1. IndE 311
Problem Session
March 2nd, 2007

2. Pr )
Queuing model: M/ M/ s

3. Case 17.1 a) Status quo at the presses:
Queuing model: M/ M/ s
In-process inventory: 7.52
Status quo at the inspection station:
Queuing model: M/ D/ 1
In-process inventory: 3.94
Inventory cost = ( )($8/hour) = $91.68 / hour
Machine cost = (10)($7/hour) = $70 / hour
Inspector cost = $17/hour
Total cost = $178.68/ hour

4. Case 17.1 (cont’d) b) Proposal 1: reduce power at presses
Queuing model for the presses: M/ M/ s
In-process inventory at the presses : 11.05
In-process inventory at the inspection station: 3.94
Inventory cost = ( )($8/hour) = $ / hour
Machine cost = (10)($6.50) = $65 / hour
Inspector cost = $17 / hour
Total cost = $ / hour
The total cost is higher than for the status quo (Slowing down the machines won’t change in-process inventory for the inspection station.)

5. c) Proposal 2: Substitute a younger inspector, with original speed
In-process inventory at the presses : 7.52
Queuing model for the inspection station: M / E2 / 1
In-process inventory at the inspection station: 4.15
Inventory cost = ( )($8/hour) = $93.36 / hour
Machine cost = (10)($7/hour) = $70 / hour
Inspector cost = $19 / hour
Total cost = $ / hour
The total cost is higher than for the status quo due to the increase in the service rate variability and the resulting increase in the in-process inventory.

6. d) Queuing model for the presses: M / M / s
They should consider increasing power to the presses (increasing there cost to $7.50 per hour but reducing their average time to form a wing section to 0.8 hours).
Queuing model for the presses: M / M / s
In-process inventory at the presses : 5.69
With original inspector: In-process inventory at the
inspection station:3.94
Inventory cost = ( )($8/hour) = $77.04 / hour
Machine cost = (10)($7.50/hour) = $75 / hour
Inspector cost = $17 / hour
Total cost = $ / hour

7. Case 2. The PCB production machine repair shop
Queuing model: M / M / s / / / N

8. a) (cont’d)

9. E(# of PCB machines available) = 10 – L = 4.782
Case 2 (cont’d)
E(# of PCB machines available) = 10 – L = 4.782
E [revenue] = E [# of machines available] * 50 * 180 * 30 = $ 1,291,075 per month
E [cost] = E [# of machines available] * 30 * 50 * ,000*3 = $ 1,130,717 per month
E [profit] = E [revenue] - E [cost] = $160,358 per month

10. b) Consider leasing another repair station
Queuing model: M / M / s / / / N
L = 4.311
E [profit] = $152,010 < = $160,358
No, they should not lease an additional repair station
c) Consider a new type of repair station
Queuing model: M / M / s / / / N

11. s=3
E [cost] = $1,315,800/month
L = 4.49
E [profit] = $171,900 / month