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 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 45
= 32 Trips
i
Number of tones a truck can carry / day A tuck can carry 80 tones in one trip Number of round up trips / truck can make in a day = 32 Number of tones in a day = load in each truck x Total trips 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 depletion or wastage wastage of time and final product supplying supplying to the customer, 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 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 truck In 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 = 2 Maximum 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 tones Total 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. Recommendation. OPT (Opt (Optim imiz ized ed Prod Produc ucti tion on techn technol ology ogy)) cite citedd by (Gal (Gallo lowa way, y, Rowb Rowbot otha ham m and Azhashemi, 2000 P-251), which places the primary focus upon the bottleneck process or process processes, es, Suggest Suggest that that all non-bot non-bottle tlenec necks ks must be optimi optimized zed to help help reliev relievee the bottleneck. bottleneck. To relieve relieve the bottleneck company may emphasize emphasize on Bottleneck Bottleneck allocation 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 more flex flexib ible le produ product ctio ionn plan planni ning ng resul results ts.. In this this case case the the crushe crushers rs are creat creatin ingg 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 = 3 Maximum 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
I 15
H9
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 22 Workstation 1
B7 C12 E5 24
F13
H9
Workstation 2
22
Workstation 4
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:
Process time = 22, Cycle time = 27 vii
G7
I15
22
Workstation 5
= [(22-27) / 27] x 100 = 18.51% Workstation 4:
Process time = 22, Cycle time = 27 = [(22-27) ÷ 27] x 100 = 18.51%
Problem (3) Statistical Process Control viii
Measurement of abandon call rates nation-wide taken every hour 16 %
1 7%
1 8%
17 %
13%
1 6%
16%
17 %
15%
Abandon call rates Avg. of 9 measurement 16.1% 16.8% 15.5% 16.5%
Hour 1 2 3 4
Hour 5 6 7 8
Avg. of 9 measurement 1 6 .5 % 1 6 .4 % 1 5 .2 % 1 6 .4 %
a)
µ = 16.1 σ = 1 E( x ) = µ E( x ) = 16.1
σx σ
x
=
σ/√ n
= 1/√12
= 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) ix
Hour 9 10 11 12
Avg. of 9 measurement 1 6. 3 % 1 4. 8 % 1 4. 2 % 1 7. 3 %
= 16.1 - 3(0.288) 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+17.3 12 X = 16
Graph shows 3 outlines Sample no
X
10 11 12
1 4. 8 1 4. 2 1 7. 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. S.no
Worker
Machine 1
Machine 2
1
Unloading M1 = 0.25
Unloading M1 = 0.25
Running time = 1.65
2 3 4 5
Reloading M1 = 0.30 Internal time M1= 0.40 Internal time M2 = 0.52 Idle time = 0.30
Reloading M1 = 0. 0 .3 Running time M1 = 1.90
6
Unloading M2 = 0.30
7
Reloading M2 = 0.38 Total cycle titime = 2.45 % Utilization = 87.76 % Idle time = 12.24
Idle time = 0.12 Unloading M2 = 0.30
Total cy cycle titime = 2.45 % Utilization = 100 % Idle time = 0
Reloading M2 =0.38 Total cy cycle time = 2. 2.45 % Utilization = 95.10 % Idle time = 4.90
@ 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 Ther Th eref efor ore, e, new new exte extern rnal al time time M1
= xi
(100 (100 / 115) 115) x 2.45 2.45 Mi Minn
=
2.13 min
=
2.45-2.13 = 0.32
=
2.13-1.90
=
0.23 min
Reduced from
=
0.55 min
The new external time M2
=
2.13-1.65
The new external time M1
Reduced from
=
0.48 min
=
0.68
New Process Cycle 115% of Standard pay. S.no 1
2 3 4 5 6 7 8
Worker Reducing Unloading time by M1 = 0.25 Reloading
0.23
M1 = 0.30 Internal time M1= 0.40 Internal time M2 = 0.52 Idle time = 0.30 Unloading Reducing in M2 = 0.30 Reloading time by 0.48
Machine 1 Unloading Reduction M1 = 0.25 Reloading M1 = 0.3
Running time = 1.65
time by 0.23
Running time M1 = 1.90 Unloading M2 = 0.30 Reloading
M2 = 0.38 Total cycle time = 2.45
Machine 2
M2 =0.38 Total cycle time = 2. 2 .45
Reduction the time by 0.48
Total cycle time = 2.45
(ii) 130% of Standard pay; (i.e 45% bonus) If the bonus of 30% is to be paid, items must be processed 100 / 130 of the standard time (2.45 min). Therefor fore, new new external time M1
=
(100 / 130) x 2.4 2.45 Min
=
1.88- 1.90
xii
Critical Path a. Network Diagram A2
C4
F4
D4
G7
H4
J3 L6
xiii
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 Over tim timee Mana Manage geme ment nt 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. Acti Action on aga again inst st act activ ivit ityy F If the activities outside the critical path speed up or slow 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 Data D= annual demand in units for inventory item = 40x52= 2080 units S = setup or ordering cost for each order = $10 xiv
H= holding or carrying cost per year =$20 x 23%= $4.60
(a)
E.O.Q
=
2x D x S H
= = (b)
(c)
2 x 2080 x $10 $4.60 ~ 96 units
Demand / Order Quantity
Total annual cost
~ 22 order / year
=
Setup cost + Holding cost
=
D
=
=
=
2080 units 96 units
=
=
(d) E.O.Q
=
Q
S +
Q 2
0 100 + 269
H 269 2
$ 2 6 9. 5
2xDxS H
2 x 500 unit x $100 $1 xv
1
=
3 1 6. 2 3
=
~ 317 units
Demand / Order Quantity =
500 units 317 units
=
1. 5 8
=
~ 2 order per year
Total annual cost (500 units)
=
Setup cost + Holding cost
=
D S + Q
=
=
(e)
500 100 317
Q 2
H 317 2
1
$ 3 1 6. 2 3
The sensiti sensitivit vityy level level is low with with these these demand demandss varia variable ble.. There There would would be no doubt doubt that the costs are being cheaper with the increase of quantity (demand).
xvi
