Handbook of Civil Engineering Calculations

IA = $1750; IB = $1480 and l.5IB = $2220; /c = $0690 and 18/c = $1.242. Also, when7V 5
= 401.5, then 1.5I 3 = $2.214. The total profit can be increased by reducing NA by 1 unit,
reducing Nc by 1.8 units, and increasing NB by 2(1.5) = 3 units, with the increase in prof-
it being approximately $2.220 + $2.214 - ($1.750 + $1.242) = $1.442. Thus, the arbitrary
set of N values given above does not yield the maximum profit.
So the total profit is maximum when IA = .5IB = 1.81/c, or 4.75 - 0.010OTVx, = 3.90 -
0.00427V 5 = 4.05 - 0.0036TV 0 Eq. b.

5. Find the production that will maximize profit
   Applying Eq. Z?, express TV 5 and Nc in terms OfTV 4. Substitute these expressions in Eq. a,
   and solve the resulting equation for NA. Then calculate NB and Nc. The results are NA =
   301,TV 5 = 514,andTVc = 641.


6. Devise a semigraphical method of solution
   In Fig. 10, plot the straight lines that represent IA l.5IB, and 1.8/c Pass an arbitrary hori-
   zontal line L through these lines to obtain a set of TV values at which T 4 = l.5IB — 1.8/c
   Scale the TV values, and determine whether they satisfy Eq. a. Now displace the horizontal
   line until the TV values do satisfy this equation.

DYNAMIC PROGRAMMING TO MINIMIZE

COST OF TRANSPORTATION

A firm must ship merchandise by truck from town A to town E, and the trip will last 4
days. The driver will stop in district B the first night, district C the second night, and dis-
trict D the third night. The number of towns in each district is: district B, three; district C,
two; district D, three. The driver can stay overnight in any of these towns, and the cost of

Number of units produced monthly, N

FIGURE 10. Plotting of indicated multiples of incremental profit.

Multiple

of

incremental profit,

$

IA

,1.5I

,1.8IB

C