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Productivity and Unit Cost

Productivity measures output per unit of resource (m³ poured per hour, m² formed per day). Unit cost is the total cost to produce one unit of output (cost per m³, per m of excavation). To find how long a job takes, divide total quantity by productivity. To find cost, multiply unit cost by quantity. Watch units carefully — a crew's productivity is different from one worker's productivity. Manhours = number of workers × hours worked, a measure of labor input regardless of schedule. If you know output and manhour requirement, you can find crew size or duration.

$$\text{Duration}=\frac{\text{Total Quantity}}{\text{Productivity}}, \qquad \text{Unit Cost}=\frac{\text{Total Cost}}{\text{Total Output}}$$
$$\text{Manhours}=\text{Workers}\times\text{Hours}, \qquad \text{Productivity}_{\text{crew}}=\frac{\text{Output}}{\text{Manhours}}$$

Formwork Crew Duration

A crew installs 120 square meters of formwork in 10 hours. How long will the same crew need for 360 square meters?

$$\text{Productivity}=\frac{120}{10}=12\text{ m}^2/\text{hr}$$
$$t=\frac{360}{12}=30\text{ hr}$$

Final answer: 30 hours.

★ Unit Cost from Labor and Output

A masonry crew of 5 workers earns P800/worker-day. They can lay 15 m² of masonry per day. Find the unit cost of masonry per square meter.

Total labor cost per day = 5 workers × P800 = P4,000/day. Output = 15 m²/day.

$$\text{Unit Cost}=\frac{\text{Cost per day}}{\text{Output per day}}=\frac{4000}{15}=\text{P}266.67\text{ per m}^2$$

Final answer: P266.67 per m².

★★ Manhours Required

A concrete pouring job requires 240 manhours. If a crew of 8 workers works 10-hour days, how many days will the job take?

The crew produces 8 workers × 10 hrs/day = 80 manhours/day.

$$\text{Days}=\frac{240\text{ manhours}}{80\text{ manhours/day}}=3\text{ days}$$

Final answer: 3 days.

★★ Determining Crew Size for a Deadline

An excavation of 500 m³ must be completed in 5 days. One crew member excavates 4 m³ per 8-hour day. How many workers are needed? Work is done 8 hours per day.

Total output needed = 500 m³ in 5 days = 100 m³/day. Each worker outputs 4 m³/day.

$$n=\frac{100\text{ m}^3/\text{day}}{4\text{ m}^3/\text{worker-day}}=25\text{ workers}$$

Final answer: 25 workers are needed.

★★★ Combined Crew Output and Cost Comparison

Crew A can do a job in 10 days at P15,000/day. Crew B can do the same job in 15 days at P9,000/day. If both crews work together, how long will the job take and which is the cheaper option: both crews working together vs. just Crew A alone?

Work rates: A does 1/10 job/day; B does 1/15 job/day.

$$\text{Combined rate}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{1}{6}\text{ job/day}$$

Together: 6 days, at a daily cost of P15,000+P9,000 = P24,000.

$$\text{Cost together}=6\times24{,}000=\text{P}144{,}000$$
$$\text{Cost A alone}=10\times15{,}000=\text{P}150{,}000$$

Final answer: Together takes 6 days costing P144,000, which is P6,000 cheaper than Crew A alone. Use both crews together.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: w204

MSTE - Construction Methods / Construction Methods / MSTE May 2022

Estimate the actual bucket load in bank cubic meters whose heaped capacity is 3.82 m3. The soil's bucket fill factor is 0.90 and its load factor is 0.80.

  1. 2.55 BCM
  2. 2.75 BCM
  3. 3.32 BCM
  4. 2.57 BCM
Multiply the heaped capacity by the bucket fill factor and load factor: $$\text{Bucket load}=3.82(0.90)(0.80)=2.75\text{ BCM}.$$

Question Bank: w205

MSTE - Construction Methods / Construction Methods / MSTE May 2022

A crawler tractor weighing 36 t is towing a rubber-tired scraper weighing 45.5 t up a grade of 4%. What is the total resistance in kg of the combination if the rolling resistance factor of the scraper is 50 kg/t? 1 t = 1000 kg.

  1. 4,556 kg
  2. 4,655 kg
  3. 5,355 kg
  4. 5,535 kg
Total resistance is the sum of grade and rolling resistance. The grade resistance is $$(36{,}000+45{,}500)(0.04)=3{,}260\text{ kg}.$$ The scraper rolling resistance is $$50\frac{\text{kg}}{\text{t}}(45.5\text{ t})=2{,}275\text{ kg}.$$ Hence $$R_T=3{,}260+2{,}275=5{,}535\text{ kg}.$$

Question Bank: w206

MSTE - Construction Methods / Construction Methods / MSTE May 2022

A power-shift crawler tractor has a rated blade capacity of 7.65 LCM. The dozer is excavating loose common earth and pushing it a distance of 61 m. Maximum reverse speed in third range is 8 km/h. Estimate the production of the dozer if job efficiency is 50 min/h.
Table 4, fixed cycle time: power-shift transmission 0.05 min; direct-drive transmission 0.10 min; hard digging 0.15 min.
Table 5, operating speeds: loose materials hauled over 30 m: 4.0 km/h; return over 30 m: maximum reverse speed in third range.

  1. 275 LCM/h
  2. 251 LCM/h
  3. 269 LCM/h
  4. 243 LCM/h
Use $$\text{Production}=\frac{\text{Capacity}}{\text{Cycle time}}\times\text{Efficiency}.$$ The cycle time is the fixed time plus dozing and return times: $$t_c=0.05+\frac{0.061}{4}(60)+\frac{0.061}{8}(60)=1.4225\text{ min}.$$ Thus $$P=\frac{7.65}{1.42}(50)=269.4\text{ LCM/h},$$ or 269 LCM/h.

Question Bank: w207

MSTE - Construction Methods / Construction Methods / MSTE May 2022

Find the expected production in loose cubic meters (Lm3) per hour of a small hydraulic excavator. Heaped bucket capacity is 0.57 m3. The material is sand and gravel with a bucket fill factor of 0.95. Job efficiency is 50 min/h. Average depth of cut is 4.3 m. Maximum depth of cut is 6.1 m, and average swing is 90°. Hint: Production (LCV/h) = C × S × V × B × E; cycle output = 250 cycles/60 min; swing-depth factor = 1.00.

  1. 113 LCM/h
  2. 111 LCM/h
  3. 131 LCM/h
  4. 133 LCM/h
Use $C=250/60$ cycles/min, swing-depth factor $S=1.00$, bucket volume $V=0.57$ m³, fill factor $B=0.95$, and job efficiency $E=50$ min/h. Then $$P=\frac{250}{60}(1.00)(0.57)(0.95)(50)=112.8\text{ LCM/h}.$$ Therefore, the expected production is 113 LCM/h.

Question Bank: w228

MSTE - Construction Methods / Construction Methods / MSTE May 2022

The bucket volume contained within the bucket when the following outline of the bucket sides is referred to as:

  1. Heap volume
  2. Bucket load capacity
  3. Water line capacity
  4. Plate line capacity
The correct answer is Plate line capacity. The plate line is the outline of the bucket sides that determines the contained bucket volume. Heap volume includes material piled above the plate line, water-line capacity refers to liquid capacity up to a designated line, and bucket load capacity is a more general load term.