Question

question 1

The following table provides a description for the project Z.

Activity	Duration	Predecessor	Cost to crash by 1 day	Max Days to Crash
A	5	-	$       300	1
B	3	-	$       100	1
C	4	A	$       100	1
D	3	A	$       200	2
E	4	B	$       400	3
F	2	D, E	$       500	2
G	4	E	$       300	3
H	3	C	$       100	2

1. What is the critical path of this project?

2. What is the duration of this project? (before crashing)

3. If your task is to crash this project by 2 days, what is the most efficient cost of doing it? (Just input the number with no decimals or dollar signs

question 2

The expected duration of the project (average) is 30 days and variance is 16.

1. What is the probability that the project will be completed on day 32 or earlier?

2. Suppose the official deadline for the project is 34 days. What is the probability that the project will be delayed?

question 3

A restaurant has tracked the number of meals served at lunch over the last four weeks. The data show little in terms of trends, but do display substantial variation by day of the week. Use the following information to determine the seasonal (daily) indices for this restaurant.

	Week
Day	1	2	3	4
Sunday	40	35	39	43
Monday	54	55	51	59
Tuesday	61	60	65	64
Wednesday	72	77	78	69
Thursday	89	80	81	79
Friday	91	90	99	95
Saturday	80	82	81	83

Answer 1

1. ALL POSSIBLE CHAINS

ACH = 12

ADF = 10

BEG = 11

BEF = 9

CRITICAL PATH = LONGEST PATH = ACH = 12

CRASHING

WE WILL CRASH FROM THE LOWEST CRASH PER PERIOD ACTIVITY THAT IS ON THE CRITICA PATH

ACTIVITY	NORMAL TIME	COST / PERIOD	TIMES CRASHED	# CRASH	CRASH COST
A	5	300			0
B	3	100	1	2ND	100
C	4	100			0
D	3	200			0
E	4	400			0
F	2	500			0
G	4	300			0
H	3	100	2	1ST AND 2ND	200

MOST EFFICIENT COST = 200 + 100 = 300

2. A. VARIANCE = 16

STANDARD DEVIATION = SQRT(VARIANCE) = SQRT(16) = 4

EXPECTED TIME = 30

DUE TIME = 32

Z = (DUE TIME - EXPECTED TIME) / STANDARD DEVIATION OF CRITICAL PATH)

Z = (32 - 30) / 4 = 0.5

PROBABILITY FOR A Z VALUE OF 0.5 = NORMSDIST(0.5) = 0.6915 OR 69.15%

B. VARIANCE = 16

STANDARD DEVIATION = SQRT(VARIANCE) = SQRT(16) = 4

EXPECTED TIME = 30

DUE TIME = 34

Z = (DUE TIME - EXPECTED TIME) / STANDARD DEVIATION OF CRITICAL PATH)

Z = (34 - 30) / 4 = 1

PROBABILITY FOR A Z VALUE OF 1 = NORMSDIST(1) = 0.8413

MORE THAN 34 = 1 - 0.8413 = 0.1587 OR 15.87%

3. AVERAGE OVERALL DEMAND = AVERAGE OF ALL DEMAND VALUES IN ALL SEASONS = 69.7143

AVERAGE PERIODIC DEMAND = SUM OF DEMAND IN A PARTICULAR PERIOD / N, WHERE N = NUMBER PERIODS TAKEN INTO CONSIDERATION

SEASONAL INDEX = AVERAGE PERIODIC DEMAND / AVERAGE DEMAND

PERIOD	AVG PERIOD DEMAND	AVG DEMAND	SEASONAL INDEX
1	40 + 35 + 39 + 43 / 4 = 38	69.7143	38 / 69.7143 = 0.545
2	54 + 55 + 51 + 59 / 4 = 53.3333	69.7143	53.3333 / 69.7143 = 0.765
3	61 + 60 + 65 + 64 / 4 = 62	69.7143	62 / 69.7143 = 0.889
4	72 + 77 + 78 + 69 / 4 = 75.6667	69.7143	75.6667 / 69.7143 = 1.085
5	89 + 80 + 81 + 79 / 4 = 83.3333	69.7143	83.3333 / 69.7143 = 1.195
6	91 + 90 + 99 + 95 / 4 = 93.3333	69.7143	93.3333 / 69.7143 = 1.339
7	80 + 82 + 81 + 83 / 4 = 81	69.7143	81 / 69.7143 = 1.162

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