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1. (10 points) Consider the following project management graph:
(b) Determine the total slack and free float (free slack) of each activity.(a) Determine the EST, EFT, LST, LFT of each activity.
Solution:time reduction for activity Let ti the start time for activity i. i, and let xi be the number of units of min s.t. (^300) tA โx Atstart + 200 โฅ x 0 B + 350xC + 260xD + 320xE t tBC โโ starttA โฅ 20 โฅ (^0) โ xA t tDE (^) โโ (^) ttBC โฅโฅ (^2550) โโ xxCB t tEf inish โ tD โ โฅ tE (^40) โฅ โ 30 x Dโ xE t tstartf inish = 0 โค 90 00 โคโค xxAB โคโค (^55) 00 โคโค xxCD โคโค (^55) 0 โค xE โค 5
(b) Suppose that instead of meeting the deadline exactly, there is a penalty of $290 for eachday the project extends beyond 90 days. In addition, there is a reward of $250 for everyday the project finishes under 90 days. Write two (or one) linear programming problems to determine by how much to expedite the duration of each activity in order to minimizethe net cost. Solution: We use two linear programs:
the objective function min Solution: Let ti the start time for activity tf inish โ tstart and remove the constraint that i, and let xi be the number of units of tstart = 0. time reduction for activity i. min s.t. ttf inishA โ t (^) startโ tstart โฅ 0 (xsa) t tBC โโ starttA โฅ 20 โฅ (^0) (x(xacsb)) t tDE โโ (^) ttCB โฅโฅ 2550 ((xxcebd)) t tEf inish โ tD โ โฅ tE (^40) โฅ 30 (x de()xef )
(b) Write the dual problem. Solution: max s.t. (^20) โx xacsa + 25 โ xsbxbd = + 50 โ 1 xce + xde xsa โ xac = 0 x xsbac โโ xxbdce = 0= 0 x xbdce โ+ xxdede โ= 0 xef = 0 x xefsa (^) , x= 1sb, xac, xbd, xce, xde, xef โฅ 0 (c) Which problem does the dual correspond to? Solution: After multiplying each flow balance constraint with โ1, we obtain: max s.t. (^20) xsax ac+ + 25xsb = 1xbd + 50xce + xde xac โ xsa = 0 x xbdce โโ xxsbac = 0= 0 x xdeef โโ xxbdce = 0โ xde = 0 x^ โsa^ x, xefsb^ =, x^ โac^1 , xbd, xce, xde, xef โฅ 0
This is the formulation of the longest path problem from start to finish.
- (20 points) Consider the project given below. ActivityA Duration 4 Predecessorsโ Workforce requirement 9 BCD 222 โ โโ (^364) EFG 323 BCF 872 (a) Draw the AON network. Determine the EST, EFT, LST, LFT of each activity.H^3 E,G^1 (b) Give the schedule diagram and the resource loading diagram when all activities are sched-uled at their earliest start time. (c) Give the schedule diagram and the resource loading diagram when all activities are sched-uled at their latest start time. (d) Give the schedule diagram and the resource loading diagram for a hybrid schedule suchthat no day requires more than 15 workers.
s
t
Augmenting path: s โ 1 โ t with bottleneck capacity ฮด = 20.
s
t
Augmenting path: s โ 2 โ t with bottleneck capacity ฮด = 15.
s
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Augmenting path: s โ 3 โ t with bottleneck capacity ฮด = 10.
s
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No more augmenting paths. Flow is optimal. Problem 5 Initial flow:
s
t
Augmenting path: s โ 1 โ 3 โ t with bottleneck capacity ฮด = 2.
s
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Augmenting path: s โ 2 โ 4 โ t with bottleneck capacity ฮด = 3.
- Add arc (can carry.j, t) for each car j with capacity equal to the number of people the car We wish to find the maximum flow from s to t.
s
Hat Mon McC Cp
c c c c
t
(b) Solve the formulation of part (a) with the Ford-Fulkerson algorithm. Show the flow aftereach iteration of the algorithm. Solution:non-zero flow. Due to space constraints, we will only draw the middle arcs that have Initial flow.
s
Hat Mon McC Cp
c c c c
t
0/4^ 0/
We consider the augmenting paths sin the augmenting path algorithm, we would consider them one by one). โ Cp โ c 4 โ t at once. Since none of the arcs overlap we can safely do this (note that s โ Hat โ c 1 โ t, s โ M on โ c 2 โ t, s โ M cC โ c 3 โ t,
s
Hat Mon McC Cp
c c c c
t
2/4^ 2/
We consider the augmenting paths sin the augmenting path algorithm, we would consider them one by one). โ Cp โ c 3 โ t at once. Since none of the arcs overlap we can safely do this (note that s โ Hat โ c 2 โ t, s โ M on โ c 1 โ t, s โ M cC Note that โ c 4 โ t, the augmenting paths have bottleneck capacity 1 or 2.
s
Hat Mon McC Cp
c c c c
t
Since all sink-adjacent arcs are saturated, this flow is optimal.
- (a) Page 431, problem 16. Formulate the problem as a maximum flow problem. Solution: We represent each month and project as node. We then add source s and sink โข t.Add arc (s, i) for each month i with capacity 8. This represents the number of - workers available in each month.Add arc (i, j) for each month i and project j with capacity 6 if the month i is
s
Augmenting path: s โ m 2 โ p 3 โ t with bottleneck capacity ฮด = 6.
Augmenting path: s โ m 3 โ p 1 โ t with bottleneck capacity ฮด = 6.
Augmenting path: s โ m 1 โ p 1 โ t with bottleneck capacity ฮด = 2. Finally, we use augmenting pathscomplete the demand for project 2. s โ m 2 โ p 2 โ t, s โ m 3 โ p 2 โ t, s โ m 4 โ p 2 โ t to
- start, 0 A, - B, - C, - D, - E, - F, - G, - H, - finish, - StartAB 002 028 002 028 000 Solution: Activity EST EFT LST LFT Total slack Free float - CDE 826 12106 10108 141214 404 - GHF^1286 1497 131213 141414 507 - Finish
- ActivityA Current Duration 20 Predecessorsโ Maximum reduction 5 Unit expediting cost 2. (10 points) Consider the project given below.
- (a) Formulate an LP to minimize the expediting cost such that the project finishes in 90 days.E^30 C,D^5
- 6/80/ t
- 0/80/ - 0/100/ - 6/ - 6/
- 6/ t
- 6/
- 0/80/ - 0/ - 0/ - 12/ - 6/ - 6/
- 6/ t
- 6/
- 6/
- 0/ - 0/106/ - 12/ - 6/ - 6/ - 6/
- 8/86/ t
- 6/80/ - 8/ - 0/ - 12/ - 6/ - 6/ - 6/ - 2/
- 8/88/ t
- 8/
- 6/ - 10/108/ - 12/ - 6/ - 6/66/ - 2/ - 2/ - 2/ - 6/
