

time, thus total penalty for two weeks worth $650, breaking down further to activity A

manager it saves $5 against $ 325 and one week of completion, furthermore if the

manager pays $275 for activity D, which manager can actually saves $50 for another

week, therefore the project will be finished with in 24 weeks and manager spends $595,

instead of $650, which saves extra $55

d. Action against activity F

If the activities outside the critical path speed up or slow down (within limits), total

project time does not change, similarly activity F (slack time), does not descend under

critical path, thus there wouldn’t be any delay in project, there fore manager will not take

any actions against activity F

Problem 6

xiv

Data

D= annual demand in units for inventory item = 40x52= 2080 units

S = setup or ordering cost for each order = $10

H= holding or carrying cost per year =$20 x 23%= $4.60

(a) E.O.Q =

=

= ~ 96 units

(b) Demand / Order Quantity = 2080 units

96 units

= ~ 22 order / year

(c) Total annual cost = Setup cost + Holding cost

=

=

= $269.5

(d) E.O.Q =

xv

2x D x S H

2 x 2080 x $10 $4.60

HQ

2DQ

S +

1269

2

0

269100 +

2xDxS H

=

= 316.23

= ~ 317 units

Demand / Order Quantity = 500 units 317 units

= 1.58

= ~ 2 order per year

Total annual cost (500 units) = Setup cost + Holding cost

=

=

= $316.23

(e) The sensitivity level is low with these demands variable. There would be no doubt

that the costs are being cheaper with the increase of quantity (demand).

xvi

2 x 500 unit x $100 $1

HQ

2

D

QS +

1317

2

500

317100