INTRODUCTIONThis case is concerned with the topic Simulation. Simulation is one of the most widely used

techniques in OR. It is used in all types of business decision making which covers all the

functional areas. Due to its flexibility, power and intuitive approach its popularity is growing

rapidly, with the increase in availability of computer power. Simulation is performed frequently

for risk analysis, project management, resource planning etc. Simulation is widely used for

systems which involves continuous decision making for a long time.

Facts of the Case1. Carl Schilling is the foreman of a large manufacturing firm which manufactures castings

through planer machines.2. Carl is called by the management to give solution for the problem they are facing. The

planer department is a problem. They cannot match the output with the workload. Work pieces were stuck up for want of free planner. This had disrupted the production schedule of next in line processes which increased the WIP costs and idle time costs which resulted into loss of production.

3. This meeting with the management helped Carl to sort out the major problems which he could see. He wanted a new planner so that bottlenecks could be eased. At one point of time there is a possibility that there would be overload and bottlenecks may occur and on the other hand the machines remain idle. Therefore Carl wanted to have uniform flow of work.

Given Information1. Company has 2 planers for cutting flat smooth surfaces in castings.2. The planers are used for 2 purposes:a. To make top surface of platen for large hydraulic lifts.b. To form mating surface of final drive housing for a large piece of earth moving

equipment.3. For each platen, the time required by a planer has an Erlang distribution with a mean of 25

minutes and shape parameter (k) as 4.4. For each housing the time required has an exponential distribution of minimum time of 10

minutes.5. The castings of both types arrive at one time.6. Casting for platens arrive randomly with mean rate of 2 per hour.7. Casting for housing have a uniform inter arrival time distributed in the time range of 20 to

40 minutes.

8. Estimated costs are:a. Platen casting cost = $200b. Housing casting cost = $100

9. All castings reserved on the FCFS basis.

To evaluate/find/ recommend1. Acquire additional planer whose total incremental cost is $30 per hour.2. Eliminate variability in inter arrival times of platen castings with castings arriving at

uniform manner of one every 30 minutes. Incremental cost = $40 per hour.3. Eliminate variability in inter arrival times of housing castings with castings arriving at

uniform manner of one every 30 minutes. Incremental cost = $20 per hour.4. To evaluate and recommend best solution.

Assumptions Here I have assumed that the manufacturing plant starts its operations at 8.00 AM and ends at 4.00 PM. The complete day schedule is broken in 4 shifts of 2 hours each; and I am considering the first shift from: 8.00 AM to 10.00 AM.

Solution I have found random numbers with the help of Excel function = rand ().

0.036565

0.577524

0.134957

0.196681 0.37766

0.072979

0.738164

0.970452

0.906679

0.031117

0.900105 0.34588

0.063045

0.730949

0.424865

0.968318

0.410758 0.1858

0.603859

0.299077

0.230593

0.439609

0.852662

0.467995

0.924451

0.049473 0.74776

0.384825

0.940334

0.502012

0.104424

0.112969

0.816350.91279

80.72655

60.22311

90.20360

50.42294

60.54554

90.24941

40.65270

50.39447

60.42507

20.86062

50.26698

30.19009

8 0.133440.66186

70.91268

1 0.279310.14292

50.52002

80.71021

90.63644

10.99463

30.28343

80.81187

20.43784

60.68051

60.25378

70.40187

40.85090

10.21385

70.31941

70.31095

10.50243

80.50648

30.91905

30.73772

40.88108

80.13433

10.31945

70.64124

10.49498

2 0.652320.31841

30.53548

30.77471

60.92609

10.13614

30.36575

20.12366

30.85974

20.09045

40.55259

10.28384

20.96668

80.53357

40.63594 0.65369 0.77838 0.97132 0.87239 0.67358 0.06782 0.13165

3 8 3 3 6 10.54986

90.92004

60.86347

80.64185

60.23046

70.72299

10.43344

7 0.443460.05901

10.72311

10.05183

20.93670

10.70796

90.41476

50.80812

50.32869

10.63269

20.71934

10.14647

50.12974

70.64861

80.74395

80.10381

90.71748

40.32241

80.18816

50.98660

60.18883

50.30498

80.50209

8 0.701330.22502

90.27262

60.66372

20.36871

6 0.038930.67763

50.80326

6 0.01190.55858

40.08133

60.30492

70.67033

90.22652

80.53570

70.08345

80.21953

60.23675

80.98777

70.40171

20.64055

90.23682

60.24186

60.64600

50.52060

20.27433

80.78943

90.45746

40.20844

70.78045

4 0.762120.78521

80.33856

30.23835

60.98248

90.99884

60.54989

10.43570

10.94980

60.93097

70.06071

90.73009

80.49748

20.35367

10.75271

40.94246

60.30312

30.80392

10.00430

70.88612

90.67426

30.02882

80.79231

3 0.867490.71864

90.05268

70.51134

80.23335

50.13710

1 0.817890.01639

70.84520

80.85852

10.67681

20.71739

10.65129

40.52088

50.72643

20.30648

90.57974

9 0.511960.31680

90.54301

30.52587

7 Now as there are two machines which do the required job, I have assigned one machine for Platen casting and the other machine for Housing casting. Platen Casting: - Arrivals:- The arrivals for platen casting at the planer machine are random (given) Hence the formula for calculating these random arrivals

t= ln (Rn )− λ

. Where ln = log, Rn = random number.

Here, lambda (λ) = 2 per hour (given); and

All the values are given in minutes, hence I have multiply the equation with 60 as in 1hr = 60

minutes.

Hence, using the above formula I got following values

Serial random Time time Arrival

number number (in minutes) (round off) (clock time)8.00 am

1 0.0366 43.09 43 8.43 am2 0.4108 11.59 12 8.55 am3 0.3110 15.21 15 9.10 am4 0.3658 13.10 13 9.23 am5 0.0590 36.87 37 10.00 am

Platen Casting: - Service:- The service rate for platen casting has Erlang distribution with mean (µ) = 25 and k = 4... (Given) Hence we can say that, µ = 1/25 = 0.04 and k = 4. Now the formula for finding service time for platen castings is –

t = ln∏i=1

k

Rn

−K μ

As the constant parameter k = 4, we shall take 4 random numbers for ln (Rn) value.

Hence, using the above formula I got following values

Housing Casting: - Arrivals:-

π(Rn)service Rounded

random numbers(Rn1*Rn2*Rn3*Rn4) time(min) off(min)0.6327 0.3224 0.2726 0.9878 0.0553 7.85 80.7894 0.9825 0.4975 0.6743 0.2602 3.54 40.5775 0.1858 0.9128 0.3945 0.0394 8.77 90.2793 0.4378 0.5024 0.4950 0.0304 9.48 90.1237 0.6537 0.9200 0.7231 0.0538 7.93 100.7193 0.1882 0.6637 0.3049 0.0274 9.76 100.4017 0.4575 0.9988 0.3537 0.0645 7.44 70.1350 0.9001 0.6039 0.7478 0.0549 7.87 80.7266 0.4251 0.1429 0.6805 0.0300 9.51 100.5065 0.6523 0.8597 0.7784 0.2210 4.09 40.1465 0.9866 0.3687 0.6703 0.0358 9.03 90.6406 0.2084 0.5499 0.7527 0.0552 7.86 80.2231 0.8606 0.5200 0.2538 0.0254 10.32 100.9191 0.3184 0.9713 0.6419 0.1824 4.62 50.2265 0.2368 0.7805 0.4357 0.0124 11.91 12

The inter arrival time is uniformly distributed time between 20 to 40 minutes (given)

Now the formula to find the arrivals of housing castings, we use the following formula –

t = a + (b – a) Rn, where, a = 20 min (least time taken) and b 40 min (max time taken).

Hence, (b-a) = 20.

Random Time Round off TimeNumber (min) (min) (clock)

8.00 am0.0813 21.62 22 8.22 am0.0311 20.62 21 8.43 am0.0495 20.99 21 9.04 am0.0288 20.57 20 9.24 am0.0518 21.03 21 9.45 am

Housing Casting: - Service:-

And the service rate is exponentially distributed with a mean of 10 minutes (given)

The formula for finding the service rate of housing castings is –

t = 10 + ln(Rn)−λ

, where λ = 1/10 = 0.1

Random Time Round offNumber (minutes) (minutes)0.9425 10.25 100.8675 10.61 110.8452 10.73 110.5797 12.37 120.3777 14.23 140.0630 22.00 220.2306 16.37 160.9403 10.27 100.2036 16.91 17

The current situation at the plant where Carl Schilling works is: -

PARTICULARS

ARRIVAL SERVICE ARRIVAL

SERVICE ARRIVAL SERVICE WAITING IDLE IDLE

TIME TIME TIME @ TIME @ TIME @ TIME @ TIME TIME @ TIME @PLANER

1PLANER

1PLANER

2PLANER

2 PLANER 1PLANER

2Housing 8.22 am 10 8.22 am 8.32 am 22Platen 8.43 am 8 8.43 am 8.51 am 43

Housing 8.43 am 11 8.43 am 8.54 am 11Platen 8.55 am 4 8.55 am 8.59 am 4

Housing 9.04 am 11 9.04 am 9.15 am 10Platen 9.10 am 9 9.10 am 9.19 am 11Platen 9.23 am 9 9.23 am 9.32 am 4

Housing 9.24 am 12 9.24 am 9.36 am 9Housing 9.45 am 14 9.45 am 9.59 am 9

Platen10.00

am 1010.00

am10.10

am 28TOTAL 0 61 90

Thus, the present condition of the work done is:-

1. Idle time on Planer Machine 1 is 61 minutes = 1 hour and 1 minute.2. Idle time on Planer Machine 2 is 90 minutes = 1 hour and 30 minutes.

Now for the convenience, I have used the following abbreviations: -

For Proposal No. 1 I have assumed that machine 3 would be used alternatively for housing as well as planer castings. Incremental cost = $ 30 per hour (given)

P AL-TE SE-TE

AL-TE

(M 1)

SE-TE (M 1)

AL-TE (M 2)

SE-TE (M 2)

AL-TE

(M 3)

SE-TE (M 3)

WTG-TE

IDL-TE (M

IDL-TE

(M2

IDL-TE

(M3

P = PARTICULARSP C = PLATEN CASTINGH C = HOUSING CASTING

AL-TE = ARRIVAL TIMESE-TE = SERVICE TIME

AL-TE (M 1) = ARRIVAL TIME AT PLANER MACHINE 1SE-TE (M 1) = SERVICE TIME AT PLANER MACHINE 1AL-TE (M 2) = ARRIVAL TIME AT PLANER MACHINE 2SE-TE (M 2) = SERVICE TIME AT PLANER MACHINE 2AL-TE (M 3) = ARRIVAL TIME AT PLANER MACHINE 3SE-TE (M 3) = SERVICE TIME AT PLANER MACHINE 3

WTG-TE = WATING TIME FOR PLANER MACHINESIDL-TE (M 1) = IDLE TIME FOR PLANER MACHINE 1IDL-TE (M2) = IDLE TIME FOR PLANER MACHINE 2IDL-TE (M3) = IDLE TIME FOR PLANER MACHINE 3

1) ) )HC

8.22 am 10

8.22 am

8.32 am 22

PC

8.43 am 8

8.43 am

8.51 am 43

HC

8.43 am 11

8.43 am

8.54 am 11

PC

8.55 am 4

8.55 am

8.59 am 4

HC

9.04 am 11

9.04 am

9.15 am 10

PC

9.10 am 9

9.10 am

9.19 am 11

PC

9.23 am 9

9.23 am

9.32 am 83

HC

9.24 am 12

9.24 am

9.36 am 9

HC

9.45 am 14

9.45 am

9.59 am 13

PC

10.00 am 10

10.00 am

10.10 am 41

TOTA

L 0 52 99 96 As seen earlier, there was a lot of waiting time on machines 1 and 2, the introduction of additional machine will further increase the idle time as well as additional cost of $ 60 per 1 hour 36 minutes of idle time on machine 3.

For Proposal No. 2 – If platens arrive at uniform rate of one every 30 minutes (given) Then the service rate also should be uniform (assumed) Hence, the service rate would be 1 every 15 minutes (assumed) Incremental cost = $ 40 per hour (given)

PARTICULARS ARRIVAL SERVICE ARRIVAL SERVICE ARRIVAL SERVICE WAITING IDLE IDLE

TIME TIME TIME @ TIME @ TIME @ TIME @ TIME TIME @ TIME @

PLANER 1 PLANER 1PLANER

2PLANER

2PLANER

1PLANER

2PLATEN 8.00 AM 15 8.00 AM 8.15 AM 0

HOUSING 8.22 AM 10 8.22 AM 8.32 AM 22PLATEN 8.30 AM 15 8.30 AM 8.45 AM 15

HOUSING 8.43 AM 11 8.43 AM 8.54 AM 11

PLATEN 9.00 AM 15 9.00 AM 9.15 AM 15HOUSING 9.04 AM 11 9.04 AM 9.15 AM 10HOUSING 9.24 AM 12 9.24 AM 9.36 AM 9PLATEN 9.30 AM 15 9.30 AM 9.45 AM 15

HOUSING 9.45 AM 14 9.45 AM 9.59 AM 9PLATEN 10.00 AM 15 10.00 AM 10.15 AM 15

TOTAL 0 60 61 Hence from the above table we can see that a uniform arrival of platen castings, reduces the idle time by 30 minutes. And the incremental cost remains at $ 40.

For Proposal No. 3 – If housings arrive at a uniform rate of one every 30 minutes (given) Then the service rate also should be uniform (assumed) Hence, the service rate for housing castings would one every 45 minutes (assumed)

PARTICULARSARRIVA

L SERVICE ARRIVAL SERVICE ARRIVAL SERVICE WAITING IDLE IDLETIME TIME TIME @ TIME @ TIME @ TIME @ TIME TIME @ TIME @

PLANER 1

PLANER 1

PLANER 2

PLANER 2

PLANER 1

PLANER 2

Housing 8.00 am 45 8.00 am 8.45 am 0Platen 8.43 am 8 8.43 am 8.51 am 43

Housing 8.30 am 45 8.30 am 9.15 am 15Platen 8.55 am 4 8.55 am 8.59 am 4

Housing 9.00 am 45 9.00 am 9.45 am 15Platen 9.10 am 9 9.10 am 9.19 am 11Platen 9.23 am 9 9.23 am 9.32 am 4

Housing 9.30 am 45 9.30 am10.15

am 15

Platen10. 00

am 1010.00

am10.10

am 28TOTAL 45 0 90

Hence, In this case, the waiting time is 45 minutes for housing castings, which makes the whole process to stretch. The idle time for planer number 2 is 90 minutes which is 1 and ½ hours whose cost comes as $ 40.

Conclusion Hence from the above calculations, I want to conclude that Proposal Number 2 can be accepted.