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Muhammad Usman Bc240440613 MTH601 Dr. Muhammad Usman Dr. Juniad Zaidi

The document presents solutions to three questions involving project management and inventory calculations. It includes activity time estimates, critical path analysis, and economic order quantity (EOQ) calculations for inventory management. Key results include an expected project completion time of 21.83 units and EOQ values of 632 and approximately 2,190 units for different scenarios.

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UsmAn KhAn SwAti
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13 views3 pages

Muhammad Usman Bc240440613 MTH601 Dr. Muhammad Usman Dr. Juniad Zaidi

The document presents solutions to three questions involving project management and inventory calculations. It includes activity time estimates, critical path analysis, and economic order quantity (EOQ) calculations for inventory management. Key results include an expected project completion time of 21.83 units and EOQ values of 632 and approximately 2,190 units for different scenarios.

Uploaded by

UsmAn KhAn SwAti
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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Student Name:

Muhammad Usman
VU ID: Bc240440613
Subject: MTH601
Section Incharge

Dr. Muhammad Usman


Submitted to

Dr. Juniad Zaidi

Solution Of Question No 1
Table 1: Activity Time Estimates and Calculations

Activity From (i) To (j) to tm tp Expected Tine (tₑ) Variance

A 1 2 2 4 10 tₑ = (2+4×4+10)/6 = 4.67 [(10-2)/6]² = 1.78

B 1 3 3 5 9 tₑ = (3+4×5+9)/6 = 5.17 [(9-3)/6]² = 1.00

C 2 4 1 3 5 tₑ = (1+4×3+5)/6 = 3.00 [(5-1)/6]² = 0.44

D 3 4 2 6 12 tₑ = (2+4×6+12)/6 = 6.33 [(12-2)/6]² = 2.78

E 2 5 2 4 8 tₑ = (2+4×4+8)/6 = 4.33 [(8-2)/6]² = 1.00

F 4 6 3 5 11 tₑ = (3+4×5+11)/6 = 5.67 [(11-3)/6]² = 1.78

G 5 6 4 6 12 tₑ = (4+4×6+12)/6 = 6.67 [(12-4)/6]² = 1.78

H 6 7 2 4 6 tₑ = (2+4×4+6)/6 = 4.00 [(6-2)/6]² = 0.44


Now we analyze the paths from node 1 to node 7:

Path 1:

A → C → F → H = 4.67 + 3.00 + 5.67 + 4.00 = 17.34 time units

Path 2:

A → E → G → H = 4.67 + 4.33 + 6.33 + 4.00 = 19.33 time units

Path 3:

B → D → F → H = 5.17 + 6.33 + 5.67 + 4.00 = 21.17 time units

Path 4:

B → D → G → H = 5.17 + 6.33 + 6.33 + 4.00 = 21.83 time units

Critical Path: B → D → G → H

Expected Project Completion Time: 21.83 units

Project Network Diagram:


Solution of Question No 2
Given:

Demand = 24,000 units/year

Ordering cost = Rs. 200

Holding cost = Rs. 2/month = Rs. 24/year

EOQ = √(2 × D × S / H) = √(2×24000×200 / 24) = √400000 = 632 units

Time between orders = (EOQ / D) × 12 = (632 / 24000) × 12 ≈ 0.32 months

Orders per year = 24000 / 632 ≈ 38

Solution of Question No 3
Given:

Demand = 15,000 units/year

Setup cost = Rs. 500

Holding cost = Rs. 2.5/year

Shortage cost = Rs. 10/year

Production rate = 30,000 units/year

EOQ = √[(2 × D × S / H) × (Cs / (Cs + H))]

EOQ = √[(2 × 15000 × 500 / 2.5) × (10 / (10 + 2.5))] = √[4,800,000] ≈ 2,190 units

Max Inventory = EOQ × (Cs / (Cs + H)) × (1 - D / P) = 2190 × 0.8 × 0.5 = 876 units

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