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Problem 6.0) Assuming that you are the BOSS program, fill in the values in the table below. Determine the corresponding interarrival times depending on the following desired distributions:

            For Uniform Distribution:                      ARRIVE{ TIME = CUNIFORM(12,14)};

            For Normal Distribution:                       ARRIVE{ TIME = 0 MAX NORMAL(12,3)};

            For Exponential Distribution:                 ARRIVE{ TIME = EXPD(0.125)};

I.e. a continuous uniform distribution between 12 and 14, a normal distribution with a mean of 12 and a sigma of 3, and an exponential distribution with a mean of 0.125.

You will note that I have generated several random numbers and the corresponding answers for you to check by. I have also generated some random numbers for you to use, and several answers have also been filled in the table. If a random number is already shown, please use it. If there is no random number listed, generate one of your own. If the “answer” is shown, tell me the random number that “gave” me that answer.

Random Number	Uniform Distribution Interarrival Time	Random Number	Normal Distribution Interarrival Time	Random Number	Exponential Distribution Interarrival Time
0.254	12.508	0.409	11.310	0.013	0.001636
0.633	13.266	0.937	16.590	0.337	0.05137
0.500		0.500		0.500
0.812		0.254		0.896
0.211		0.607		0.254
0.637		0.633		0.904
	12.5		16.0		0.2
	13.0		18.0		0.4
	13.5		19.0		1.0
*		*		*
*		*		*
*		*		*

  * Generate your own random numbers for these cells.

Given: A construction firm has supply centers located in Atlanta, Chicago, and New York City. These centers have available 40, 20, and 30 units of a particular resource, respectively. The firm’s job sites have asked for the following number of units: Cleveland, 25; Louisville, 10; Memphis, 20; Pittsburgh, 30; and Richmond, 15. The shipping cost per unit in dollars between each center and job site is given in the following table:

	To Cleveland	To Louisville	To Memphis	To Pittsburgh	To Richmond
From Atlanta	55	30	40	50	40
From Chicago	35	30	100	45	60
From New York	40	60	95	35	30

Required: Solve for the minimum total shipping cost. If you cannot supply all of their requests, ship what you have as economically as possible.

Learning Objectives – Class 30

After today’s lecture, and after working the homework problems, the student should be able to

• Compute interarrival times based on any of several probability distributions
  • Zero-one Uniform
  • Uniform
  • Normal
  • Exponential
• Simulate a simple engineering system

Topics covered in today’s class

Where the data comes from, how you know which distribution fits your data, mean, sigma, 3 sigma, 6 sigma, …

 

ARRIVE {TIME = CUNIFORM(15.6, 19.9)};
Time = low value + randomnumber*(highnumber – lownumber)
For ARRIVE{TIME = CUNIFORM(15.6,19.9)}; with a random number = 0.234:
Interarrival time = 15.6 + (19.9 – 15.6)*0.234 = 16.6062 seconds (or whatever time units are being used.)

ARRIVE{TIME = 0 MAX NORMAL(15.6, 2.34)};
Time = mean + Zrand*sigma
For ARRIVE{TIME = 0 MAX NORMAL(15.6, 2.34)}; with random number = 0.281:
Interarrival time = MEAN + Ztable value for the random number * SIGMA
Get Z0.281 = -0.58, so the Interarrival Time = 15.6 – 0.58*2.34 = 14.2428 seconds
If the random number was greater than 0.5, then the Z table value will be positive, telling you that you must go to the right of the mean.

ARRIVE { TIME = EXPD(0.2) };
Time = interarrival time = -Ln(1 – Random Number)/Lambda
Lambda = 5 trucks per minute (measured in field as total trucks/total time)
For ARRIVE{TIME = EXPD(0.2)}; (i.e. lambda = 5 cars/minute), with a random number = 0.393469:
Interarrival time = -Ln(1-0.393469)/5 = 0.10000 seconds

Z tables – Note that 0 is to the RIGHT of the page.

Z	0.09	0.08	0.07	0.06	0.05	0.04	0.03	0.02	0.01	0
-3.5	0.0002	0.0002	0.0002	0.0002	0.0002	0.0002	0.0002	0.0002	0.0002	0.0002
-3.4	0.0002	0.0003	0.0003	0.0003	0.0003	0.0003	0.0003	0.0003	0.0003	0.0003
-3.3	0.0003	0.0004	0.0004	0.0004	0.0004	0.0004	0.0004	0.0005	0.0005	0.0005
-3.2	0.0005	0.0005	0.0005	0.0006	0.0006	0.0006	0.0006	0.0006	0.0007	0.0007
-3.1	0.0007	0.0007	0.0008	0.0008	0.0008	0.0008	0.0009	0.0009	0.0009	0.0010
-3	0.0010	0.0010	0.0011	0.0011	0.0011	0.0012	0.0012	0.0013	0.0013	0.0013
-2.9	0.0014	0.0014	0.0015	0.0015	0.0016	0.0016	0.0017	0.0018	0.0018	0.0019
-2.8	0.0019	0.0020	0.0021	0.0021	0.0022	0.0023	0.0023	0.0024	0.0025	0.0026
-2.7	0.0026	0.0027	0.0028	0.0029	0.0030	0.0031	0.0032	0.0033	0.0034	0.0035
-2.6	0.0036	0.0037	0.0038	0.0039	0.0040	0.0041	0.0043	0.0044	0.0045	0.0047
-2.5	0.0048	0.0049	0.0051	0.0052	0.0054	0.0055	0.0057	0.0059	0.0060	0.0062
-2.4	0.0064	0.0066	0.0068	0.0069	0.0071	0.0073	0.0075	0.0078	0.0080	0.0082
-2.3	0.0084	0.0087	0.0089	0.0091	0.0094	0.0096	0.0099	0.0102	0.0104	0.0107
-2.2	0.0110	0.0113	0.0116	0.0119	0.0122	0.0125	0.0129	0.0132	0.0136	0.0139
-2.1	0.0143	0.0146	0.0150	0.0154	0.0158	0.0162	0.0166	0.0170	0.0174	0.0179
-2	0.0183	0.0188	0.0192	0.0197	0.0202	0.0207	0.0212	0.0217	0.0222	0.0228
-1.9	0.0233	0.0239	0.0244	0.0250	0.0256	0.0262	0.0268	0.0274	0.0281	0.0287
-1.8	0.0294	0.0301	0.0307	0.0314	0.0322	0.0329	0.0336	0.0344	0.0351	0.0359
-1.7	0.0367	0.0375	0.0384	0.0392	0.0401	0.0409	0.0418	0.0427	0.0436	0.0446
-1.6	0.0455	0.0465	0.0475	0.0485	0.0495	0.0505	0.0516	0.0526	0.0537	0.0548
-1.5	0.0559	0.0571	0.0582	0.0594	0.0606	0.0618	0.0630	0.0643	0.0655	0.0668
-1.4	0.0681	0.0694	0.0708	0.0721	0.0735	0.0749	0.0764	0.0778	0.0793	0.0808
-1.3	0.0823	0.0838	0.0853	0.0869	0.0885	0.0901	0.0918	0.0934	0.0951	0.0968
-1.2	0.0985	0.1003	0.1020	0.1038	0.1056	0.1075	0.1093	0.1112	0.1131	0.1151
-1.1	0.1170	0.1190	0.1210	0.1230	0.1251	0.1271	0.1292	0.1314	0.1335	0.1357
-1	0.1379	0.1401	0.1423	0.1446	0.1469	0.1492	0.1515	0.1539	0.1562	0.1587
-0.9	0.1611	0.1635	0.1660	0.1685	0.1711	0.1736	0.1762	0.1788	0.1814	0.1841
-0.8	0.1867	0.1894	0.1922	0.1949	0.1977	0.2005	0.2033	0.2061	0.2090	0.2119
-0.7	0.2148	0.2177	0.2206	0.2236	0.2266	0.2296	0.2327	0.2358	0.2389	0.2420
-0.6	0.2451	0.2483	0.2514	0.2546	0.2578	0.2611	0.2643	0.2676	0.2709	0.2743
-0.5	0.2776	0.2810	0.2843	0.2877	0.2912	0.2946	0.2981	0.3015	0.3050	0.3085
-0.4	0.3121	0.3156	0.3192	0.3228	0.3264	0.3300	0.3336	0.3372	0.3409	0.3446
-0.3	0.3483	0.3520	0.3557	0.3594	0.3632	0.3669	0.3707	0.3745	0.3783	0.3821
-0.2	0.3859	0.3897	0.3936	0.3974	0.4013	0.4052	0.4090	0.4129	0.4168	0.4207
-0.1	0.4247	0.4286	0.4325	0.4364	0.4404	0.4443	0.4483	0.4522	0.4562	0.4602
0	0.4641	0.4681	0.4721	0.4761	0.4801	0.4840	0.4880	0.4920	0.4960	0.5000

Z tables – Note that 0 is to the LEFT of the page.

Z	0	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0	0.5000	0.5040	0.5080	0.5120	0.5160	0.5199	0.5239	0.5279	0.5319	0.5359
0.1	0.5398	0.5438	0.5478	0.5517	0.5557	0.5596	0.5636	0.5675	0.5714	0.5753
0.2	0.5793	0.5832	0.5871	0.5910	0.5948	0.5987	0.6026	0.6064	0.6103	0.6141
0.3	0.6179	0.6217	0.6255	0.6293	0.6331	0.6368	0.6406	0.6443	0.6480	0.6517
0.4	0.6554	0.6591	0.6628	0.6664	0.6700	0.6736	0.6772	0.6808	0.6844	0.6879
0.5	0.6915	0.6950	0.6985	0.7019	0.7054	0.7088	0.7123	0.7157	0.7190	0.7224
0.6	0.7257	0.7291	0.7324	0.7357	0.7389	0.7422	0.7454	0.7486	0.7517	0.7549
0.7	0.7580	0.7611	0.7642	0.7673	0.7704	0.7734	0.7764	0.7794	0.7823	0.7852
0.8	0.7881	0.7910	0.7939	0.7967	0.7995	0.8023	0.8051	0.8078	0.8106	0.8133
0.9	0.8159	0.8186	0.8212	0.8238	0.8264	0.8289	0.8315	0.8340	0.8365	0.8389
1	0.8413	0.8438	0.8461	0.8485	0.8508	0.8531	0.8554	0.8577	0.8599	0.8621
1.1	0.8643	0.8665	0.8686	0.8708	0.8729	0.8749	0.8770	0.8790	0.8810	0.8830
1.2	0.8849	0.8869	0.8888	0.8907	0.8925	0.8944	0.8962	0.8980	0.8997	0.9015
1.3	0.9032	0.9049	0.9066	0.9082	0.9099	0.9115	0.9131	0.9147	0.9162	0.9177
1.4	0.9192	0.9207	0.9222	0.9236	0.9251	0.9265	0.9279	0.9292	0.9306	0.9319
1.5	0.9332	0.9345	0.9357	0.9370	0.9382	0.9394	0.9406	0.9418	0.9429	0.9441
1.6	0.9452	0.9463	0.9474	0.9484	0.9495	0.9505	0.9515	0.9525	0.9535	0.9545
1.7	0.9554	0.9564	0.9573	0.9582	0.9591	0.9599	0.9608	0.9616	0.9625	0.9633
1.8	0.9641	0.9649	0.9656	0.9664	0.9671	0.9678	0.9686	0.9693	0.9699	0.9706
1.9	0.9713	0.9719	0.9726	0.9732	0.9738	0.9744	0.9750	0.9756	0.9761	0.9767
2	0.9772	0.9778	0.9783	0.9788	0.9793	0.9798	0.9803	0.9808	0.9812	0.9817
2.1	0.9821	0.9826	0.9830	0.9834	0.9838	0.9842	0.9846	0.9850	0.9854	0.9857
2.2	0.9861	0.9864	0.9868	0.9871	0.9875	0.9878	0.9881	0.9884	0.9887	0.9890
2.3	0.9893	0.9896	0.9898	0.9901	0.9904	0.9906	0.9909	0.9911	0.9913	0.9916
2.4	0.9918	0.9920	0.9922	0.9925	0.9927	0.9929	0.9931	0.9932	0.9934	0.9936
2.5	0.9938	0.9940	0.9941	0.9943	0.9945	0.9946	0.9948	0.9949	0.9951	0.9952
2.6	0.9953	0.9955	0.9956	0.9957	0.9959	0.9960	0.9961	0.9962	0.9963	0.9964
2.7	0.9965	0.9966	0.9967	0.9968	0.9969	0.9970	0.9971	0.9972	0.9973	0.9974
2.8	0.9974	0.9975	0.9976	0.9977	0.9977	0.9978	0.9979	0.9979	0.9980	0.9981
2.9	0.9981	0.9982	0.9982	0.9983	0.9984	0.9984	0.9985	0.9985	0.9986	0.9986
3	0.9987	0.9987	0.9987	0.9988	0.9988	0.9989	0.9989	0.9989	0.9990	0.9990
3.1	0.9990	0.9991	0.9991	0.9991	0.9992	0.9992	0.9992	0.9992	0.9993	0.9993
3.2	0.9993	0.9993	0.9994	0.9994	0.9994	0.9994	0.9994	0.9995	0.9995	0.9995
3.3	0.9995	0.9995	0.9995	0.9996	0.9996	0.9996	0.9996	0.9996	0.9996	0.9997
3.4	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9997	0.9998
3.5	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998	0.9998

 

I have measured the interarrival times (the times between successive arrivals) for my CVEN 422 class – i.e. I stood in the doorway and every time a student entered the room I wrote down the time.  I then entered these times into Excel, and took their differences to get a list of the interarrival time (the times between arrivals) for the students.  My raw data looks like the following:

Student Number	Student Arrival Times (Seconds)	Difference in Arrival Times (Interarrival Times) (seconds)
1	0
2	6.43	6.43
3	12.61	6.18
4	18.93	6.32
5	25.22	6.29
6	31.80	6.58
7	37.78	6.07
etc.

After getting all the above values computed I converted the “Equations” to “Values” in Excel, and sorted them, and found that when grouped into similar time ranges, I had the following groupings:

Interarrival Times (Seconds)	Total Number of Students Falling Within That Range
5.9 to 6.0	1
6.0 to 6.1	32
6.1 to 6.2	36
6.2 to 6.3	33
6.3 to 6.4	35
6.4 to 6.5	35
6.5 to 6.6	34
6.6 to 7.0	1

After plotting these I notice that they closely follow a CONTINUOUS UNIFORM distribution, and make the assumption that I can indeed model student arrival times with a continuous uniform distribution, with a low value = 6.0 seconds between students and high value = 6.6 seconds.  The corresponding BOSS command is:

            ARRIVE {TIME = CUNIFORM(6, 6.6)};

meaning “Please born me a student every 6 to 6.6 seconds, in a uniform continuous random fashion, and run him/her through my system to see how the system will perform.”

            On the next day I was measuring how often cars arrived at a traffic light, and measured the following data:

Truck Number	Arrival Times (Seconds)	Difference (Interarrival Times) (seconds)
1	0
2	3.5	3.5
3	14.0	10.5
4	24.5	10.5
5	34.5	10
6	45.0	10.5
7	62.0	17.0
etc.

            Notice that in this case the differences (the interarrival times) do not seem to be uniform – there are many more 10’s than 3.5’s.  In fact, after tabulating all my readings, I get the following number of interarrival times:

Interarrival Time (Seconds)	Number of Trucks
3-4	2
4-5	3
5-6	14
6-7	25
7-8	46
8-9	66
9-10	76
10-11	76
11-12	64
12-13	43
13-14	26
14-15	12
15-16	6
16-17	2

Now when I plot this, it looks very much like a normal distribution curve, with a mean of about 10 seconds, and a sigma of around 14/6 seconds.  To check this I first run some statistical tests to check the goodness of fit, and it does indeed reasonably approximate a normal distribution.  Thus I should be able to use my calculator and its random number generator (uniform distribution between 0 & 1) and a book of Z tables to accurately simulate when the cars will arrive in the system.  The corresponding BOSS command is:

            ARRIVE {TIME = 0 MAX NORMAL(10,2.333)};

Note that the “0 MAX” is to prevent the ARRIVE statement from (rarely) selecting a negative time, which can happen on any NORMAL curve, regardless of how unlikely.  Since that would crash the simulation you need to avoid it.  Note also the you don’t need “0 MAX” on a CUNIFORM or EXPD distribution.

Finally, the next day I was measuring the time between trucks at my borrow pit, and measured the following:

Number of trucks arriving	After Time	But Before Time
50	0 seconds	0.1 seconds
30	0.1 seconds	0.2 seconds
18	0.2 seconds	0.3 seconds
12	0.3 seconds	0.4 seconds
5	0.4 seconds	0.5 seconds
3	0.5 seconds	0.6 seconds
1	0.6 seconds	0.7 seconds
2	0.7 seconds	0.8 seconds

            When plotted, this appears to be an exponential distribution with lambda = 5 trucks per minute.  The corresponding BOSS command is:

            ARRIVE { TIME = EXPD(0.2) };

Examples for Continuous Uniform Distribution Curves:

As seen from the figure above, the total area under the curve is 1.0, and t represents that time which gives the appropriate shaded area under the curve. Thus assume that the distribution of interarrival times for your problem is uniform continuous, ranging from lowval = 10 to highval = 25. If you generate a random 0-1 number of 0.726, that says that the next arrival time (how long you must wait for the next entity arrival) should be “t”, where t will cause 0.726 of the area under the curve to be shaded. Thus the next entity will arrive at t = 10 + 0.726*(25-10) = 20.89 seconds (or minutes, or hours) after the preceding arrival.

To get the interarrival time for a continuous uniform distribution curve starting at 0 and going to 1,  simply press the rand or random button on your calculator. Thus, t = random number. To get the interarrival time for a continuous uniform distribution with a low value of 15.6 and a high number of 19.9, the time will equal:

            Time = low value + randomnumber*(highnumber – lownumber)

For ARRIVE{TIME = CUNIFORM(15.6,19.9)}; with a random number = 0.234:

            Interarrival time = 15.6 + (19.9 – 15.6)*0.234 = 16.6062 seconds (or whatever, time units.)

Examples for Normal Distribution Curves:

The same rules apply here. If your curve of interarrival times plots as a normal curve, having a mean, and approximately 3 sigma on either side beyond which there are very few arrivals, you can use a 0-1 random number generator to properly predict the next interarrival time. Again, the random number generated denotes what percentage of the curve should be shaded.

To get an interarrival time for a normal distribution curve, get a random number. Then enter that 0-1 number into the Z tables, and use the equations below. For example: If the mean time to deliver a load of dirt from a hill to a pit is 22 minutes, and if sigma = 4 minutes, how long is it likely to take the next truck to deliver its dirt?

Draw a normal distribution curve centered around 22 minutes, with the tail on the left end almost, but not quite touching the x axis at 22 – 3*4 = 10 minutes, and with the tail on the right almost, but not quite touching the x axis at 22 + 3*4 = 34 minutes. Then, get a random 0-1 number from your calculator (say 0.242) and shade the left 24.2% of the area under the curve. Then the right edge of the shaded area will predict the time it will likely take the next truck to travel from the hill to the pit. In this case, entering the Z tables at 0.242 (in the middle of the tables) gives Z (on the outside edge of the tables = -0.7. This is saying that you will have shaded 24.2% of the curve, if you will move to the left of center (the mean) by 0.7 sigma, draw a line and shade everything to the left of that line. The corresponding time will then be

        t = mean + Zrand*sigma = 22 + (-0.7)(4) = 20.4 minutes travel time.

If the random number was greater than 0.5, then the Z will be positive, telling you that you must go to the right of the mean.

For ARRIVE{TIME = 0 MAX NORMAL(15.6, 2.34)}; with random number = 0.281:

            Interarrival time = MEAN + Ztable value for the random number * SIGMA

            Get Z_0.281 = -0.58, so the Interarrival Time = 15.6-0.58*2.34 = 14.2428 seconds

Examples for Exponential Distribution Curves:

The figure below shows a graphical representation of an exponential distribution for the interarrival times measured between successive trucks at an intersection. In this case the average number of trucks arriving at the intersection is 5 per minute, or 12 seconds between truck arrivals. To determine the time it will take the next truck to arrive, get a 0-1 random number (say 0.4) and shade that much area under the curve. Then the corresponding interarrival time for the next truck will be given by:

Lambda = 5 trucks per minute (measured in field as total trucks/total time)

Shaded area = random number = 0.4 (from your calculator)

t = interarrival time = -ln(1 – Random Number)/Lambda = 0.102 minutes before the next truck will arrive.

For ARRIVE{TIME = EXPD(0.2)}; (i.e. lambda = 5 cars/minute) with random number = 0.393469:

            Interarrival time = -ln(1-0.393469)/5 = 0.10000

Given: An “unbalanced” transportation problem, where the supply of aggregate is not equal to the demand.

You are in charge of a large construction project that involves two sites. You currently need 50 tons of aggregate at site 1 and 22 tons of aggregate at site 2. Two suppliers are available, and their limits are listed below. Restrictions on the type of work to be done require that at least 20 tons used at site 1 have to come from supplier B and no more than 10 tons of aggregate from supplier A can be used at site 2. Aggregate delivery costs ($/ton) are as follows:

Delivery Costs:	To Site1	To Site 2
From Supplier A	$1/ton	$2/ton
From Supplier B	$1.5/ton	$1.25/ton

Required:

a) Solve this problem as an “unbalanced” transportation problem where the supply exceeds demand, assuming that Supplier A can deliver up to 31 tons, and Supplier B can deliver up to 43 tons.

b) Solve as an “unbalanced” transportation problem where the available supply is less than the demand, assuming that Supplier A can deliver a maximum of 27 tons, and Supplier B can deliver a maximum of 43 tons.

TEXAS A&M UNIVERSITY

CIVIL ENGINEERING DEPARTMENT
STRUCTURAL SYSTEMS 979/845-4395
COLLEGE STATION, TEXAS 77843-3136 Fax 979/845-6156

August 26, 1990

Mr. Jim Ridgeway Re: Problem 5.64
Bryan Construction Co., Inc.
1400 South East Bypass
Bryan, Texas 77801

Solution (perhaps) is here.  Work it first before you check your solution.

Dear Jim:

I have received your request for an optimum solution to your problem regarding bids made to your company from various contractors.

As I understand it, you have a list of contractors who are bidding on one of your jobs. Each contractor may bid on the entire job, as well as being permitted to bid on individual parts of the total job. For example, contractor A may bid 16 million to do the entire job, consisting of parts 1,2,3,4,5,6,7 and 8. He may also bid on just parts 1,2 and 3 (7 million), or on parts 4,5,6,7 and 8 (9 million), or on parts 1,3 and 8 (3 million). A list of the contractors you gave me over the phone, and their bids are shown below:

Contractor	Job Part Bid								Amount
A0	1	2	3	4	5	6	7	8	16
A1	1	2	3						7
A2				4	5	6	7	8	10
A3	1		3					8	3
B0	1			4	5		7		4
B1		2	3						2
C0			3		5				4
C1	1					6			4
C2		2	3	4				8	8
C3						6	7		2
D0	1	2	3		5				2
D1				4	5		7		3

What a mess! I think that you should just make them bid on each piece individually and forget it. However, I do understand that bidding in groups of things that they know how to do best would save them (and you) a lot of money. The problem looks rather like an assignment problem, except that in this case not every machine is going to get to do a job, and there’s no way that some of the jobs aren’t going to be assigned twice. Some practical solution to this will have to be suggested. Also we have the problem of making absolutely sure that every job gets done at least once. Then of course we need to reduce your costs as much as possible.

Additionally, I understand that since this is a government job, you must assign at least 20% of the total monetary award to minority contractors. You mentioned that bidders B and C were minority owned. Also, contractor C, although a real whiz at this type of work, will be unable to be bonded for over 10 million dollars, because of his small size. Also, contractor B has limited personnel, and could only handle a single job, even if he got two of them. Finally, you mentioned that regardless of the costs, bidder D is a subsidiary of your company, and that you wish to assign at least one of the jobs to him to keep your people working.

As far as I can tell, we should be able to handle all of these requirements using linear programming, and we will proceed at once. I look forward to working with you on this project.

Very truly yours,

Lee L. Lowery, Jr., P.E.

Note to class:

Note that it may be cheaper to let two contractors do one job twice! Example:

Contractor	Do job 1?	Do job 2?	Do job 3?	Bid Price ($)
A	Yes	Yes	Yes	800,000
B	Yes	Yes	No	200,000
C	No	Yes	Yes	300,000

As you see, it would actually be less expensive to give B and C the jobs, even though job 2 would be done twice. Actually, you can always renegotiate with the two contractors to see which will knock off the most money to get out of doing Job 2. Thus you should not try to force each job be done only once, as that may not be in your best economic interest!

Learning Objectives – Class 28

After today’s lecture, and after working the homework problems, the student should be able to

• Work problems similar to those listed today as Boss Practice Problems

Topics covered in today’s class

See main sheet