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The engineer of a concrete products plant is scheduling production of a certain size concrete pipe over the next 5 months. No pipe is on hand at the beginning of the first month. The engineer wants to meet the projected demand over the 5-month period at minimum cost. Constraints are imposed on the monthly regular rate processing time and the amount of raw materials available each month, as shown in the table below:

	Jul	Aug	Sep	Oct	Nov	Dec
Pipe Required (Number of 20 foot stands)	20	30	40	20	40	30
Cost of labor during regular time $/hr	$15	$15	$15	$18	$18	$18
Cost of labor during overtime $/hr	$17	$17	$17	$20	$20	$20
Labor available (persons)	30	30	40	40	20	20
Material available (stands)	40	50	50	10	30	20

It takes 75 hours to produce one stand of pipe, and $20 to store one stand of pipe for one month.

1. a)Assume that no materials can be carried over from month to month.
2. b)Assume that materials can be carried over from month to month at no cost.

As head of the engineering division of HP&L you have been asked to determine the least expensive way to add Bryan, Texas to the power grid, from Houston, Texas. Local substations lead from Houston to Bryan, and are currently on the grid, but if you use any of them them they will have to be modernized and upgraded at the following costs:

	Location	Cost to upgrade a station ($ million) if you go through that station
Station 1	Houston	1
Station 2	Somerville	4
Station 3	Hearne	7
Station 4	Tidwell	11
Station 5	Athens	8
Station 6	Pawnee	5
Station 7	Bryan	9

Costs for right-of-way acquisition, tower construction, and new line costs for connecting through any of the possible locations (in millions) are as shown below. If there is no station pair shown, you cannot acquire the right-of-way to connect between those stations. For example, it will cost you $6 million if you decide to string lines between Houston and Somerville (1-2), and you cannot acquire right-of-way between Houston and Tidwell, since 1-4 is not listed.

Line costs in $millions:

Between Station	And Station	Cost ($millions)
1	2	6
2	4	5
4	5	3
5	7	11
1	7	32
1	6	8
6	7	9
1	3	12
3	5	10
2	3	6

Determine the least expensive way to connect Houston to Bryan.

Given: The manager of a paper company receives an order for low, medium, and high grade paper. The company consists of two factories which are each capable of producing each style of paper, but in differing quantities per day. Pertinent data are given below. Note that if you decide to run Factory 1 for one complete day, it will spew out 8 tons of low grade paper, 1 ton of medium grade paper, and 2 tons of high grade paper. That’s just what it does. It is not necessary that your solution be based on “whole” days – a part of a day is acceptable.

Factory outputs, costs to run, and order quantities are as follow:

	Required amount to satisfy order	Factory 1	Factory 2
Low grade	16 tons	8 tons/day	2 tons/day
Medium grade	5 tons	1 ton/day	1 ton/day
High grade	20 tons	2 tons/day	7 tons/day
Daily cost to run plant		$1000/day	$2000/day

Required:

1. Solve the problem to fill the order at a minimum cost.
2. Solve the problem to fill the order while running the two factories for as few total days as possible – i.e. get the order filled while running the two factories for a combined minimum total number of days. For example, I don’t care if factory 1 works 11.2 days and factory 2 works 6.8 days, for a total of 18 days, but I want that total number of days in the factories (18) minimized.
3. Solve the problem to fill the order in a minimum time – i.e. to get the order finished and out the door and to the customer as quickly as possible.

Hint:

Dr. Lowery,

I have a question about web problem 5.18 that is due tomorrow. It appears to me that numbers 2) and 3) under what is required are asking the same thing. This is what they say:

2. Solve the problem to fill the order while running the two factories for as few total days as possible.

3. Solve the problem to fill the order in a minimum time.

Aren’t “the fewest days” and “minimum time” the same thing?

Joe:

Sorry I am just getting back to you. Too many emails.

Not really.

The optimum answer for cost might be A = 10 days, B = 30 days. (Note A+B = 40 days, cost = $100)

The optimum answer for minimum total time = A+B might be A = 14 days, B = 20 days. (A+B now = 34 days, cost = $120)

In the first case the customer would not get his order until 30 days.

In the second case he would not get his order until 20 days.

Do you see a way to get the customer his order quicker than 20 days?

L^3

Given: A company has four machines that can accomplish three jobs. Each job can be assigned to one, and only one machine. The objective is to assign the jobs to the machines in the most cost-efficient manner. Costs for each job-machine combination are given in the following table.

	Machine 1	Machine 2	Machine 3	Machine 4
Job 1	19	23	28	31
Job 2	7	14	16	19
Job 3	10	15	20	22

Required: Solve for the minimum cost.

Given: A rapid transit system serves a metropolitan area consisting of five bus stations spread around in the suburbs. Each morning, busses are dispatched from their night storage and repair barns to these stations from three bus repair barns near the center of town. Data for bus availability at each bus barn, and bus requirements to pick up passengers along each station’s route, along with the costs of dispatching a bus from the bus barns to the stations are given below.

Bus Stations where busses first pick up riders and finish their routes.	Busses Required at 5:00 am
1	30
2	20
3	10
4	10
5	15

 

Bus repair barns where busses return at 2:00 am for cleaning and repair until leaving at 5:00 am for the rider stations.	Number of bus cleaning/repair stations available at barns.
A	40
B	20
C	25

Given: In order to keep to a construction schedule, I must place 1000 tons of pavement a day. Because of other projects in the area, I will be unable to purchase all of my pavement from one batch plant. In fact, I will have to purchase from three plants in the area because of the high demand.

Plant G can supply me with up to 275 tons/day, total, of either Topping or Binder – my choice. Thus you can get up to 275 tons of whatever you want – 275 tons of Topping and no Binder, or 100 tons of Topping and 175 tons of Binder, or 2 tons of Topping and 273 tons of Binder, or 10 tons of Topping and no Binder – anything you like, but no more than 275 tons per day.

Plant H can supply up to 350 tons/day of either Topping or Binder, same idea as for Plant G.
Plant I can supply me with up to 595 tons/day of either Topping or Binder, same idea as for Plant G.

I have two pavers, one laying Topping and one laying Binder. The Topping Paver lays 575 tons/day and the Binding Paver lays 425 tons/day, which as you might imagine is the exact daily amount required to construct the road.

The cost to transport paving material from Plant G to the Topping Paver is $32/ton. Cost from Plant G to the Binder Paver is $36/ton, from Plant H to Topping Paver, $30/ton, from Plant H to Binder paver, $33/ton, I to Topping paver, $31/ton and I to Binder paver, $35/ton.

Required: Solve for the minimum cost of acquiring the necessary material.

Three cities (A, B, and C) bring in refuse of 100 tons/day, 50 tons/day, and 30 tons/day, respectively. The refuse can be shipped in several directions, including to each other, or to two transfer stations D and E, using local trucks. Once the refuse is shipped to the transfer stations at D or E, it is transferred to highway trucks and shipped to one of three landfills F, G, or H. The landfills have the following maximum capacities per day: F = 100 tons/day, G = 50 tons/day, and H = 60 tons/day.

The following unit costs per ton are charged for any and all refuse that comes into, through, or to the cities, transfer stations, or landfills. It is a “handling” charge applied to any material transported through the node:

City A	$4/ton
City B	$6/ton
City C	$4/ton
Transfer Station D	$3/ton
Transfer Station E	$4/ton
Land Fill F	$8/ton
Land Fill G	$12/ton
Land Fill H	$7/ton

Thus, the 100 tons/day entering City A will incur a $400 charge, simply because the refuse entered City A.

The costs to transport material between locations are as listed:

	A	B	C	D	E	F	G	H
City A	–	$2/ton	–	$3/ton	$4/ton	–	–	–
City B	$2/ton	–	$3/ton	$5/ton	$4/ton	–	–	–
City C	–	$3/ton	–	$7/ton	$9/ton	–	–	–
Transfer Station D	–	–	–	–	$3/ton	$10/ton	$8/ton	$12/ton
Transfer Station E	–	–	–	$3/ton	–	$6/ton	$4/ton	$13/ton

SEE TABLE ABOVE FOR THE NUMBERS SHOWN ON THE MAP BELOW

Total hauling costs per ton are shown in the table, and on the schematic above. Set up and solve the LP solution to determine the least cost way to get the refuse from the cities to the landfills. Note that it is possible to haul refuse between the cities as shown, and also between the transfer stations if that proves to be more economical.

Print out this page and bring it to class with you.

A contractor wants to hire various sub-contractors to work for him on a job. Each of the subs have tendered the sealed bids as follows:

Job Part	1	2	3	4	5	6	7	8	Cost	Special conditions
Bid ID
A1	Y	Y	Y				Y		8	None
A2				Y	Y	Y		Y	10
A3	Y	Y	Y	Y	Y	Y	Y	Y	16
B1	Y		Y	Y					4	Brother-in-law (at least one job) and minority contractor (10%)
B2						Y	Y		5
C1	Y							Y	2	Small contractor with small work force (maximum 2 jobs)
C2		Y	Y	Y					3
C3					Y	Y	Y		2
C4			Y	Y	Y	Y	Y		4
D1	Y					Y			3	Minority contractor (minorities must get at least 10% of award)
D2			Y				Y	Y	5
E1	Y	Y	Y			Y			10	New contractor (maximum bond – $18)
E2				Y			Y		6
E3	Y				Y			Y	8

A different problem, unrelated to the previous problem:

Job Part	1	2	3	Cost
Bid ID
A	Y	Y	Y	20
B		Y	Y	6
C	Y	Y		8

Assume that you are an engineer with the planning department for a county that contains several major metropolitan areas. The cities are located in a flood plain of a broad, meandering river with mountains overlooking the valley. The cities are under the jurisdiction of a single sewage authority that provides service for the entire area. There are five regional wastewater treatment plants that treat sewage for this metropolitan area. The treatment plants process the waste, and then ship sludge to any of five disposal sites. The disposal sites include two landfills (LF), two land disposal (LD) sites, and one incinerator (INC).

The treatment plants generate the following sludge (lb/day):

WT1 – 2500, WT2 – 6000, WT3 – 4000, WT4 – 6800, WT5 -3000 lb/day

The land fill, disposal sites, and incinerator can accept the following amounts of sludge up to (lb/day):

LF1 – 6700, LF2 – 7300, LD1 – 3000, LD2 – 5000, INCINERATOR – 2000 lb/day

Distances from the wastewater treatment plants to the disposal sites (in miles):

	LF1	LF2	LD1	LD2	INC
WT1	2	6	4	7	9
WT2	6	4	2	3	6
WT3	8	6	3	5	4
WT4	9	7	4	5	3
WT5	8	5	6	3	4

Transportation Costs (cents per lb per mile):

	LF1	LF2	LD1	LD2	INC
WT1	20	40	35	15	18
WT2	20	38	34	16	20
WT3	35	17	19	40	22
WT4	18	33	27	17	15
WT5	25	23	20	26	16

Once you get it to the disposal site, the costs for burying, or incineration will be (in cents/lb):

at LF1	at LF2	at LD1	at LD2	at INC
80	73	26	52	105

Required: Solve for the minimum sludge disposal costs.