Exam 2014, answers PDF

Preview

Summary

Download Exam 2014, answers PDF

Description

Solutions Question 1

a. The capacity of BIG: The travel time for one round trip is (500*2 nm) / 20 nm/hr = 50 hours. Now including the 3hr operating time, a full trip will take 50 + 3 = 53 hours. Therefore, the capacity is 2 tons*24hr/53hr *30 days = 27.17 tons/month

Similarly, the capacity of MEDIUM is 1 tons * 24/69.67 *30 days = 10.33 tons /month, and capacity of SMALL is 0.5 tons * 24/86.33 *30 days = 4.17 tons /month Therefore, the fishing company’s total capacity is 51.42 +10.33+4.17= 41.67 ton/month b. Avg. Inventory for BIG is FR*FT= (2/(1000/(20*.8)+3))*(500/(20*.8))= 0.954 ton Avg. Inventory for MEDIUM is (1 /(1000/(15*.8)+3) * (500/(15*.8))=0.482ton Avg. Inventory for SMALL is (0.5/(1000/(12*.8)+3) * (500/(12*.8))=0.243 ton

c. (NOTE This LP has many equivalent forms) Let b1 to b6 denote the fish caught by BIG Let m1 to m6 denote the fish caught by MEDIUM Let s1 to s6 denote the fish caught by SMALL fi denote the fish population at the beginning of month i.

MAX ∑ 󰇛     󰇜 Subject to (month 1)        (month 2)        (month 3)        (month 4)       

(month 5)        (month 6)          700     󰇛     󰇜 ∗ 1.2     󰇛     󰇜 ∗ 1.2     󰇛     󰇜 ∗ 1.2     󰇛     󰇜 ∗ 1.2     󰇛     󰇜 ∗ 1.2    󰇛     󰇜 ∗ 1.2      for i=1 to 6    for i=1 to 6    for i=1 to 6   0 for i=1 to 6   0 for i=1 to 6   0 for i=1 to 6

Question 2 a. 3 machines are needed to make sure the utilization is less than one. b. For each check point (with one machine)   40 ⁄ ,   60⁄  ,   1,   1,   0.5,   1. (1)  

 



 . 

(2)    ∙

 

∙

  

 1.25

(2)       2.25 (2)    ∙   1.5 c. For the check point (with all three machines)   120 ⁄ ,   60⁄  ,   1,   1,   0.5,   3. 

 



 . 



   ∙

󰇛󰇜 

 

∙



 0.298

      1.298    ∙   2.596

d. (1) The centralized plan has lower flow time and less passengers waiting. (2) The main cause is the pooling effect: make sure no machines are idle if any passengers are waiting. Thus, the utilization is maximized, and passengers wait less. e. Open question. In reality, no airports use centralized system, because passengers may need to travel too far if they are asked to get through one specific check point. Or, the centralized plan may make terrorist attacks easier.

Question 3

0

7

7

0

0

7

7

2

0

3

0

1

13

1

1

2

1

1

16

2

1 H, 1

F ,3

B ,2 1

1 End

17

16

7 1

0

1

17 G,4

6 C,4

Sta 0

1

13 D,6

A,7

1

1

1 E,1

3

4

1

The critical paths are ADG and ADFH The earliest project completion time is 17 days. Activity start and finish times are indicated above. Activities that can be delayed are B,C,E

Question 4 Open ended question. Quality Management – students may discuss challenges such as controlling costs of quality, continuous improvement, employee involvement, integrating product and process design. Ways of addressing these challenges may include implementing PDCA cycle, six sigma quality, implementing statistical quality control at all levels of the organization, and using frameworks such ISO and Quality Awards. Supply Chain Management – challenges include facility location, managing inventory, bullwhip effect. Various tactics to address these challenges may be discussed – consolidation via centralized warehouses, postponement to reduce inventory, information sharing and partnering, lead time reduction. Students are expected to use examples to support their answers.

Question 5 a. S = Ordering Cost;

H = Unit Holding Cost

CYL: (36000/1500)*S + (1500/2) *H = TC SBL: (36000/1000)*S + (1000/2) * H = TC – 9500 OR 24S + 750H = TC

(1)

36S + 500H = TC -9500

(2)

Subtracting (2) from (1) we get, -12S + 250H = 9500 OR S = [250H -9500]/12

(3)

EOQ = 600 = SQRT (2*36000*S/H) from which we get, S = 5H

(4)

Combining (3) and (4) 5H = [250H -9500]/12, from which we get H = 50; S = 250

Cost (CYL) = 24*250 + 750*50 = $ 43,500 Cost (SBL) = 36*250+500*50 = $ 34,500 Cost (EOQ) = (36000/600) *250 + (600/2) * 50 = $30,000

b. µD = 72,000

P = 2 weeks (given)

S = $250

D = 14,000

L = 1 week

H = $12.5

(P+L) = SQRT [(2+1)/52] * 14000 = 3362.7 Z = 2.33 (given) SS = 2.33 * 3362.7 = 7835.1 Total Cost = [52/2]* 250 + [1/2] * [72000/52] * 2 * 12.5 + 7835.1 * 12.5 = 6500 + 17307.7 + 97938.4 = $ 121,746.1...