Section C: Problem solving (55 marks)

Question 1 (10 marks)

The Queen City Nursery manufactures bags of potting soil from compost and topsoil. Each cubic foot of compost costs 12 cents and contains 4 pounds of sand, 3 pounds of clay, and 5 pounds of humus. Each cubic foot of topsoil costs 20 cents and contains 3 pounds of sand, 6 pounds of clay, and 12 pounds of hu-mus. Each bag of potting soil must contain at least 12 pounds of sand, 12 pounds of clay, and 10 pounds of humus.

Explain how this problem meets the conditions of a linear programming problem. (2.5)

The condition of non-negativity has been met because the variables X1, X2 and X3 can only take positive values. In addition, the problem is infeasible because there is vector x, where the satisfaction of the constraints of the problem is achieved. Another condition that has been achieved by the problem is that it is unbounded. This means that all the constraints do not restrict the cost function. Therefore, many feasible solutions can be developed to improve the cost function. Lastly, there is an optimal solution to the problem. This means that the function established has a unique minimum point and/or maximum point.

Plot the constraints and identify the feasible region. Graphically or with corner points find the best combination of compost and topsoil that meets the stated conditions at the lowest cost per bag. Identify the lowest cost possible. (7.5)

Let X1 = cost of sand

X2= cost of clay

X3= cost of humus

Cost of Compost

4X1 + 3X2 + 5X3 ≤ 12 cents

X1 ≥ 0

X2 ≥ 0

X3 ≥ 0

Cost of Topsoil

3X1 + 6X2 + 12X3 ≤ 20 cents

X1 ≥ 0

X2 ≥ 0

X3 ≥ 0

Each potting soil costs: 12 X1 + 12 X2 + 10 X3

Minimize p= 12 X1 + 12 X2 + 10 X3

Subject to

4X1 + 3X2 + 5X3 ≤ 12 cents……………..i

3X1 + 6X2 + 12X3 ≤ 20 cents…………….ii

Therefore; putting the dummy variables gives

p= 12 X1 + 12 X2 + 10 X3

4X1 + 3X2 + 5X3 + S1 + S2 = 12 cents……………..i

3X1 + 6X2 + 12X3 + S1 + S2 = 20 cents…………….ii

Tableau 1

X1

X2

X3

S1

S2

P

rhs

4

3

5

1

0

0

12

3

6

12

0

1

0

20

-12

-12

-10

0

0

1

0

Tableau 2

X1

X2

X3

S1

S2

P

rhs

4

6

5

1

0

0

24

3

0

12

0

1

0

40

-12

0

-10

0

0

1

0

(Not complete)

Question 2 (10 marks)

Gibson Products produces cast bronze valves for use in offshore oil platforms. Currently, Gibson produces 1600 valves per day. The 20 workers at Gibson work from 7 a.m. until 4 p.m., with 30 minutes off for lunch and a 15 minute break during the morning work session and another at the afternoon work session. Gibson is in a competitive industry, and needs to increase productivity to stay competitive. They feel that a 20 percent increase is needed.

Gibson’s management believes that the 20 percent increase will not be possible without a change in working conditions, so they change work hours. The new schedule calls on workers to work from 7:30 a.m. until 4:30 p.m., during which workers can take one hour off at any time of their choosing. Obviously, the number of paid hours is the same as before, but production increases, perhaps because workers are given a bit more control over their workday. After this change, valve production increased to 1800 units per day.

Calculate labor productivity for the initial situation (3)

Output = 1600 valves per day

There are 20 workers, and this means that each worker produces 80 valves (1600 valves / 20) in a day. In one day, the workers operate for 8 hours. Therefore, the productivity is 8 vales per hour per worker (80 valves/ 10 hours).

Calculate labor productivity for the hypothetical 20 percent increase (3)

The 20% percent increase means that the output should be 1920 valves. Therefore, each worker should produce 96 valves (1920 valves / 20). In a day, the workers will produce 10 valves.

After adopting the new strategy, the productivity of workers is: 1800/20 = 90 valves. Daily productivity is 90/10 = 9 valves.

d. Write a short paragraph analyzing these results. (4)

The results show that there was an increase in the productivity of the employees after the implementation of the new policy. This is indicated by the increased from productivity of 8 valves to 9 valves. However, the employees did not increase productivity as expected. This is because the management speculated that the workers would produce 10 valves per day, but the actual productivity was 9 valves per day. However, there is a small deviation from the expected performance of the workers.

Question 3 (10 marks)

The annual demand for an item is 40,000 units. The cost to process an order is $40 and the annual inventory holding cost is $3 per item per year.

What is the optimal order quantity, given the following price breaks for purchasing the item? (5)

What is the total annual cost at the optimal behavior? (5)

Quantity

Price

1-1,499

$2.50 per unit

1,500 – 4,999

$2.30 per unit

5,000 or more

$2.25 per unit

Question 4 ( 15 marks)

A network consists of the following list. Times are given in weeks.

Activity

Preceding

Optimistic

Probable

Pessimistic

A

—

5

11

14

B

–

3

3

9

C

—

6

10

14

D

A, B

3

5

7

E

B

4

6

11

F

C

6

8

13

G

D, E

2

4

6

H

F

3

3

9

Draw the network diagram and find the total project time.(5)

Calculate the expected duration and variance of the critical path.(5)

Calculate the probability that the project will be completed in less than 28 weeks. (5)

Question 5 ( 10 marks)

Consider the following requirements for a certain product.

Period

1

2

3

4

5

6

7

8

Gross requirements

0

200

200

500

0

400

0

400

Beginning inventory = 500 units

Setup cost = $500 per setup

Lead time = 1 week

Holding cost = $3 per unit per week

Develop the lot-for-lot MRP table.

Calculate the total relevant cost.

A

Lead time

1

Safety Stock

0

Lot size

1

Minimum quantity

0

Period 0

Period 1

Period 2

Period 3

Period 4

Period 5

Period 6

Period 7

Period 8

Gross requirements

Scheduled receipts

On Hand Inventory

NET POQ Req

Planned receipts

Planned orders