7-Solutions Review - Solutions de quelques exercices du manuel PDF

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Solutions de quelques exercices du manuel...

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Review – Solutions Chapter 3 Problem 7

month 1 2 3 4 5 6 7 8 9 10

A 31 34 33 35 37 36 38 40 40 41

a. F 31 31 31.9 32.2 33.1 34.2 34.8 35.7 37 37.9 38.8 MAD

c. |A – F| 0 3 1.1 2.77 3.939 1.7573 3.23011 4.26108 2.98275 3.08793

S 30 31.9 33.119 34.562 36.211 37.136 38.282 39.658 40.653 41.5901

b. T 1 1.27 1.2547 1.3111 1.4125 1.2664 1.2302 1.2741 1.1901 1.1143

F 31 33.17 34.374 35.873 37.623 38.403 39.512 40.933 41.843 42.704

2.61282

c. |A – F| 3 0.17 0.6263 1.1273 1.6234 0.4028 0.4879 0.9326 0.8429 1.0237

c. Regarding the MAD, the best forecasting model is the exponential smoothing with trend forecast.

Problem 18 month Apr May Jun Jul Aug Sep

F 250 325 400 350 375 450

A 200 250 325 300 325 400

A–F -50 -75 -75 -50 -50 -50

|A – F| 50 75 75 50 50 50

RSFEt -50 -125 -200 -250 -300 -350

MADt 50 62.5 66.6667 62.5 60 58.3333

TS -1 -2 -3 -4 -5 -6

No, the model used is very very bad! The tracking signal jump under -4.

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Problem 22

month 1 2 3 4 5 6 7 8 9 10 11 12

A 62 65 67 68 71 73 76 78 78 80 84 85

a. F

64.7 66.7 68.7 70.7 73.3 75.7 77.3 78.7 80.7 83 MAD

e. |A – F|

3.3333 4.3333 4.3333 5.3333 4.6667 2.3333 2.6667 5.3333 4.3333 4.07407

b. F

65.4 67.1 69.3 71.4 74.1 76.4 77.6 79 81.6 83.7

e. |A – F|

2.6 3.9 3.7 4.6 3.9 1.6 2.4 5 3.4 3.45556

c. F 61 61.3 62.41 63.787 65.051 66.836 68.685 70.879 73.016 74.511 76.158 78.51 80.457

e. |A – F| 1 3.7 4.59 4.213 5.9491 6.16437 7.31506 7.12054 4.98438 5.48907 7.84235 6.48964 5.40479

S 60 62.76 65.49 67.84 70.4 72.85 75.47 77.96 79.7 81.37 83.61 85.52

d. T 1.8 2.088 2.282 2.302 2.379 2.399 2.467 2.472 2.254 2.078 2.128 2.061

F 61.8 64.848 67.7753 70.1446 72.7801 75.2448 77.938 80.4288 81.9538 83.4455 85.7395 87.5788

e. |A – F| 3.2 2.152 0.22472 0.8553992 0.2198887 0.7552414 0.0620165 2.4288224 1.9538156 0.5545326 0.7395316 1.195088

e. Regarding the MAD, the best forecasting model is the exponential smoothing with trend forecast.

Chapter 5 Problem 10 a.

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b. Activity A B C D E F G H

Activity A B C D E F G H

a 1 1 1 2 3 3 1 2

I.P. A A B C, D D, E F, G

Times (daysx) m 3 2 2 3 4 4 4 4

Durée 3 2 2 3 5 4 3.8333 3.8333

b 5 3 3 4 11 5 6 5

Mean time

Variance

3 2 2 3 5 4 3.8333 3.8333

0.4444 0.1111 0.1111 0.1111 1.7778 0.1111 0.6944 0.25

ES LS 0 3 0 2 3 5 3 6 2 7 6 10 7 10.8333 10.8333 14.6667

EF 0.8333 0 4.8333 3.8333 2 6.8333 7 10.8333

LF 3.8333 2 6.8333 6.8333 7 10.8333 10.8333 14.6667

I.S. C, D E F F, G G H H -

Slack 0.8333 0 1.8333 0.8333 0 0.8333 0 0

Critical path: B-E-G-H c. 14.67 days d. First, critical path ET = 14.6667 days, critical path variation = 0.1111 + 1.7778 + 0.6944 + 0.25 = 2.8333

 16 − 14.6667  P (D ≤ 16 ) = P  Z ≤  = P (Z ≤ 0.79 ) = 78.52% 2.8333   Problem 15 a. Critical path = A-C-D-F-G, early completion time = 25 days b. Project duration: 25 days

Initial total cost = $35,000

Paths: A-B-D-F-G

24 days

A-B-D-E-G

22 days

A-C-D-F-G

25 days  critical path

A-C-D-E-G

23 days

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Activity max crash period Crash cost by day A 1 $1,000 B 1 $2,000 C 1 $1,200 D 1 $1,500 E 1 $1,000 F 2 $1,500 G 1 $3,000 First step Paths: A-B-D-F-G (24 days), A-B-D-E-G (22 days), A-C-D-F-G (25 days; critical path), A-C-D-E-G (23 days) Options: Activity max crash period Crash cost by day A 1 $1,000 B 1 $2,000 C 1 $1,200 D 1 $1,500 E 1 $1,000 F 2 $1,500 G 1 $3,000 Reduce A of 1 day for $1,000 Paths update: A-B-D-F-G (23 days), A-B-D-E-G (21 days), A-C-D-F-G (24 days; critical path), A-C-D-E-G (22 days) Project duration: 25 – 1 = 24 days Total cost = $35,000 + $1,000 = $36,000 Second step Paths: A-B-D-F-G (23 days), A-B-D-E-G (21 days), A-C-D-F-G (24 days; critical path), A-C-D-E-G (22 days) Options: Activity max crash period Crash cost by day B 1 $2,000 C 1 $1,200 D 1 $1,500 E 1 $1,000 F 2 $1,500 G 1 $3,000

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Reduce C of 1 day for $1,200 Paths update: A-B-D-F-G (23 days; critical path), A-B-D-E-G (21 days), A-C-D-F-G (23 days; critical path), AC-D-E-G (21 days) Project duration: 24 – 1 = 23 days Total cost = $36,000 + $1,200 = $37,200 Third step Paths: A-B-D-F-G (23 days; critical path), A-B-D-E-G (21 days), A-C-D-F-G (23 days; critical path), A-C-D-E-G (21 days) Options: For A-B-D-F-G:

For A-C-D-F-G:

Activity max crash period Crash cost by day B 1 $2,000 D 1 $1,500 E 1 $1,000 F 2 $1,500 G 1 $3,000

Activity max crash period Crash cost by day B 1 $2,000 D 1 $1,500 E 1 $1,000 F 2 $1,500 G 1 $3,000

Reduce D (or F) of 1 day for $1,500 (reduce both critical path at the lowest cost) Paths update (with D reduction): A-B-D-F-G (22 days; critical path), A-B-D-E-G (20 days), A-C-D-F-G (22 days; critical path), A-C-D-E-G (20 days) Project duration: 23 – 1 = 22 days Total cost = $37,200 + $1,500 = $38,700 Fourth step Paths: A-B-D-F-G (22 days; critical path), A-B-D-E-G (20 days), A-C-D-F-G (22 days; critical path), A-C-D-E-G (20 days) Options: For A-B-D-F-G:

For A-C-D-F-G:

Activity max crash period Crash cost by day B 1 $2,000 E 1 $1,000 F 2 $1,500 G 1 $3,000

Activity max crash period Crash cost by day B 1 $2,000 E 1 $1,000 F 2 $1,500 G 1 $3,000

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Reduce F of 1 day for $1,500 (reduce both critical path at the lowest cost) Paths update: A-B-D-F-G (21 days; critical path), A-B-D-E-G (20 days), A-C-D-F-G (21 days; critical path), AC-D-E-G (20 days) Project duration: 22 – 1 = 21 days Total cost = $38,700 + $1,500 = $40,200

Chapter 7 Problem 18 Arrival: λ = 10 customers/h Service: 1/µ = 5 min/customer  µ = 1/5 = 0.2 customer/min = 12 customers/h S = 1 machine, constant time  model 2 (because the machine gives the service) a. Lq =

102 λ2 = = 2.0833 customers 2µ ( µ − λ ) 2 × 12( 12 − 10)

b. Ls = Lq + c. Wq =

Lq

d. Ws =

Ls

=

2.0833 = 0.20833 h/customer = 12.5 min/customer 10

=

2.9167 = 0.29167 h/customer = 17.5 min/customer 10

λ

λ

10 λ = 2.0833 + = 2.9167 customers µ 12

e. If the arrival rate increases to more than 12 customers/h, it is impossible to manage because there is an overflow i.e. more customers arriving than the capacity of the system.

Problem 21 Arrival: λ = 2 customers/h Service: 1/µ = 20 min/customer  µ = 1/20 = 0.05 customer/min = 3 customers/h S = 1 server, model 1 a. Lq =

b. Wq =

2 λ2 2 = = 1.3333 customer µ ( µ − λ ) 3(3 − 2)

Lq

λ

=

1.3333 = 0.6667 h/customer = 40 min/customer 2

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c. Ls =

2 λ = = 2 customers µ − λ 3− 2

d. ρ =

λ 2 = = 66.67% µ 3

Ws =

Ls

λ

=

2 = 1 h/customer 2

Problem 23 Arrival: λ = 6 customers/h Service: 1/µ = 6 min/customer  µ = 1/6 = 0.1667 customer/min = 10 customers/h S = 1 server, model 1 a. average customer in line: Lq =

λ2 62 = = 0.9 customer µ ( µ − λ) 10 (10 − 6 )

average customer in the camera department: Ls = average time in the system: Ws = b. ρ =

Ls

λ

=

λ µ −λ

=

6 = 1.5 customer 10 − 6

1.5 = 0.25 h/customer = 15 min/customer 6

λ 6 = = 60% µ 10

c. P> 2 = 1 − P0 − P1 − P2 = 1 − ( (1 − ρ ) ρ 0 ) − ( (1 − ρ ) ρ 1 ) − ( (1 − ρ ) ρ 2 ) = 1 − 0.4 − 0.24 − 0.1440 = 21.60% d.

S = 2 servers  model 3

ρ = 0.6  table  Lq = 0.059 customer average customer in the camera department: Ls = Lq + average time in the system: Ws =

© M.-C. Bolduc 2022.

Ls

λ

=

6 λ = 0.059 + = 0.659 customer 10 µ

0.659 = 0.1098 h/customer = 6.59 min/customer 6

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Chapter 10 Problem 25 n=4

In the table: A2 = 0.7286

D4 = 2.2821

D3 = 0

R-chart: Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1010 995 990 1015 1013 994 989 1001 1006 992 996 1019 981 999 1013

Readings 991 985 996 1009 1003 1015 1020 1009 1019 1005 1001 994 992 982 986 996 989 1005 1007 1006 1006 997 996 991 991 989 993 988 1002 1005

986 994 1008 998 993 1005 1020 996 1007 979 989 1011 1003 984 992

R max – min = 1010 – 985 = 25 15 25 22 26 11 38 15 18 28 17 28 22 15 21 moyenne = ฀ ฀ = ฀฀฀฀, ฀฀฀฀฀฀

LCSR = D4 R = 2.2821× 21.7333 = 49.5976 LCIR = D3 R = 0 × 21.7333 = 0

The range is under control. © M.-C. Bolduc 2022.

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฀฀-chart: Sample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

1010 995 990 1015 1013 994 989 1001 1006 992 996 1019 981 999 1013

Readings 991 985 996 1009 1003 1015 1020 1009 1019 1005 1001 994 992 982 986 996 989 1005 1007 1006 1006 997 996 991 991 989 993 988 1002 1005

986 994 1008 998 993 1005 1020 996 1007 979 989 1011 1003 984 992

฀฀ (1010 + 991 + 985 + 986)/4 = 993 998,5 1004 1010,5 1007,5 998,5 995,75 994,75 1001,8 996 997 1004,3 991 991 1003 moyenne = ฀ ฀ = ฀฀฀฀฀฀, ฀฀

LC x = X ± A2R = 999.1 ± 0.7286 × 21.7333 = [LCL = 983.2651;UCL = 1014.9349]

The mean is under control. The process is under control.

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Problem 26 a. Z1 – α/2 = 97.75% = 2 Sample Dec 1 Dec 2 Dec 3 Dec 4 Dec 5 Dec 6 Dec 7 Dec 8 Dec 9 Dec 10 Sum p=

Sample size Errors (n) (np) 1000 74 1000 42 1000 64 1000 80 1000 40 1000 50 1000 65 1000 70 1000 40 1000 75 10,000 600

p 74 / 1000 = 0,074 0,042 0,064 0,08 0,04 0,05 0,065 0,07 0,04 0,075

600 = 0.06 10,000

sp =

0.06(1− 0.06) p (1 − p ) = = 0.0075 n 1000

LC = p ± Z (1−α 2 ) s p = 0.06 ± 2 × 0.0075 = [ LCL = 0.045; UCL = 0.075 ]

b. The process is out of control because samples 2, 5 and 9 are under the LCL and sample 4 is over the UCL. We should stop the production and correct the problem.

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Problem 30 AQL = 0.03

LTPD = 0.10

α = 5%

β = 10%

a. DS =

LTPD 0.10 = = 3.333 AQL 0.03

In the table (this value or more):

c=5 n × AQL = 2.613

n × AQL = 2.613 n × 0.03 = 2.613 n=

2.613 = 87.1 ≈ 88 (arrondi supérieur) 0.03

Sampling plan is (n = 88, c = 5)

b. To accept the shipment, it should contain 5 defectives or less.

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