Solution of Case 14.pdf

7/27/2019 Solution of Case 14.pdf

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I..Case No. 1Max. duration = 18 working daysThe Equipment resources which we need I shiftWe will need ( 4 shovels + 3 2 trucks for 12 days)

(2 shovels + 16 trucks for the remaining 6 days)Human Resources CostWe need only one day shift crew for the whole project durationDay shift of crew costs 110 L.E./dayEquipment Resources Cost/dayShovels cost = 4 x 600 = 2,400 L.E.& = 2 X 600 = 1,200 L.E.Trucks cost = 32 x 150 =4,800 L.E.& = 16 X 150 = 2,400 L.E.Transportation cost= (4 x 300) + (32 x 75) = 3,600 L.E.

........ phase (1)..... . .. phase (2)

. .... ...phase (1)

. .... ... phase (2)

. . . .....phase (1)

........phase (2)

The direct cost= [ (2,400 +4,800 + 110) x 12] + [ (1 ,200 +2,400 + 110) x 6] +3,600 = 113,580 L.E.Loan duration= 18 + 3 + 14 = 35 daysAc tual cost with the loan Interest= 113,580 x [ 1+ (0.0005 x 35)] =115,567.65 L.E

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. 'I No. of shovels= volume ofexcavation I 400. No . of rucks for one shovel= Shovel capacity x soil factor/truck capacity=400 x 1.1 I 55= 8 trucks. Cost of shovels = no. of shovels x 600. Cost of rucks = no. of rucks x 150.If the excavated volume is greater than 1 600 m3 then day and night shifts will be used, else day shift onlywill be used. The transportation cost for every duration will be calculated as follows:Total no. ofpower shovel x 300 +Total no. oftrucks x 75 Crew cost for each phase will be calculated as follows:no. ofday shifts x 110 + no. of night shifts x 165 The direct cost is calculated as follows:

[(Cost ofpower shovels/day+ cost of trucks/day+ cost of crew/day) x no. ofdays/1st phase+(Cost of power shovels/day+ cost of rucks/day+ cost ofcrew/day) x no. ofdays/2nd phase+Transportation cost].

Loan duration = completion duration (including weekends) + 2 calendar weeks Total cost= direct cost x (1 + 0.0005 x no. of days)

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''.

Case No.2Min. duration = 12 working daysThe Equipment resources which we need I shiftWe will need ( 5 shovels + 40 trucks for 12 days)Human Resources CostWe need one day shift and one night shift crews for the whole project durationCrew cost/day= 110 + 165 = 275 L.E./dayEquipmentResources Cost/dayShovels cost= 5 x 600 = 3,000 L.E.Trucks cost = 40 x 150 = 6,000 L.E.Transportation cost= (5 x 300) + (40 x 75) = 4,500 L.E.The direct cost= [ (3,000 + 6,000 + 275) x 12] + 4,500 = 115,800 L.E.Loan duration= 12 + 2 + 14 = 28 days

Actual cost with the loan Interest= 115,800 x [ 1+ (0.0005 x 28)] = 117,421.2 L.E

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I..Case No.3Optimum duration = 15 working daysThe Equipment resources which we need I shiftWe will need (4 shovels+ 32 trucks for 15 days)Human Resources CostWe need only one day shift crew for the whole project durationCrew cost/day = 11 0 L.E./dayEquipment Resources Cost/dayShovels cost = 4 x 600 = 2,400 L.E.Trucks cost = 32 x 150 =4,800 L.E.Transportation cost= ( 4 x 300) + (32 x 75) = 3,600 L.E.The direct cost= [ (2,400 +4,800 + 110) x 15 ] + 3,600 = 113,250 L.E.Loan duration= 15 + 2 + 14 =31 daysActual cost with the loan Interest= 113,250 x [ 1+ (0.0005 x 31)] = 115,005.4 L.E

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'f Case No . 4

Max. duration = 13 working daysThe Equipment resources which we need I shiftWe will need ( 5 shovels + 40 trucks for 8 days)& (4 shovels + 32 trucks for the remaining 5 days)Human Resources CostCrew cost= 110 + 165 = 275 L.E./dayCrew cost = 110 L.E./dayEquipment Resources Cost/dayShovels cost= 5 x 600 = 3,000 L.E.& = 4 X 600 = 2,400 L.E.Trucks cost = 40 x 150 = 6,000 L.E.& =32 X 150 = 4,800 L.E.Transportation cost= (5 x 300) + (40 x 75) = 4,500 L.E.

. ....... phase (1)

.. . . .. .. phase (2)

.. . . .. .. phase (1)

........ phase (2)

. .. .. ...phase (1)

. .. .. ...phase (2)

. .. . .. ..phase (1)

. .. . p h a s e (2)

--

The direct cost= [ (3,000 + 6,000 + 275) x 8] + [ (2,400 + 4,800 + 110) x 5] + 4,500 = 115,250 L.E.Loan duration= 13 + 2 + 14 = 29 daysActual cost with the loan Interest= 115,250 x [ 1+ 0.0005 x 29)] = 116,921.13 L.E

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I..Case No.5Max. duration = 14 working daysThe Equipment resources which we need I shiftWe will need (5 shovels+ 40 trucks for 4 days)& (4 shovels + 32 trucks for the remaining 10 days)Human Resources CostCrew cost= 110 + 165 = 275 L.E./dayCrew cost= 110 L.E./dayEquipment Resources Cost/dayShovels cost= 5 x 600 = 3,000 L.E.& = 4 X 600 = 2,400 L.E.Trucks cost = 40 x 150 = 6,000 L.E.& = 32 X 150 = 4,800 L.E.Transportation cost= (5 x 300) + (40 x 75) = 4,500 L.E.

....... . phase (1). . ...... phase (2)

. . ... . .. phase (1)

........ phase (2)

. ......phase (1)

........phase (2)

........phase (1)

..... ...phase (2)

The direct cost= [ (3,000 + 6,000 + 275) x 4] + [ (2,400 + 4,800 + 110) x 10] + 4,500 = 114,700 L.E.Loan duration= 14 + 2 + 14 = 30 daysActual cost with the loan Interest= 114,700 x [ 1+ 0.0005 x30)] =116,420.5 L.E

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Case No.6Max. duration = 16 working daysThe Equipment resources which we need I shiftWe will need (4 shovels+ 32 trucks for 14 days)& (2 shovels + 16 trucks for the remaining 2 days)Human Resources CostCrew cost= 110 L.E./dayCrew cost= 110 L.E./dayEquipment Resources Cost/dayShovels cost = 4 x 600 = 2,400 L.E.& = 2 X 600 = 1,200 L.E.Trucks cost = 32 x 150 = 4,800 L.E.& = 16 X 150 = 2,400 L.E.Transportation cost= (4 x 300) + (32 x 75) = 3,600 L.E.

........ phase (1)........ phase (2)

........ phase (1)

........ phase (2)

. .......phase (1)

. . .. . ...phase (2)

. .......phase (1)

. .......phase (2)

The direct cost= [ (2,400 +4,800 + 110) x 14] + [ (1,200 + 2,400 + 110) x 2] + 3,600 = 113,360 L.E.Loan duration= 16 + 2 + 14 =32 daysActual cost with the loan Interest= 113,360 x [ 1 + (0.0005 x 32)] = 115,173.76 L.E

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Case No.7Max. duration = 17 working daysThe Equipment resources which we need I shiftWe will need (4 shovels + 3 2 t rucks for 13 days)& (2 shovels + 16 trucks for the remaining 4 days)Human Resources CostCrew cost= 110 L.E./dayCrew cost = 11 0 L.E./dayEquipment Resources Cost/dayShovels cost = 4 x 600 = 2,400 L.E.& = 2 X 600 = 1,200 L.E.Trucks cost = 32 x 150 = 4,800 L.E.& = 16 X 150 = 2,400 L.E.Transportation cost= (4 x 300) + (32 x 75) == 3,600 L.E.

........ phase (1)

..... ... phase (2)

........ phase (1)

........ phase (2)

.. ......phase (1)

........phase (2)

.. ......phase (1)

. ........phase (2)

rhc direct cost= [ (2,400 + 4,800 + 110) X 13] + [ (1,200 + 2,400 + 110) X 4] + 3,600 = 113,470 L.E.l "oa11 duration = 17 +2 + 14 = 33 daysAolunl ctml with the loan Interest= 113,470 x [ 1 + (0.0005 x 33)] =115,342.26 L.E

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Solution of Case 14.pdf