MRE 5001 Assignment 1: Hints for those who made mistakes
1
Question 2.
| B | |
| Investment | -10,000.00 |
| Annual receipts | 3,000.00 |
| Salvage Value | 2,000.00 |
| MARR 15% | |
| PW Receipts | 10,056.47 |
| PW Salvage | 994.35 |
| Total PW of cash inflow | 11,050.82 |
| PW of investment | -10000.00 |
| NET PW | 1,050.82 |
| Annual Receipts | 3,000.00 |
| Annual Costs | -2,686.52 |
| NET AW | 313.48 |
| FW Receipts | 20,227.14 |
| FW Salvage | 2,000.00 |
| Total FW of cash inflow | 22,227.14 |
| FW of Investment | -20,113.57 |
| NET FW | 2,113.57 |
Question 3
M ct =0.5hr
MTBMu = 2.0hrs
M pt = 2.0hrs
.
MTBMs =1000hrs
MTBM= MTBMu MTBMs
1 1
1
= 1.996hrs
Failure rate is assumed constant
MTBF = MTTF + MTTR = 2.5
= MTBF
1
= 0.400
MRE 5001 Assignment 1: Hints for those who made mistakes
2
pt
| = |
MTBM s
1
= 0.0010
M = ( )
( )( ) ( )( )
pt
M ct pt M pt
= 0.5037
MTBM M
MTBM
= 79.8011% (Achieved Availability).
Question 4.
M N
M
ct
ct
=61.48min
For 10% Risk, Z (Standardised normal Distribution Function) = 1.28
1
50
1
2
( )
N
i
Mctj Mct
= 16.63
Upper Limit = N
Z
ct
___
M
,=64.49 min. is less than the specified limit of 65 min and therefore the
system has passed the test.