<< Chapter < Page Chapter >> Page >

Quantitative decision making

Industrial engineers are often tasked by firms to make decisions that affect manufacturing operations. In making decisions, industrial engineers must weigh a variety of factors. These factors include cost, safety, regulatory constraints, ergonomics and others.

In determining which design alternatives for manufacturing are the best, industrial engineers often form mathematical models. With such models, decisions can be made that are optimal in terms of various criteria. The following exercise illustrates the sort of decision making that might be involved in a hypothetical manufacturing situation.

Example 7: A manufacturing firm produces commercial aircraft. The firm plans to expand its manufacturing operations to produce aircraft by opening a second plant at another site. An industrial engineer oversees the manufacturing operations for the firm and is charged with making decisions that impact the operations at the new site.

The industrial engineer has a deep understanding of two processes that could be used to govern manufacturing at the new site. For convenience let us refer to these two processes as process A and process B. The industrial engineer knows that manufacturing cost is a very important issue with the firm. In researching the two options, the industrial engineer has determined mathematical expressions that express the manufacturing cost for each of the two options.

In order to produce n aircraft during a 30-day production run, the cost in millions of dollars associated with process A is known to be

8 n 2 + n + 2 size 12{8`n rSup { size 8{2} } +n+2} {}

The cost in millions of dollars associated with manufacturing n aircraft over the same span of time for process B is known to be

7 n 2 + 2 n + 14 size 12{7`n rSup { size 8{2} } +2`n+"14"} {}

Which process has the lowest associated cost?

We begin the process of determining which process is most cost effective by forming the difference between the cost associated with process A and the cost associated with process B. Let us designate this difference by the variable D

D = ( 8 n 2 + n + 2 ) ( 7 n 2 + 2 n + 14 ) = n 2 n 12 size 12{D= \( 8`n rSup { size 8{2} } +n+2 \) - \( 7`n rSup { size 8{2} } +2`n+"14" \) =`n rSup { size 8{2} } - n - "12"} {}

Whenever D is greater than zero, one may conclude that the cost associated with process A is greater than that associated with process B. Whenever D is less than zero, one may conclude that the cost associated with process B is greater than that associated with process A. The cost is the same for each process whenever D equals zero.

D is a quadratic polynomial, so we begin by solving its roots using the equation below

n 2 + n 12 = 0 size 12{n rSup { size 8{2} } +n - "12"=0} {}

We can solve for the roots of this equation using the method commonly known as solution by completing the square. Let us rearrange the equation with the terms involving n2 and n on the left hand side and the constant on the right.

n 2 + n = 12 size 12{n rSup { size 8{2} } +n="12"} {}

If we add ¼ to the left and the right sides we obtain

n 2 + n + 1 / 4 = 12 . 25 size 12{n rSup { size 8{2} } +n+1/4="12" "." "25"} {}

The left hand side can be expressed as the square ( n + ½) 2 . We reflect this below

( n + 1 / 2 ) 2 = 12 . 25 size 12{ \( n+1/2 \) rSup { size 8{2} } ="12" "." "25"} {}

If we take the square root of each side, we are left with

n + 1 / 2 = ± 3 . 5 size 12{n+1/2= +- 3 "." 5} {}

This yields the pair of roots 3 and -4.

Let us recall that the variable n represents the number of aircraft produced during a 30-day production run. The root (-4) is physically impossible for it is impossible to produce a negative number of aircraft. The remaining root (3) tells us that the cost associated with process A is equal to the cost associated with process B when 3 aircraft are produced during a 30-day production run.

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Math 1508 (laboratory) engineering applications of precalculus' conversation and receive update notifications?

Ask