Handbook of Civil Engineering Calculations

(singke) #1

the average inventory = Q.5AB = QAN parts. By taking the product of the carrying cost
and the average inventory, the annual cost of carrying the inventory = $1.40(0.4TV) =
0.567V. With a $5 per unit regular cost, the annual regular cost = $5(41,600 units) =
$208,000. Then the annual total cost C = (22,880,000/7V) + 0.567V+ 208,000.



  1. Find the economical lot size
    To minimize C, set the derivative of C with respect to TV equal to zero; solve for TV to find
    the economical lot size. Thus, dCIdN= -(22,880,000/7V^2 ) + 0.56 = O; TV= 6392 parts per
    lot.

  2. Compute the total annual cost
    Substitute the number of parts from step 5 in the total annual-cost equation in step 4. Or,
    C = 3580 + 3580 + 208,000 = $215,160.


EFFECT OF QUANTITY DISCOUNT


ON OPTIMAL INVENTORY LEVEL


Given the data from the previous calculation procedure, the firm finds that it can obtain
quantity discounts if the parts are produced in lots of 7500 or more. These discounts re-
duce the regular production cost from $5 to $4.80 per part; this saving reduces the interest
cost on inventory by $0.02 per part. Determine the most economical lot size under these
conditions.


Calculation Procedure:


  1. Determine the number of parts for minimum cost
    Assume that TV ^ 7500. Proceeding as before, express C in terms of TV, and set dC/dN= O.
    The annual cost of carrying the inventory = $1.38(0.4TV) = 0.552TV. Also, the annual
    regular cost = 41,600($4.80) = $199,680. Also, C = (22,880,000/TV) + 0.552TV+ 199,680.
    And dCldN = -(22,880,000/TV2) + 0.552 = O. For minimum cost, TV- 6438 parts. Howev-
    er, since discounts are not obtained until TV reaches 7500, the last calculation lacks signif-
    icance for this situation.

  2. Compute the economical lot size
    Set N= 7500 and substitute in the cost equation above. Then C = 3051 + 4140 + 199,680
    = $206,871. Since this result is less than the value of $215,160 computed in the previous
    calculation procedure corresponding to 6392 parts, the economical lot size is 7500 parts.


PROJECT PLANNING BY THE


CRITICAL-PATH METHOD


Table 16 lists the activities performed in preparing a building site and installing the utili-
ties. Assuming that the estimated durations are precise, determine the minimum time
needed to complete the project. Identify the critical path. Upon completion of the project,
it was found that each activity was undertaken at the earliest possible date and that its du-
ration coincided with the estimate, except for the following: activity D was started 3 days
late; activity H required 9 days instead of 6; activity I required 5 days instead of 4. Deter-
mine the duration of the project.

Free download pdf